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A MODERN FORMAL
LOGIC PRIMER
Paul Teller
UC Davis
UC Davis
A Modern Formal Logic Primer
Paul Teller
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This text was compiled on 03/08/2024
TABLE OF CONTENTS
Licensing
1: Preface- A Guide to the Primer
2: Volume I- Sentence Logic

2.1: Basic Ideas and Tools
2.1.1: Logic As the Science of Argument
2.1.2: Sentences and Connectives
2.1.3: Truth Tables and the Meaning of '~', 'and', and 'v'
2.1.4: Truth Functions
2.1.5: Compounding Compound Sentences
2.1.6: Rules of Formation and Rules of Valuation
2.1.S: Basic Ideas and Tools (Summary)
2.2: Transciption Between English and Sentence Logic
2.2.1: Grouping in Transcription
2.2.2: GROUPING IN TRANSCRIPTION
2.2.3: Adequacy of Transcriptions
2.3: Logical Equivalence, Logical Truths, and Contradictions
2.3.1: Logical Equivalence
2.3.2: Substitution of Logical Equivalents and Some More Laws
2.3.3: Logical Truths and Contradictions
2.3.4: Disjunctive Normal Form and the Sheffer Stroke
2.4: Validity and Conditionals
2.4.1: Validity
2.4.2: Invalidity and Counter Examples
2.4.3: Soundness
2.4.4: The Conditional
2.4.5: The Biconditional
2.5: Natural Deduction for Sentence Logic - Fundamentals
2.5.1: The Idea of Natural Deduction
2.5.2: Subderivations
2.5.3: The Complete Rules of Inference
2.6: Natural Deduction for Sentence Logic - Strategies
2.6.1: Constructing Correct Derivations
2.6.2: Recognizing the Main Connective
2.6.3: Derivations- Overview, Definitions, and Points to Watch out for
2.7: Natural Deduction for Sentence Logic - Derived Rules and Derivations without Premises
2.7.1: Derived Rules
2.7.2: Argument by Cases
2.7.3: Further Derived Rules
2.7.4: Derivations without Premises
2.8: Truth Trees for Sentence Logic - Fundamentals
2.8.1: Proving Validity with Truth Trees
2.8.2: The General Strategy for the Rules
2.8.3: Proving Invalidity with Truth Trees
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2.8.4: The Complete Rules for the Connectives
2.8.5: In Which Order Should You Work on the Sentences in a Tree?
2.9: Truth Trees for Sentence Logic - Applications
2.9.1: Application of the Rules to Complex Sentences
2.9.2: Other Uses for Truth Trees
3: Volume II- Predicate Logic

3.1: Predicate Logic - Syntax
3.1.1: We Need More Logical Form
3.1.2: Quantifiers and Variables
3.1.3: The Sentences of Predicate Logic
3.2: Predicate Logic - Semantics and Validity
3.2.1: Interpretations
3.2.2: Truth in an Interpretation
3.2.3: Validity in Predicate Logic
3.3: More about Quantifiers
3.3.1: Some Examples of Multiple Quantification
3.3.2: Quantifier Scope, Bound Variables, and Free Variables
3.3.3: Correct Definitions of Substitution Instance and Truth in an Interpretation
3.3.4: Some Logical Equivalences
3.4: Transcription
3.4.1: Restricted Quantifiers
3.4.2: Transcribing from English into Logic
3.4.3: Transcription Strategies
3.5: Natural Deduction for Predicate Logic - Fundamentals
3.5.1: Review and Overview
3.5.2: The Universal Elimination Rule
3.5.3: The Existential Introduction Rule
3.5.4: The Existential Elimination and Universal Introduction Rules- Background in lnformal Argument
3.5.5: The Universal Introduction Rule
3.5.6: The Existential Elimination Rule
3.6: More on Natural Deduction for Predicate Logic
3.6.1: Multiple Quantification and Harder Problems
3.6.2: Some Derived Rules
3.6.3: Logical Truth, Contradictions, Inconsistency, and Logical Equivalence
3.7: Truth Trees for Predicate Logic - Fundamentals
3.7.1: The Rule for Universal Quantification
3.7.2: The Rule for Existential Quantification
3.7.3: Applying the Rules
3.7.4: Negated Quantified Sentences
3.8: More on Truth Trees for Predicate Logic
3.8.1: Contradictions, Logical Truth, Logical Equivalence, and Consistency
3.8.2: Truth Trees with Multiple Quantifiers
3.8.3: Three Shortcuts
3.8.4: Infinite Trees
3.9: Identity, Functions, and Definite Descriptions
3.9.1: Identity
3.9.2: Inference Rules for Identity
3.9.3: Functions
3.9.4: Definite Descriptions
3.10: Metatheory - The Basic Concepts
3.10.1: Object Language and Metalanguage
3.10.2: Syntax and Semantics
3.10.3: Soundness and Completeness
3.10.4: Some Further Notation and Definitions
3.11: Mathematical Induction
3.11.1: Informal Introduction
3.11.2: The Principle of Weak Induction
3.11.3: Strong Induction
3.12: Soundness and Completeness for Sentence Logic Tree
3.12.1: Preliminaries
3.12.2: Soundness and Completeness of the Tree Method- Informal Statement
3.12.3: Soundness and Completeness for Sentence Logic Trees- Formal Details
3.13: Soundness and Completeness for Sentence Logic Derivation
3.13.1: Soundness for Derivations- Informal Introduction
3.13.2: Soundness for Derivations- Formal Details
3.13.3: Completeness for Derivations- Informal Introduction
3.13.4: Completeness for Derivations- Formal Details
3.14: Compactness, and Generalization to Infinite Sets of Premises
3.14.1: Koenig's Lemma
3.14.2: Compactness and Infinite Sets of Premises
3.15: Interpretations, Soundness, and Completeness for Predicate Logic
3.15.2: Soundness and Completeness for Tree
3.15.3: Soundness for Predicate Logic Derivations
3.15.4: Completeness for Predicate Logic Derivations
3.15.5: Compactness, Identity, and Functions
3.15.6: Conclusion
Index
Glossary
Detailed Licensing
Licensing
A detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.
1: Preface- A Guide to the Primer
This text is a primer in the best sense of the word: A book which presents the basic elements of a subject. In other respects, I have
sought to write a different kind of text, breaking with what I regard as an unfortunate tradition in teaching formal logic. From truth
tables through completeness, I seek to explain, as opposed to merely presenting my subject matter. Most logic texts (indeed, most
texts) put their readers to sleep with a formal, dry style. I have aimed for a livelier lecture style, which treats students as human
beings and not as knowledge receptacles. In a text, as in the classroom, students need to be encouraged and to hear their difficulties
acknowledged. They need variation in pace. They need shifts in focus among "I," "we," and "you," just as most of us speak in the
classroom. From time to time students simply need to rest their brains.
One fault of logic textbooks especially bothers me: Some authors feel so concerned to teach rigor that they end up beating their
students over the head with it. I have not sacrificed rigor. But I have sought to cultivate it rather than rubbing it in.
Now to the contents of the Primer. Volume I presents sentence logic. Volume 11, Part I lays out predicate logic, including identity,
functions, and definite descriptions; Part II introduces metatheory, including mathematical induction, soundness, and completeness.
The text includes completely independent presentations of Fitch-style natural deduction and the tree method as developed by
Richard Jeffrey. I have presented the material with a great deal of modularity.
I have presented the text in two volumes to maximize flexibility of use in a variety of courses. Many introductory courses cover a
mix of informal and formal logic. Too often I have heard instructors express dissatisfaction with what they find available for the
formal portion of such a course. Volume I provides a new option. Using it in tandem with any of the many available inexpensive
'informal texts, instructors can combine the best of both subjects. Volume I will present a serious-minded introduction to formal
logic, which at the same time should prove accessible and encouraging to those students who will never again take another logic
course. The relatively small numbers who continue to a second course, devoted exclusively to formal logic, need only purchase
Volume II to build on the foundation already laid.
The Primer incorporates a number of unusual features. Chapters 1, 3, and 4 emphasize the concept of a truth function. Though the
idea is simple once you get it, many students need several passes. The optional section 34, on disjunctive normal form and the
Scheffer stroke, serves the didactic function of providing yet more drill on truth functionality.
Following Richard Jeffrey, I have thoroughly presented '&', 'v', and '-' before treating '3' and I='. '&', 'v', and '-' are much less
controversial correlates of their English counterparts than is '3'. Using '&', 'v' and '-' as a vehicle for introducing the idea of a truth
function, I can deal honestly with the difficulties of giving a truth functional formulation of conditionals. In turn, this honest
examination provides further drill with the concept of a truth function.
Sentences in English and logic often do not correspond very accurately. Consequently, I speak of transcription, not translation
between logic and English. I treat sentence logic transcription quite briefly in chapter 1 of Volume I and further in the short,
optional chapter 2. Predicate logic transcription gets a minimal introduction in chapter 1 of Volume II and then comes in for a
thorough workout in chapter 4, also optional. There I deal with the subject matter of domains and the traditional square of
opposition by using the much more general method of restricted quantifier subscripts and their elimination. This technique provides
an all-purpose tool for untangling complicated transcription problems. Chapter 4 of Volume II also examines quantificational
ambiguity in English, which most logic texts strangely ignore.
Training in metatheory begins in Volume I, chapter 1. But the training is largely implicit: I use elementary ideas, such as
metavariables, and then call attention to them as use makes their point apparent. After thorough preparation throughout the text,
chapter 10 of Volume II brings together the fundamental ideas of metatheory.
Standard treatments of sentence logic present sentence logic semantics, in the form of truth tables, before sentence logic derivation
rules. Only in this way do students find the rules clearly intelligible, as opposed to poorly understood cookbook recipes. Often texts
do not follow this heuristic for predicate logic, or they do so only half-heartedly. Presumedly, authors fear that the concept of an
interpretation is too difficult. However, one can transparently define interpretations if one makes the simplifying assumption of
including a name for each object in an interpretation's domain, in effect adopting a substitutional interpretation of the quantifiers. I
further smooth the way by stressing the analogy of form and function between interpretations and truth value assignments in
sentence logic.
This approach is ample for fixing basic ideas of semantics and for making predicate logic rules intelligible. After introducing
predicate logic syntax in Volume 11, chapter 1, and semantics in chapters 2 and 3, tree rules are almost trivial to teach; and
derivation rules, because they can be better motivated, come more easily. I have clearly noted the limitation in my definition of an
1.1 https://human.libretexts.org/@go/page/1940
interpretation, and I have set students thinking, in an exercise, why one may well not want to settle for a substitutional
interpretation. Finally, with the ground prepared by the limited but intuitive definitions of chapters 2 and 3 of Volume 11, students
have a relatively easy time with the full characterization of an interpretation in chapter 15.
No one has an easy time learning-r teaching-natural deduction quantifier rules. I have worked hard to motivate them in the context
of informal argument. I have made some minor modifications in detail of formulation, modifications which I believe make the rules
a little easier to grasp and understand. For existential elimination, I employ the superficially restrictive requirement that the
instantiating name be restricted to the sub-derivation. I explain how this restriction works to impose the more complex and
traditional restrictions, and I set this up in the presentation so that instructors can use the more traditional restrictions if they prefer.
For the proof of completeness of the natural deduction system I have fashioned my own semantic tableau proof. I believe that on its
own it is at least as accessible as the Henkin and other more familiar proofs. In addition, if you do tree completeness first, you can
explain the natural deduction completeness proof literally in a few minutes.
I have been especially careful not to dive into unexplained proofs of soundness and completeness. Instructors will find, in separate
sections, informal and intuitive explanations of the sentence logic proofs, unencumbered with formal details, giving an
understanding of how the proofs work. These sections require only the first short section of the induction chapter. Instructors
teaching metatheory at a more elementary level may want to conclude with some of these sections. Those ready for the tonic of
rigor will find much to satisfy them in the succeeding sections. In some chapters I have worked as hard on the exercises as on the
text. I have graded the skill problems, beginning with easy comprehension checkers, through skill builders, to some problems
which will test real skiU mastery. I think few will not find enough problems.
Exercises should exercise understanding as well as skills. Any decent mathematics text puts problems to this task, as well as uses
them to present auxihy material. Too few logic texts fall in this tradition. I hope that students and instructors will enjoy my efforts
in some of the exercises to introduce auxiliary material, to lay foundations for succeeding material, to engage creative
understanding, and to join in the activity of conceptual exploration.
For teaching plans the key word is "modularity." Those using just Volume 1 in an informal/formal course may teach chapters 1, 2
(optional), 3, and 4 to introduce sentence logic. Then, as taste and time permit, you may do natural deduction (chapters 5, 6, and 7)
or trees (chapters 8 and 9), or both, in either order.
Volumes I and II together provide great flexibility in a first symbolic logic course. Given your introduction of sentence logic with
chapters 1, 3, and 4 of Volume I and grounding of predicate logic with chapters 1, 2, and 3 of Volume II you can do almost
anything you want. I have made treatment of derivations and trees completely independent. You can run through the one from
sentence to predicate logic, and then go back and do the other. Or you can treat both natural deduction and trees for sentence logic
before continuing to predicate logic. You can spend up to two weeks on transcription in chapter 2 of Volume I and chapter 4 of
Volume 11, or you can rely on the minimal discussion of transcription in the first - chapters of Volumes I and II and omit chapter 2
of Volume I and chapter 4 of Volume II altogether.
If you do both trees and natural deduction, the order is up to you. Trees further familiarize students with semantics, which helps in
explaining natural deduction rules. On the other hand, I have found that after teaching natural deduction I can introduce trees
almost trivially and still get their benefit for doing semantics and metatheory.
Your only limitation is time. Teaching at an urban commuter university, in one quarter I cover natural deduction (Volume I,
chapters 1, 2,3, 4, 5, 6, 7; Volume 11, chapters 1, 2, 3, 5, and perhaps 6), or trees and sentence logic natural deduction (Volume I,
chapters 1, 2, 3, 4, 8, 9; Volume 11, chapters I, 2, 3, 7, 8; Volume I, chapters 5, 6, and 7). A semester should suffice for all of
Volume I and Volume II through chapter 8, and perhaps 9. Again, you may want to follow the chapter sequencing, or you may want
to do natural deduction first, all the way through predicate logic, or trees first.
If you do just natural deduction or just trees you have more time for identity, functions, definite descriptions, and metatheory.
Chapter 10 of Volume 11, basic metatheoretical concepts, can provide a very satisfying conclusion to a first course. A two quarter
sequence may suffice for all of the metatheory chapters, especially if you do not do both natural deduction and trees thoroughly. To
this end the metatheory chapters cover soundness and completeness for both natural deduction and trees independently. Or, you
may choose to end with the sections presenting the informal explanations of induction and the soundness and completeness proofs.
The text will provide a leisurely full year course or a faster paced full year course if you supplement it a bit at the end of the year.
I want to mention several features of my usage. I use single quotes to form names of expressions. I depart from logically correct
use of quotation marks in one respect. In stating generalizations about arguments I need a formulation which makes explicit use of
metavariables for premise and conclusion. But before chapter 10 of Volume 11, where I make the metalanguagd object language
distinction explicit, I do not want to introduce a special argument forming operator because I want to be sure that students do not
mistake such an operator for a new symbol in the object language. Consequently I use the English word 'therefore'. I found,
however, that the resulting expressions were not well enough set off from their context. For clarity I have used double quotes when,
for example, I discuss what one means by saying that an argument, "X Therefore Y." is valid.
Throughout I have worked to avoid sexist usage. This proves difficult with anaphoric reference to quantified variables, where
English grammar calls for constructions such as 'If someone is from Chicago he likes big cities.' and ' Anyone who loves Eve loves
himself.' My solution is to embrace grammatical reform and use a plural pronoun: 'If someone is from Chicago they like big cities.'
and 'Anyone who loves Eve loves themself.' I know. It grates. But the offense to grammar is less than the offense to social attitudes.
As this reform takes hold it will sound right to all of us.
I thank the many friends and family who have actively supported this project, and who have born with me patiently when the toil
has made me 'hard to live with. I do not regard the project as finished. Far from it. I hope that you-instructors and students-will
write me. Let me know where I am still unclear. Give me your suggestions for further clarification, for alternative ways to explain,
and for a richer slate of problems. Hearing your advice on how to make this a better text will be the best sign that I have part way
succeeded.
Paul Teller
Contributors and Attributions

Paul Teller (UC Davis). The Primer was published in 1989 by Prentice Hall, since acquired by Pearson Education. Pearson
Education has allowed the Primer to go out of print and returned the copyright to Professor Teller who is happy to make it
available without charge for instructional and educational use.
1: Preface- A Guide to the Primer is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
2: Volume I- Sentence Logic
Volume I presents sentence logic.

2: Volume I- Sentence Logic is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
1
SECTION OVERVIEW
2.1: Basic Ideas and Tools


2.1: Basic Ideas and Tools is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
2.1.1 https://human.libretexts.org/@go/page/1661
Adam is happy, or so I tell you. If you don't believe me, I try to convince you with an argument: Adam just got an 'A' on his logic
exam. Anyone who gets an 'A' on an exam is happy. So Adam is happy. A logician would represent such an argument in this way:
(1) Premises a) Adam just got an 'A' on his logic exam.
b) Anyone who gets an 'A' on an exam is happy.
Conclusion c) Adam is happy.
We ordinarily think of an argument as an attempt to convince someone of a conclusion by offering what a logician calls premises,
that is, reasons for believing the conclusion. But in order to study arguments very generally, we will characterize them by saying:
An Argument is a collection of declarative sentences one of which is called the conclusion and the rest of which are called the
premises.
An argument may have just one premise, or it may have many.
By declarative sentences, I mean those, such as 'Adam is happy.' or 'Grass is green.', which we use to make statements. Declarative
sentences contrast with questions, commands, and exclamations, such as 'Is Adam happy?', 'Cheer up, Adam!' and 'Boy, is Adam
happy!' Throughout this text I will deal only with declarative sentences; though if you continue your study of logic you will
encounter such interesting topics as the logic of questions and the logic of commands.
For an argument to have any interest, not just any premises and conclusion will do. In any argument worth its name, we must have
some connection or relation between the premises and conclusion, which you can think of intuitively in this way:
Ordinarily, the premises of an argument are supposed to support, or give us reasons, for believing the conclusion.
A good way of thinking about logic, when you are beginning to learn, is to say that logic is the study of this reason-giving
connection. I like to say, more generally, that logic is the science of arguments. Logic sets out the important properties of
arguments, especially the ways in which arguments can be good or bad. Along the way, logicians also study many things that are
not themselves arguments or properties of arguments. These are things which we need to understand in order to understand
arguments clearly or things which the study of arguments suggests as related interesting questions.
In order to see our subject matter more clearly, we need to distinguish between inductive and deductive arguments. Argument (1) is
an example of a deductive argument. Compare (1) with the following:
(2) a) Adam has smiled a lot today.
b) Adam has not frowned at all today.
c) Adam has said many nice things to people today, and no unfriendly things.
d) Adam is happy today.
The difference between arguments (1) and (2) is this: In (I), without fail, if the premises are true, the conclusion will also be true. I
mean this in the following sense: It is not possible for the premises to be true and the conclusion false. Of course, the premises may
well be false. (I, for one, would suspect premise (b) of argument (I).) But in any possible situation in which the premises are true,
the conclusion will also be true.
In argument (2) the premises relate to the conclusion in a different way. If you believe the second argument's premises, you should
take yourself to have at least some fairly good reasons for believing that the conclusion is true also. But, of course, the premises of
(2) could be true and the conclusion nonetheless false. For example, the premises do not rule out the possibility that Adam is
merely pretending to be happy.
Logicians mark this distinction with the following terminology:
Valid Deductive Argument: An argument in which, without fail, if the premises are true, the conclusion will also be true.
Good Inductive Argument: An argument in which the premises provide good reasons for believing the conclusion. In an
inductive argument, the premises make the conclusion likely, but the conclusion might be false even if the premises are true.
What do we mean by calling an argument 'deductive' or 'inductive', without the qualifiers 'valid' or 'good'? Don't let anyone tell you
that these terms have rigorous definitions. Rather,
2.1.1.1 https://human.libretexts.org/@go/page/1655
We tend to call an argument 'Deductive' when we claim, or suggest, or just hope that it is deductively valid. And we tend to call an
argument 'Inductive' when we want to acknowledge that it is not deductively valid but want its premises to aspire to making the
conclusion likely.
In everyday life we don't use deductively valid arguments too often. Outside of certain technical studies, we intend most of our
arguments as inductively good. In simple cases you understand inductive arguments clearly enough. But they can be a bear to
evaluate. Even in the simple case of argument (2), if someone suggests that Adam is just faking happiness, your confidence in the
argument may waver. How do you decide whether or not he is faking? The problem can become very difficult. In fact, there exists
a great deal of practical wisdom about how to evaluate inductive arguments, but no one has been able to formulate an exact theory
which tells us exactly when an inductive argument is really good.
In this respect, logicians understand deduction much better. Even in an introductory formal logic course, you can learn the rules
which establish the deductive validity of a very wide and interesting class of arguments. And you can understand very precisely
what this validity consists in and why the rules establish validity. To my mind, these facts provide the best reason for studying
deductive logic: It is an interesting theory of a subject matter about which you can, in a few months, learn a great deal. Thus you
will have the experience of finding out what it is like to understand a subject matter by learning a technical theory about that
subject matter.
Studying formal logic also has other, more practical, attractions. Much of what you learn in this book will have direct application in
mathematics, computer science, and philosophy. More generally, studying deductive logic can be an aid in clear thinking. The point
is that, in order to make the nature of deductive validity very precise, we must leaq a way of making certain aspects of the content
of sentences very precise. For this reason, learning deductive logic can pay big dividends in improving your clarity generally in
arguing, speaking, writing, and thinking.
 Exercise 2.1.1.1
Explain in your own words what an argument is. Give an example of your own of an inductive argument and of a deductive
argument. Explain why your example of an inductive argument is an inductive argument and why your example of a deductive
argument is a deductive argument.

2.1.1: Logic As the Science of Argument is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
I have said that arguments are composed of declarative sentences. Some logicians prefer to say that arguments are composed of the
things we say with sentences, that is, statements or propositions. Sentences can be problematic in logic because sentences are often
ambiguous. Consider this sentence:
(3) I took my brother's picture yesterday.
I could use this sentence to mean that yesterday I made a photograph of my brother. Or I could use the sentence to mean that I stole
a picture that belonged to my brother. Actually, this sentence can be used to say a rather amazingly large number of different things.
Ambiguous sentences can make a problem for logic because they can be true in one way of understanding them and false in
another. Because logic has to do with the truth and falsity of premises and conclusions in arguments, if it is not clear whether the
component sentences are true or false, we can get into some awful messes. This is why some logicians prefer to talk about
statements or propositions which can mean only one thing. In a beginning course, I prefer to talk about sentences just because they
are more familiar than statements and propositions. (What are statements and propositions supposed to be, anyway?) We can deal
with the problem of ambiguity of sentences by insisting that we use only unambiguous sentences, or that we specify the meaning
which a possibly ambiguous sentence will have in an argument and then stick to that meaning.
Actually, in most of our work we will be concerned with certain facts about the logical form of sentences and we won't need to
know exactly what the sentences mean in detail. All we will need in order to avoid problems about ambiguity is that a given
sentence be either true or false (although we usually won't know which it is) and that the sentence should not change from true to
false or from false to true in the middle of a discussion. As you will see very soon, the way we will write sentences will make it
extremely easy to stick to these requirements.
In fact, by restricting our attention to sentences which are either true or false, we have further clarified and extended our restriction
to declarative sentences. Questions ('Is Adam happy?'), commands ('Cheer up, Adam!'), and exclamations ('Boy, is Adam happy!')
are not true or false. Neither, perhaps, are some declarative sentences. Many people don't think "The woman who landed on the
moon in 1969 was blond." is either true or false because no woman landed on the moon in 1969. In any case, we shall study only
sentences which are definitely one or the other, true or false.
We will initially study a very simple kind of logic called Sentence Logic. (Logicians who work with propositions instead of
sentences call it Propositional Logic.) The first fact on which you need to focus is that we won't be concerned with all the details of
the structure of a sentence. Consider, for example, the sentence 'Adam loves Eve.' In sentence logic we won't be concerned with the
fact that this sentence has a subject and a predicate, that it uses two proper names, and so on.
Indeed, the only fact about this sentence which is relevant to sentence logic is whether it happens to be true or false. So let's ignore
all the structure of the sentence and symbolize it in the simplest way possible, say, by using the letter 'A'. (I put quotes around
letters and sentences when I talk about them as opposed to using them. If this use of quotes seems strange, don't worry about it-you
will easily get used to it.) In other words, for the moment, we will let the letter 'A' stand for the sentence 'Adam loves Eve.' When
we do another example we will be free to use 'A' to stand for a different English sentence. But as long as we are dealing with the
same example, we will use 'A' to stand for the same sentence.
Similarly, we can let other capital letters stand for other sentences. Here is a transcription guide that we might use:
Transcription Guide
A: Adam loves Eve.
B: Adam is blond.
C: Eve is clever.
'A' is standing for 'Adam loves Eve.', 'B' is standing for 'Adam is blond.', and 'C' is standing for 'Eve is clever.' In general, we will
use capital letters to stand for any sentences we want to consider where we have no interest in the internal structure of the sentence.
We call capital letters used in this way Atomic Sentences, or Sentence Letters. The word 'atomic' is supposed to remind you that,
from the point of view of sentence logic, these are the smallest pieces we need to consider. We will always take a sentence letter
(and in general any of our sentences) to be true or false (but not both true and false!) and not to change from true to false or from
false to true in the middle of a discussion.
Starting with atomic sentences, sentence logic builds up more complicated sentences, or Compound Sentences. For example, we
might want to say that Adam does not love Eve. We say this with the Negation of 'A', also called the Denial of 'A'. We could write
this as 'not A'. Instead of 'not', though, we will just use the negation sign, '~'. That is, the negation of 'A' will be written as '~A', and
will mean 'not A', that is, that 'A' is not true. The negation sign is an example of a Connective, that is, a symbol we use to build
longer sentences from shorter parts.
We can also use the atomic sentences in our transcription guide to build up a compound sentence which says that Adam loves Eve
and Adam is blond. We say this with the Conjunction of the sentence 'A' and the sentence 'B', which we write as 'A&B'. 'A' and 'B'
are called Conjuncts or Components of 'A&B', and the connective '&' is called the Sign of Conjunction.
Finally, we can build a compound sentence from the sentence 'A' and the sentence 'B' which means that either Adam loves Eve or
Adam is blond. We say this with the Disjunction of the sentence 'A' and the sentence 'B', which we write as 'AvB'. 'A' and 'B' are
called Disjuncts or Components of 'AvB', and the connective 'v' is called the Sign of Disjunction.
You might wonder why logicians use a 'v' to mean 'or'. There is an interesting historical reason for this which is connected with
saying more exactly what 'v' is supposed to mean. When I say, 'Adam loves Eve or Adam is blond.', I might actually mean two
quite different things. I might mean that Adam loves Eve, or Adam is blond, but not both. Or I might mean that Adam loves Eve, or
Adam is blond, or possibly both.
If you don't believe that English sentences with 'or' in them can be understood in these two very different ways, consider the
following examples. If a parent says to a greedy child, 'You can have some candy or you can have some cookies,' the parent clearly
means some of one, some of the other, but not both. When the same parent says to an adult dinner guest, 'We have plenty, would
you like some more meat or some more potatoes?' clearly he or she means to be offering some of either or both.
Again, we have a problem with ambiguity. We had better make up our minds how we are going to understand 'or', or we will get
into trouble. In principle, we could make either choice, but traditionally logicians have always opted for the second, in which 'or' is
understood to mean that the first sentence is true, or the second sentence is true, or possibly both sentences are true. This is called
the Inclusive Sense of 'or'. Latin, unlike English, was not ambiguous in this respect. In Latin, the word 'vel' very specifically meant
the first or the second or possibly both. This is why logicians symbolize 'or' with 'v'. It is short for the Latin 'vel,' which means
inclusive or. So when we write the disjunction 'AvB', we understand this to mean that 'A' is true, 'B' is true, or both are true.
1-2 Transcribe the following sentences into sentence logic, using 'G' to transcribe 'Pudding is good.' and 'F' to transcribe
'Pudding is fattening.'
a. Pudding is good and pudding is fattening.
b. Pudding is both good and fattening.
c. Pudding is either good or not fattening.
d. Pudding is not good and not fattening.
You may well have a problem with the following transcriptions, because to do some of them right you need to know something
I haven't told you yet. But please take a try before continuing. Trying for a few minutes will help you to understand the
discussion of the problem and its solution in the next section. And perhaps you will figure out a way of solving the problem
yourself!
a. Pudding is not both good and fattening.
b. Pudding is both not good and not fattening.
c. Pudding is not either good or fattening.
d. Pudding is either not good or not fattening.
e. Pudding is neither good nor fattening
To summarize this section:

Sentence logic symbolizes its shortest unambiguous sentences with Atomic Sentences, also called Sentence Letters, which are
written with capital letters: 'A', 'B', 'C' and so on. We can use Connectives to build Compound Sentences out of shorter
sentences. In this section we have met the connectives '~' (the Negation Sign), '&' (the Sign of Conjunction), and 'v' (the Sign
of Disjunction).
2.1.2: Sentences and Connectives is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We have said that '~A' means not A, 'A&B' means A and B, and 'AvB' means A or B in the inclusive sense. This should give you a
pretty good idea of what the connectives '~', '&', and 'v' mean. But logicians need to be as exact as possible. So we need to specify
how we should understand the connectives even more exactly. Moreover, the method which we will use to do this will prove very
useful for all sorts of other things.
To get the idea, we start with the very easy case of the negation sign, '~'. The sentence 'A' is either true or it is false. If 'A' is true,
then '~A' is false. If 'A' is false, then '~A' is true. And that is everything you need to know about the meaning of '~'. We can say this
more concisely with a table, called a Truth Table:
Truth table definition of '~'
Case A ~A
1 t f
2 f t
The column under 'A' lists all the possible cases involving the truth and falsity of 'A'. We do this by describing the cases in terms of
what we call Truth Values. The case in which A is true is described by saying that A has the truth value t. The case in which A is
false is described by saying that A has the truth value f. Because A can only be true or false, we have only these two cases. We
explain how to understand '~' by saying what the truth value of '~A' is in each case. In case 1, '~A' has the truth value f; that is, it is
false. In case 2, '~A' has the truth value t; that is, it is true. Although what we have done seems trivial in this simple case, you will
see very soon that truth tables are extremely useful.
Let us see how to use truth tables to explain '&'. A conjunction has two atomic sentences, so we have four cases to consider:
Case A B A&B
1 t t
2 t f
3 f t
4 f f
When 'A' is true, 'B' can be true or false. When 'A' is false, again 'B' can be true or false. The above truth table gives all possible
combinations of truth values which 'A' and 'B' can have together.
We now specify how '&' should be understood by specifying the truth value for each case for the compound 'A&B':
Truth Table definition of '&'
Case A B A&B
1 t t t
2 t f f
3 f t f
4 f f f
In other words, 'A&B' is true when the conjuncts 'A' and 'B' are both true. 'A&B' is false in all other cases, that is, when one or both
of the conjuncts are false.
A word about the order in which I have listed the cases. If you are curious, you might try to guess the recipe I used to order the
cases. (If you try, also look at the more complicated example in Section 1.5.) But I won't pause to explain, because all that is
important about the order is that we don't leave any cases out and all of us list them in the same order, so that we can easily
compare answers. So just list the cases as I do.
We follow the same method in specifying how to understand 'V'. The disjunction 'AvB' is true when either or both of the disjuncts
'A' and 'B' are true. 'AvB' is false only when 'A' and 'B' are both false:
truth table definition of 'v'
Case A B AvB
1 t t t
2 t f t
3 f t t
4 f f f
We have defined the connectives '~', '&', and t' using truth tables for the special case of sentence letters 'A' and 'B'. But obviously
nothing will change if we use some other pair of sentences, such as 'H' and 'D'.
This section has focused on the truth table definitions of '~', '&' and 'v'. But along the way I have introduced two auxiliary notions
about which you need to be very clear. First, by a Truth Value Assignment of Truth Values to Sentence Letters, I mean, roughly, a
line of a truth table, and a Truth Table is a list of all the possible truth values assignments for the sentence letters in a sentence:
An Assignment of Truth Values to a collection of atomic sentence letters is a specification, for each of the sentence letters,
whether the letter is (for this assignment) to be taken as true or as false. The word Case will also be used for 'assignment of
truth values'.
A Truth Table for a Sentence is a specification of all possible truth values assignments to the sentence letters which occur in
the sentence, and a specification of the truth value of the sentence for each of these assignments.

2.1.3: Truth Tables and the Meaning of '~', 'and', and 'v' is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
I want to point out one more thing about the way we have defined the connectives '~', '&', and 'v'. Let us start with '~'. What do you
have to know in order to determine whether '~A' is true or false? You don't have to know what sentence 'A' actually stands for. You
don't have to know whether 'A' is supposed to mean that Adam loves Eve, or that pudding is 10 fattening, or anything like that. To
know whether '~A' is true or false, all you have to know is whether 'A' itself is true or false. This is because if you know the truth
value of 'A', you can get the truth value of '~A' by just looking it up in the truth table definition of '~'.
The same thing goes for '&' and 'v'. To know whether 'A&B' is true or false, you don't have to know exactly what sentences 'A' and
'B' are supposed to be. All you need to know is the truth value of 'A' and the truth value of 'B'. This is because, with these truth
values, you can look up the truth value of 'A&B' with the truth table definition of '&'. Likewise, with truth values for 'A' and for 'B',
you can look up the truth value for 'AvB'.
Logicians have a special word for these simple facts about '~', '&' and 'v'. We say that these connectives are Truth Functional. In
other words (to use '&' as an example), the truth value of the compound sentence 'A&B' is a function of the truth values of the
components 'A' and 'B'. In other words, if you put in truth values for 'A' and for 'B' as input, the truth table definition of '&' gives
you, as an output, the truth value for the compound 'A&B'. In this way 'A&B' is a function in the same kind of way that 'x + y' is a
numerical function. If you put in specific numbers for 'x' and 'y', say, 5 and 7, you get a definite value for 'x + y', namely, 12.
'A&B' is just like that, except instead of number values 1, 2, 3, . . . which can be assigned to 'x' and to 'y', we have just two truth
values, t and f, which can be assigned to 'A' and to 'B'. And instead of addition, we have some other way of combining the assigned
values, a way which we gave in the truth table definition of '&'. Suppose, for example, that I give you the truth values t for 'A' and f
for 'B'. What, then, is the resulting truth value far 'A&B'? Referring to the truth table definition of 'A&B', you can read off the truth
value f for 'A&B'. The truth tables for '~' and for 'v' give other ways of combining truth values of components to get truth values for
the compound. That is,'~' and 'v' are different truth functions.
Let's pull together these ideas about truth functions:
A Truth Function is a rule which, when you give it input truth values, gives you a definite output truth value. A Truth
Functional Connective is a connective defined by a truth function. A Truth Functional Compound is a compound sentence
forked with truth functional connectives.
1-3. Try to explain what it would be for a declarative compound sentence in English not to be truth functional. Give an
example of a declarative compound sentence in English that is not truth functional. (There are lots of them! You may find this
exercise hard. Please try it, but don't get alarmed if you have trouble.)

2.1.4: Truth Functions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We have seen how to apply the connectives '~', '&', and 'v' to atomic sentences such as 'A' and 'B' to get compound sentences such
as '~A', 'A&B', and 'AvB'. But could we now do this over again? That is, could we apply the connectives not just to atomic
sentences 'A', 'B', 'C', etc., but also to the compound sentences '~A', 'A&B', and 'AvB'? Yes, of course. For example, we can form
the conjunction of '~A' with 'B', giving us '~A&B'. Using our current transcription guide, this transcribes into 'Adam does not love
Eve and Adam is blond.'
As another example, we could start with the conjunction 'A&B' and take this sentence's negation. But now we have a problem.
(This is the problem you encountered in trying to work exercise 1~2, e~i.) If we try to write the negation of 'A&B' by putting a '~'
in front of 'A&B', we get the sentence we had before. But the two sentences should not be the same! This might be a little
confusing at first. Here is the problem: We are considering two ways of building up a complex sentence from shorter parts,
resulting in two different complex sentences. In the first way, we take the negation of 'A', that is, '~A', and conjoin this with 'B'. In
the second way, we first conjoin 'A' and 'B' and then negate the whole. In English, the sentence 'It is not the case both that Adam
loves Eve and Adam is blond.' is very different from the sentence 'Adam does not love Eve, and Adam is blond.' (Can you prove
this by giving circumstances in which one of these compound sentences is true and the other one is false?)
In order to solve this problem, we need some device in logic which does the work that 'both' does in English. (If you are not sure
you yet understand what the problem is, read the solution I am about to give and then reread the last paragraph.) What we need to
do is to make clear the order in which the connectives are applied. It makes a difference whether we first make a negation and then
form a conjunction, or whether we first form the conjunction and then make a negation. We will indicate the order of operations by
using parentheses, much as one does in algebra. Whenever we form a compound sentence we will surround it by parentheses. Then
you will know that the connective inside the parentheses applies before the one outside the parentheses. Thus, when we form the
negation of 'A', we write the final result as '(~A)'. We now take '(~A)' and conjoin it with 'B', surrounding the final result with
parentheses:
[(~A}\&B] \label{4}
This says, take the sentence '(~A)' and conjoin it with 'B'. To indicate that the final result is a complete sentence (in case we will use
it in some still larger compound), we surround the final result in parentheses also. Note how I have used a second style for the
second pair of parentheses-square brackets-to make things easier to read.
Contrast ??? 4)
[(A&B)] (2.1.5.1)
with which means that one is to conjoin 'A' with 'B' and then take the negation of the whole.
In the same kind of way we can compound disjunctions with conjunctions and conjunctions with disjunctions. For example,
consider
[(A&B)vC ] (2.1.5.2)
[(A&(BvC ))] (2.1.5.3)
Sentence (6) says that we are first to form the conjunctions of 'A' with 'B' and then form the disjunction with 'C'. Expression
2.1.5.3, on the other hand, says that we are first to form the disjunction of 'B' with 'C' and then conjoin the whole with 'A'. These
are very different sentences. Transcribed into English, they are 'Adam both loves Eve and is blond, or Eve is clever.' and 'Adam
loves Eve, and either Adam is blond or Eve is clever.'
We can show more clearly that Expression 2.1.5.2 and 2.1.5.3 are different sentences by writing out truth tables for them. We now
have three atomic sentences, 'A', 'B', and 'C'. Each can be true or false, whatever the others are, so that we now have eight possible
cases. For each case we work out the truth value of a larger compound from the truth value of the parts, using the truth value of the
intermediate compound when figuring the truth value of a compound of a compound:
a b c d e g h
Case
A B C (A&B) (BvC) [(A&B)vC] [A&(BvC)]
1 t t t t t t t
2 t t f t t t t
a b c d e g h
Case
A B C (A&B) (BvC) [(A&B)vC] [A&(BvC)]
3 t f t f t t t
4 t f f f f f f
5 f t t f t t f
6 f t f f t f f
7 f f t f t t f
8 f f f f f f f
Let's go over how we got this truth table. Columns a, b, and c simply give all possible truth value assignments to the three sentence
letters 'A', 'B', and 'C'. As before, in principle, the order of the cases does not matter. But to make it easy to compare answers, you
should always list the eight possible cases for three letters in the order I have just used. Then, for each case, we need to calculate
the truth value of the compounds in columns d through h from the truth values given in columns a, b, and c.
Let us see how this works for case 5, for example. We first need to determine the truth value to put in column d, for '(A&B)' from
the truth values given for this case. In case 5 'A' is false and 'B' is true. From the truth table definition of '&', we know that a
conjunction (here, 'A&B') is false when the first conjunct (here, 'A') is false and the second conjunct (here, 'B') is true. So we write
an 'f' for case 5 in column d. Column e is the disjunction of 'B' with 'C'. In case 5 'B' is true and 'C' is true. When we disjoin
something true with something true, we get a true sentence. So we write the letter 't', standing for the truth value t, in column e for
case 5.
Moving on to column g, we are looking at the disjunction of '(A&B)' with 'C'. We have already calculated the truth value of
'(A&B)' for case 5~that was column d~and the truth value of 'C' for case 5 is given in column c. Reading off columns c and d, we
see that '(A&B)' is false and 'C' is true in case 5. The sentence of column g, '[(A&B)vC]', is the disjunction of these two
components and we know that the disjunction of something false with something true is, again, true. So we put a 't' in column g for
case 5. Following the same procedure for column h, we see that for case 5 we have a conjunction of something false with
something true, which gives the truth value f. So we write 'f for case 5 in column h.
Go through all eight cases and check that you understand how to determine the truth values for columns d through h on the basis of
what you are given in columns a, b, and c.
Now, back to the point that got us started on this example. I wanted to prove that the sentences '[(A&B)vC]' and '[A&(BvC)]' are
importantly different. Note that in cases 5 and 7 they have different truth values. That is, there are two assignments of truth values
to the components for which one of these sentences is true and the other is false. So we had better not confuse these two sentences.
You see, we really do need the parentheses to distinguish between them.
Actually, we don't need all the parentheses I have been using. We can make two conventions which will eliminate the need for
some of the parentheses without any danger of confusing different sentences. First, we can eliminate the outermost parentheses, as
long as we put them back in if we decide to use a sentence as a component in a larger sentence. For example, we can write 'A&B'
instead of '(A&B)' as long as we put the parentheses back around 'A&B' before taking the negation of the whole to form '~(A&B)'.
Second, we can agree to understand '~' always to apply to the shortest full sentence which follows it. This eliminates the need to
surround a negated sentence with parentheses before using it in a larger sentence. For example, we will write '~A&B' instead of
'(~A)&B'. We know that '~A&B' means '(~A)&B' and not '~(A&B)' because the '~' in '~A&B' applies to the shortest full sentence
which follows it, which is 'A' and not 'A&B'.
This section still needs to clarify one more aspect of dealing with compound sentences. Suppose that, before you saw the last truth
table, I had handed you the sentence '(A&B)vC' and asked you to figure out its truth value in each line of a truth table. How would
you know what parts to look at? Here's the way to think about this problem. For some line of a truth table (think of line 5, for
example), you want to know the truth value of '(A&B)vC'. You could do this if you knew the truth values of 'A&B' and of 'C'. With
their truth values you could apply the truth table definition of 'v' to get the truth value of '(A&B)vC'. This is because '(A&B)vC' just
is the disjunction of 'A&B' with 'C'. Thus you know that '(A&B)vC' is true if at least one of its disjuncts, that is, either 'A&B' or 'C',
is true; and '(A&B)vC' is false only if both its disjuncts, 'A&B' and 'C', are false.
And how are you supposed to know the truth values of 'A&B' and of 'C'? Since you are figuring out truth values of sentences in the
line of a truth table, all you need do to figure out the truth value of 'C' on that line is to look it up under the 'C' column. Thus, if we
are working line 5, we look under the 'C' column for line 5 and read that in this case 'C' has the truth value t. Figuring out the truth
value for 'A&B' for this line is almost as easy. 'A&B' is, by the truth table definition of conjunction, true just in case both conjuncts
(here, 'A' and 'B') are true. In line 5 'A' is false and 'B' is true. So for this line, 'A&B' is false.
Now that we finally have the truth values for the parts of '(A&B)vC', that is, for 'A&B' and for 'C', we can plug these truth values
into the truth table definition for v and get the truth value t for '(A&B)vC'.
Now suppose that you have to do the same thing for a more complicated sentence, say
\label{8} ~{[Av~C\]&[Bv(~A&C)]}
Don't panic. The principle is the same as for the last, simpler example. You can determine the truth value of the whole if you know
the truth value of the parts. And you can determine the truth value of the parts if you can determine the truth value of their parts.
You continue this way until you get down to atomic sentence letters. The truth value of the atomic sentence letters will be given to
you by the line of the truth table. With them you can start working your way back up.
You can get a better grip on this process with the idea of the Main Connective of a sentence. Look at sentence (8) and ask yourself,
'What is the last step I must take in building this sentence up from its parts?" In the case of (8) the last step consists in taking the
sentence '[Av~C]&[Bv(~A&C)]' and applying '~' to it. Thus (8) is a negation, '~' is the main connective of (a), and '[Av~C]&
[Bv(~A&C)]' is the component used in forming (8).
What, in turn, is the main connective of '[Av~C]&[Bv(~A&C)]'? Again, what is the last step you must take in building this
sentence up from its parts? In this case you must take 'Av~C' and conjoin it with 'Bv(~A&C)'. Thus this sentence is a conjunction,
'&' is its main connective, and its components are the two conjuncts 'Av~C' and 'Bv(~A&C)'. In like manner, 'Bv(~A&C)' is a
disjunction, with 'v' its main connective, and its components are the disjuncts 'B' and '~A&C'. To summarize,
The Main Connective in a compound sentence is the connective which was used last in building up the sentence from its
component or components.
Now, when you need to evaluate the truth value of a complex sentence, given truth values for the atomic sentence letters, you know
how to proceed. Analyze the complete sentence into its components by identifying main connectives. Write out the components, in
order of increasing complexity, so that you can see plainly how the larger sentences are built up from the parts. In the case of (8),
we would lay out the parts like this:
A, B, C, ~A, ~C, Av~C, ~A&C, Bv(~A&C), [Av~C]&[Bv(~A&C)], ~[Av~C]&[Bv(~A&C)]}
You will be given the truth values of the atomic sentence letters, either by me in the problem which I set for you or simply by the
line of the truth table which you are working. Starting with the truth values of the atomic sentence letters, apply the truth table
definitions of the connectives to evaluate the truth tables of the successively larger parts.
For each of the following sentences, state whether its main connective is '~', '&', or 'v' and list each sentence's components.
Then do the same for the components you have listed until you get down to atomic sentence letters. So you can see how you
should present your answers, I have done the first problem for you.
a)
Sentence Main Connective Components
~(Av~B) ~ Av~B
Av~B v A,~B
~B ~ B
a. ~(Av~B)
b. (D&~G)v(G&D)
c. [(Dv~~B)&(DvB)]&(DvB)
d. L&[Mv[~N&(Mv~L)}}
2.1.5: Compounding Compound Sentences is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We can summarize many important points discussed so far by giving explicit rules which tell us what counts as a sentence of
sentence logic and how to determine the truth values of compound sentences if we are given the truth values of the components.
 Formation Rules
i. Every capital letter 'A', 'B', 'C' ... is a sentence of sentence logic. Such a sentence is called an Atomic Sentence or a Sentence
Letter.
ii. If X is a sentence of sentence logic, so is (~X), that is, the sentence formed by taking X, writing a '~' in front of it, and
surrounding the whole by parentheses. Such a sentence is called a Negated Sentence.
iii. If X and Y are sentences of sentence logic, so is (X&Y), that is, the sentence formed by writing X, followed by '&',
followed by Y, and surrounding the whole with parentheses. Such a sentence is called a Conjuncton, and X and Y are called
its Conjuncts.
iv. If X and Y are sentences of sentence logic, so is (XvY), that is, the sentence formed by writing X, followed by 'v', followed
by Y, and surrounding the whole with parentheses. Such a sentence is called a Disjunction, and X and Y are called its
Disjuncts.
v. Until further notice, only expressions formed by using rules i) through iv) are sentences of sentence logic.
If you wonder why I say "until further notice," I want you to digest the present and some new background material before I
introduce two new connectives, corresponding to the expressions "If . . . then" and "if and only if." When I introduce these new
connectives, the formation rules will need to be extended accordingly.
As I explained earlier, we agree to cheat on these strict rules in two ways (and in these two ways only!). We omit the outermost
parentheses, and we omit parentheses around a negated sentence even when it is not the outermost sentence, because we agree to
understand '~' always to apply to the shortest full sentence which follows it.
I should also clarify something about formation rule i). In principle, sentence logic can use as many atomic sentences as you like. It
is not limited to the 26 letters of the alphabet. If we run out of letters, we can always invent new ones, for example, by using
subscripts, as in 'A ' and 'C '. In practice, of course, we will never need to do this.
1 37
 Rules of Valuation
i. The truth value of a negated sentence is t if the component (the sentence which has been negated) is f. The truth value of a
negated sentence is f if the truth value of the component is t.
ii. The truth value of a conjunction is t if both conjuncts have truth value t. Otherwise, the truth value of the conjunction is f.
iii. The truth value of a disjunction is t if either or both of the disjuncts have truth value t. Otherwise, the truth value of the
disjunction is f.
Note that these rules apply to any compound sentence. However, they only apply if somehow we have been given a truth value
assignment to the atomic sentence letters. That is, if we have been given truth values for the ultimate constituent atomic sentence
letters, then, using the rules of valuation, we can always calculate the truth value of a compound sentence, no matter how complex.
Once again, this is what we mean when we say that the connectives are truth functional.
How does one determine the truth value of atomic sentences? That's not a job for logicians. If we really want to know, we will have
to find out the truth value of atomic sentences from someone else. For example, we'll have to consult the physicists to find out the
truth value of "light always travels at the same speed." As logicians, we only say what to do with truth values of atomic
constituents once they are given to us. And when we do truth tables, we do not have to worry about the actual truth values of the
atomic sentence letters. In truth tables, like those in the following exercises, we consider all possible combinations of truth values
which the sentence letters could have.
The truth table definitions of the connectives give a graphic summary of these rules of valuation. I'm going to restate those truth
table definitions here because, if truth be told, I didn't state them quite right. I gave them only for sentence letters, 'A' and 'B'. I did
this because, at that point in the exposition, you had not yet heard about long compound sentences, and I didn't want to muddy the
waters by introducing too many new things at once. But now that you are used to the idea of compound sentences, I can state the
truth table definitions of the connectives with complete generality.
Suppose that X and Y are any two sentences. They might be atomic sentence letters, or they might themselves be very complex
compound sentences. Then we specify that:
Truth table definition of '~'
Case X ~X
1 t f
2 f t
Truth table definition of '&'
Case X Y X&Y
1 t t t
2 t f f
3 f t f
4 f f f
Truth table definition of 'v'
Case X Y XvY
1 t t t
2 t f t
3 f t t
4 f f f
The difference between my earlier, restricted truth table definitions and these new general definitions might seem a bit nitpicky. But
the difference is important. You probably understood the intended generality of my first statement of the truth table definitions.
However, a computer, for example, would have been totally confused. Logicians strive, among other things, to give very exact
statements of everything. They enjoy exactness for its own sake. But exactness has practical value too, for example, when one
needs to write a program that a computer can understand.
1-5 Which of the following expressions are sentences of sentence logic and which are not?
a) A&~B
b) A~&B
c) Gv(~B&~H))
d) A&(C&~(DvH))
e) (A&B)v(C&D)
f) (AvB)&CvD
1~6. Construct a complete truth table for each of the following sentences. The first one is done for you: 1~7. Philosopher's
problem: Why do I use quotation marks around sentences, writing things like and '~(CV~ A)' but no quotation marks about
boldface capital letters, writing X, Y, XVY, etc. when I want to talk about sentences generally?
A B ~B ~BvA
t t f t
t f t t
f t f f
f f t t
a) ~BvA
b) ~(BvA)
c) (QvT)&(~Qv~T)
d) (D&~G)v(G&D)
e) Av(~BvC)
f) Kv[~P&(~PvM)]
g) [(Dv~~B)&(Dv~B)]&(DvB)
h) L&[Mv[~N&(~Mv~L)]}
1-7. Philosopher's problem: Why do I use quotation marks around sentences, writing things like
'B'
and
'~(Cv~A)'
but no quotation marks about boldface capital letters, writing
X,Y,XvY,etc.
when I want to talk about sentences generally?
This section has also illustrated another thing worth pointing out. When I talked about sentences generally, that is, when I wanted to
say something about any sentences, X and Y, I used boldface capital letters from the end of the alphabet. I'm going to be doing this
throughout the text. But rather than dwell on the point now, you will probably best learn how this usage works by reading on and
seeing it illustrated in practice.

2.1.6: Rules of Formation and Rules of Valuation is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
The following list gives you the important terms which have been introduced in this chapter. Make sure you understand all of them
by writing a short explanation of each. Please refer back to the text to make sure, in each case, that you have correctly explained the
term. Keep your explanation of these terms in your notebook for reference and review later on in the course.
a. Argument
b. Valid Deductive Argument
c. Good Inductive Argument
d. Deductive Argument
e. Inductive Argument
f. Atomic Sentence (also called 'Sentence Letter' or 'Atomic Sentence Letter')
g. Compound Sentence
h. Connective
i. Component d 'Sentence Letter' or
j. ~ (called the 'Negation Sign' or 'Sign of Denial')
k. Negation
l. & (called the 'Sign of Conjunction')
m. Conjunction
n. Conjunct
o. v (called the 'Sign of Disjunction')
p. Disjunction
q. Disjunct
r. Inclusive Or
s. Exclusive Or
t. Truth Value
u. Truth Table
v. Truth Table Definition
w. Assignment of Truth Values
x. Case
y. Truth Function
z. Truth Functional Connective
aa. Truth Functional Compound
ab. Main Connective

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SECTION OVERVIEW
2.2: Transciption Between English and Sentence Logic

2.2.2: GROUPING IN TRANSCRIPTION

2.2: Transciption Between English and Sentence Logic is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
As we saw in chapter 1, for many English sentences we can find corresponding sentences of sentence logic. For example, if 'A'
stands for the sentence 'Adam loves Eve.' and 'B' for the sentence 'Adam is blond.' then 'Bv~A' corresponds to 'Either Adam is
blond' or he does not love Eve.'
Many logicians use the word 'translation' to describe the relation between a sentence of English and a corresponding sentence of
logic. I think that 'translation' is the wrong word to use. If a first sentence translates a second, the two sentences are supposed to
have exactly the same meaning. But the correspondence between English and logic is often looser than having the same meaning,
as the next examples show.
Consider the sentence 'Adam loves Eve, but he left her.' This English sentence is a compound of two shorter sentences, 'Adam
loves Eve.', which we will transcribe with the sentence letter 'A', and 'He left her.' (that is, Adam left Eve), which we will transcribe
with the sentence letter 'L'. These two sentences have been connected in English with the word 'but'. So we can get a partial
transcription into logic by writing 'A but L'. We are still not finished, however, because 'but' is a word of English, not logic. What in
logic corresponds to 'but'?
If I assert the sentence 'Adam loves Eve but he left her.', what am telling you? Well, first of all, I assert that Adam loves Eve. In
asserting the original sentence, I am also telling you that Adam left Eve. In other words, so far, I seem to be saying: 'Adam loves
Eve and he left her.'
What's the difference between 'Adam loves Eve but he left her.' and Adam loves Eve and he left her.', that is, between 'A but L' and
'A&L'? Not much. In English, we tend to use the word 'but' when we want to assert two things (a conjunction), but the first thing
asserted may well lead one to expect the opposite of the second thing asserted. 'But' functions much as do the words '. . . and,
contrary to what I just said would lead you to expect. . . .': 'Adam loves Eve, and, contrary to what I just said would lead you to
expect, he left her.'
Logic has no way of expressing the idea of 'contrary to what the first conjunct would lead you to expect.' So we simply transcribe
'but' as '&'. In sentence logic we can't improve upon 'A&L' as a transcription of 'Adam loves Eve but he left her.' Several other
English words function very much like 'but', and should likewise get transcribed as '&': 'however', 'nevertheless', 'although', and
'despite (the fact that)'.
Perhaps you are starting to see why I want to talk about transcribing, instead of translating, English into logic. 'A&L' isn't a very
good translation of 'Adam loves Eve but he left her.' If it were a good translation, we would have to say that 'and' means the same
thing as 'but', which it clearly does not. However, '&' is the closest we have to 'but' in logic, so that's what we use.
A Transcrpition of an English sentence into sentence logic is a sentence of sentence logic which expresses, as closely as
possible, what the English sentence expresses.
Logicians sometimes use the words 'paraphrasing' or 'symbolizing' for what I am calling 'transcribing' English sentences in logic.
2.2.1: Grouping in Transcription is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
2.2: GROUPING IN TRANSCRIPTION
Here is another problem which comes up in transcribing English into logic. Consider the sentence.
(1) Eve is clever and Eve is dark-eyed or Adam is blond.
How do we transcribe this? Should we understand (1) as
(1a) (Eve is clever and Eve is dark-eyed) or Adam is blond.
Or should we understand it as
(1b) Eve is clever and (Eve is dark-eyed or Adam is blond).
As we know from section 1-5, the grouping makes a difference. The problem here is that (1) is bad English. In English we should
also indicate the grouping, which we can easily do with a comma. Thus (1a) corresponds to
(1c) Eve is clever and Eve is dark-eyed, or Adam is blond.
and (1b) corresponds to
(1d) Eve is clever, and Eve is dark-eyed or Adam is blond.
Using the following transcription guide
B: Adam is blond.
C: Eve is clever.
D: Eve is dark-eyed.
we get as transcriptions of (1a) and (1c):
(1e) (C&D)vB
And as transcriptions of (1b) and (1d):
Notice that by using parentheses in (1a) and (1b) I have used a mixture of English and sentence logic as an aid to figuring out what
seems to be going on. Such mixtures often help in transcription. If you don't see a correct transcription right away, transcribe part,
or features of, the English sentence. Then go to work on the parts which you did not transcribe in your first pass at the problem.
The expression 'Either . . . or ' functions in English to indicate grouping in some respects as do parentheses in logic. Anything that
goes where you see the '. . .' acts as if it had parentheses around it, even if it is quite complex. (Often something which goes where
you see the '-' also acts like it had parentheses around it, but this English device does not always work.) Thus we could write (1a)
and (1c) as
(1g) Either Eve is clever and Eve is dark-eyed, or Adam is blond.
'Both . . . and _____' serves much as does 'Either . . . or _____', although the complexities of English grammar don't let you say
things such as 'Both Eve is clever and Eve is dark-eyed, or Adam is blond.' To speak grammatical English, one has to say
(1h) Eve is both clever and dark-eyed, or Adam is blond.
which we clearly transcribe as (1e).
Notice that in (1h) we have done some collapsing of English sentence units. When transcribing into logic, you should rewrite 'Eve
is clever and dark-eyed.' as a conjunction of two atomic sentences, that is, as 'Eve is clever and Eve is dark-eyed.' or finally as
'C&D'. And, to consider a new example, you should rewrite 'Eve is clever or dark-eyed.' as a disjunction of two atomic sentences,
that is, as 'Eve is clever or Eve is dark-eyed.', or finally as 'CvD'.
2.2: GROUPING IN TRANSCRIPTION is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
It's your turn to figure out an example. Before reading on, try transcribing
(2) Adam is neither ugly nor dumb.
What did you get? 'Neither' suggests a negation, and 'nor' suggests a disjunction. But (2) is tricky. If we use 'U' for 'Adam is ugly.'
and 'D' for 'Adam is dumb.', '~(UvD)' is a correct transcription. '~Uv~D' is not. How can you tell which is correct? We want the
English sentence and the proposed transcription to say the same thing, as nearly as possible. One way to test for such agreement is
to transcribe back into English. Suppose you proposed '-Uv~D' as a transcription of (2). Transcribe '~Uv~D' back into English, as
literally as you can. '~Uv~D' is a disjunction of two negations, so we transcribe it back as
(3) Either Adam is not ugly or Adam is not dumb.
Now, do (2) and (3) say the same thing? No! Sentence (2) is stronger. It says that Adam is both not ugly and also not dumb.
Sentence (3) says that Adam is either not one or not the other (or possibly not both). It's enough for (3) to be true that Adam not be
ugly. That's not enough for (2). To make (2) true, Adam will have to fail both in being ugly and in being dumb.
If what it takes to make (2) true is that Adam not be ugly and Adam not be dumb, could we also transcribe (2) as '~U&~D'? Yes. To
test, transcribe back into English. '-U&-D' transcribes back as
(2a) Adam is not ugly and Adam is not dumb.
(Or, equally good: 'Adam is not ugly and not dumb.') Compare (2a) with ,(2) 1 hope you will see that they say the same thing.
Generalizing the moral of this example we have:
 First Tramscription Test

To check a transcription of an English sentence, transcribe back into English as literally as possible. To the extent that the
original and the retranscribed sentences seem to say the same thing, you have reason to think that you have an Adequate
Transcription.
Our example also suggests another test for adequate transcription. So far, I have relied on your intuitive understanding of when two
sentences do and don't say the same thing. But we can spell out one part of this understanding in more detail. The trouble with
transcribing (2) as '~Uv~D' is that there is a situation in which '~Uv~D' is true but in which (2) is false. A situation in which Adam
is ugly and is not dumb provides just such a case. But if a first sentence can be true while, in the same situation, a second sentence
is false, then the two sentences are not saying the same thing.
Let's make this test for adequate transcription more precise. Consider a proposed transcription. Ask yourself: Is there an assignment
of truth values to sentence letters (a case) which makes the proposed transcription true and the English sentence false, or the
transcription false and the English sentence true? If so, reject the proposed transcription. If there is no such case, the transcription is
as good as it can get. Of course, in applying this test you will have to do the best you can to determine whether or not, for a case
described in terms of truth values assigned to sentence letters, your English sentence is true. The structure of English is
complicated, so there are no simple rules for determining the truth value of arbitrary English sentences. Nonetheless, this test can
often help you to decide whether a proposed transcription is adequate.
We summarize the test by saying:
Second Transcription Test: Given a sentence of sentence logic as a proposed transcription of an English sentence, try to imagine a
case, described in terms of an assignment of truth values to sentence letters, which makes one of the sentences true and the other
false. If there is such a case, reject the proposed transcription. If there is no such case, you have an Adequate Transcription.
This second test and the last example bring out a curious fact. Look back and you will see that both '~(UvD)' and '~U&~D' seem to
be adequate transcriptions of (2), for, by our first crude test, they both seem to say the same thing as (2). Are both '~(UvD)' and
'~U&~D' adequate transcriptions of (2) according to the second test? If you think it through, you should be able to satisfy yourself
that they are. But if so, that is, if both these sentences are true in exactly the same cases as (2), then they will have to be true in
exactly the same cases as each other. Any case in which one is true is a case in which the other is true. Any case in which one is
false is a case in which the other is false.
We will say that two such sentences are logically equivalent, a notion which I won't dwell on now because it provides the subject of
the next chapter. But even this quick description of logical equivalence will help 46 Transcription between English and Sentence
Logic you pull together the ideas of the last few paragraphs. At least so far as sentence logic goes, two sentences say the same thing
if and only if they are logically equivalent. With this way of understanding "saying the same thing," our two tests for adequacy of
transcription ultimately do the same work. For if "saying the same thing" just means "being true in exactly the same cases," two
sentences say the same thing (our first test for an adequate transcription) if and only if they are true in the same cases (our second
test for an adequate transcription).
Chapter 3 will clarify your understanding of logical equivalence. For the moment, however, you will be served by an intuitive
understanding of a summary of this section:
If two sentence logic sentences are logically equivalent to each other, they provide equally good transcriptions of a given
English sentence.
2- 1. Consider the sentence I

(2*) Adam is not both ugly and dumb.
lCarry out a study of its transcription into sentence logic which is similar to the study of (2). In particular, show that this
sentence has two logically equivalent, and so equally accurate, transcriptions, both of which need carefully to be distinguished
from a somewhat similar, but inadequate, transcription. If you have trouble with this exercise, spend a minute guessing at a
transcription of (2*). Write down your guess and then reread the discussion of the transcription of (2). ~,
2-2. Using this transcription guide, transcribe the following sentences into sentences of sentence logic.
A: Adam loves Eve,
B: Adam is blond.
C: Eve is clever.
D: Eve is dark-eyed.
E: Eve loves Adam.
a) Eve is clever or Eve is dark-eyed.
b) Eve is clever or dark-eyed.
c) Eve is clever and dark-eyed.
d) Eve is clever but not dark-eyed.
e) Eve either is not clever or she is not dark-eyed.
f) Eve is either not clever or not dark-eyed.
g) Eve is dark-eyed and Adam loves her.
h) Either Adam is blond and loves Eve, or he is not blond and Eve loves him.
i) Eve is both not dark-eyed and either clever or in love with Adam.
j) Eve is dark-eyed, but Adam does not love her.
k) Adam is either blond or in love with Eve; nevertheless, Eve does not love him.
l) Although either Eve is dark-eyed or Adam is blond, Adam does not love Eve.
m) Despite Eve's being clever and not loving Adam, Adam does love Eve.
n) Adam loves Eve even though she is not dark-eyed.
o) Adam not only loves Eve, Eve also loves Adam.
p) Even though Eve is either clever or not dark-eyed, either Adam is blond or in love with Eve.
q) Eve is both in love with Adam and not dark-eyed, despite Adam's being either blond or not in love with Eve.
r) Adam does not love Eve. Also, Adam is blond, and Eve is either clever or in love with Adam.
s) Adam is either in love with Eve or not.
t) Adam is either in love with Eve or not. However, although she is clever, Eve is either dark-eyed or in love with Adam.
u) Either Adam is blond, or it is both the case that Eve loves Adam and is either dark-eyed or clever.
v) Either it is the case that both Adam is blond or not in love with Eve and Eve is dark-eyed or in love with Adam, or it
is the case that both Adam does love Eve or is not blond and Eve is clever but not dark-eyed.
2-3. Using the same transcription guide as in exercise 2-2, transcribe the following into English:
a) Bv~B
b) A&~B
c) -~AvC)
d) Bv(D&~C)
e) (Ev~C)&(~BvA)
f) [(AvE)&~C]v(C&~D)
g) {[(~BvA)&D]v~(E&B))&C (This is almost impossible to transcribe into English, but do the best you can. I'm giving
this problem not to give you a bad time but to illustrate how logic has certain capacities to state things exactly, no matter
how complex they are, while English, in practice, breaks down.)
2-4. Make up your own transcription guide and transcribe the following sentences into sentence logic. Your transcriptions
should be as detailed as possible. or example, transcribe 'Roses are red and violets are blue.' not with one sentence letter but
with two sentence letters conjoined, like this: 'R&B' (R: Roses are red, B: Violets are blue).
a) Roses are red or Teller will eat his hat.
b) Monty Python is funny but Robert Redford is not
c) Chicago is not bigger than New York wen though New York is not the largest city.
d) Either I will finish this logic course or I will die trying.
e) W. C. Fields was not both handsome and smart.
f) Uncle Scrooge was neither generous nor understanding.
g) Although Minnesota Fats tried to diet, he was very overweight.
h) Peter likes pickles and ice cream, but he does not like to eat them together.
i) Roses are red and violets are blue. Transcribing this jingle is not had to do.
j) Columbus sailed the ocean blue in 1491 or 1492, but in any case he discovered neither the South nor the North Pole.
k) Either Luke will catch up with Darth Vader and put an end to ,him or Darth Vader will get away and cause more
trouble. But eventually the Empire will be destroyed.
 CHAPTER SUMMARY EXERCISE
Give brief explanations of the following terms introduced in this chapter. Again, please refer to the text to make sure you have
the ideas right.
a) Transcription
b) Adequate Transcription
Also, give a brief description of how English marks the grouping of sentences, that is, describe how English accomplishes the
work done in logic by parentheses.
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SECTION OVERVIEW
2.3: Logical Equivalence, Logical Truths, and Contradictions


2.3: Logical Equivalence, Logical Truths, and Contradictions is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
I introduced logic as the science of arguments. But before turning to arguments, we need to extend and practice our understanding
of logic's. basic tools as I introduced them in chapter 1. For starters, let's look at the truth table for 'A', '~A', and the negation of the
negation of 'A', namely, '~~A'
A ~A ~~A
t f t
f t f
This truth table exhibits the special situation which I mentioned at the end of the last chapter: The truth value of '~~A' is always the
same as that of 'A'. Logicians say that 'A' and '~~A' are Logically Equivalent.
As we will see in a moment, much more complicated sentences can be logically equivalent to each other. To get dear on what this
means, let us review some of the things that truth tables do for us. Suppose we are looking at a compound sentence, perhaps a very
complicated one which uses many sentence letters. When we write out the truth table for such a sentence, we write out all the
possible cases, that is, all the possible assignments of truth values to sentence letters in all possible combinations. In each one of
these possible cases our original sentence has the truth value t or the truth value f.
Now suppose that we look at a second sentence which uses the same sentence letters, or perhaps only some of the sentence letters
that the first sentence uses, and no new sentence letters. We say that the two sentences are logically equivalent if in each possible
case, that is, for each line of the truth table, they have the same truth value.
Two sentences of sentence logic are Logically Equivalent if and only if in each possible case (for each assignment of truth
values to sentence letters) the two sentences have the same truth value.
What we said about the double negation of 'A' naturally holds quite generally:
The Law of Double Negation (DN): For any sentence X, X and ~~X are logically equivalent.
Here are two more laws of logical equivalence:
De Morgan's Laws (DM): For any sentences X and Y, ~(X&Y) is logically equivalent to ~Xv~Y. And ~(XvY) is logically
equivalent to ~X&~Y.
Thus 'Adam is not both ugly and dumb.' is logically equivalent to 'Either Adam is not ugly or Adam is not dumb.' And 'Adam is not
either ugly or dumb.' is logically equivalent to 'Adam is not ugly and Adam is not dumb.' You should check these laws with truth
tables. But I also want to show you a second, informal way of checking them which allows you to "see" the laws. This method uses
something called Van Diagram.
A Venn diagram begins with a box. You are to think of each point inside the box as a possible case in which a sentence might be
true or false. That is, think of each point as an assignment of truth values to sentence letters, or as a line of a truth table. Next we
draw a circle in the box and label it with the letter 'X', which is supposed to stand for some arbitrary sentence, atomic or compound.
The idea is that each point inside the circle represents a possible case in which X is true, and each point outside the circle represents
a possible case in which X is false.
Look at Figure 3-1. What area represents the sentence ~X? The area outside the circle, because these are the possible cases in
which X is false.
Now let's consider how Venn diagrams work for compound sentences built up from two components, X and Y. Depending on what
the sentences X and Y happen to be, both of them might be true, neither might be true, or either one but not the other might be true.
Not to omit any of
these eventualities, we must draw the circles representing X and Y as overlapping, as in Figure 3-2 and 3-3.
The conjunction X&Y is true in just those cases represented by points that lie inside both the X and Y circles, that is, the shaded
area in Figure 3-2. The disjunction XvY is true in just those cases represented by points that lie inside either the X or the Y circle
(or both), that is, the shaded area in Figure 3-3.
Now we can use Venn diagrams to check De Morgan's laws. Consider first a negated conjunction. Look for the area, shown in
Figure 3-4, which represents ~(X&Y). This is just the area outside the shaded lens in Figure 3-2.
Let us compare this with the area which represents ~Xv~Y. We draw overlapping X and Y circles. Then we take the area outside
the first circle (which represents ~X; see Figure 3-5), and we take the area outside the second (which represents ~Y; see Figure 3-
6).
Finally, we put these two areas together to get the area representing the disjunction ~Xv~Y, as represented in Figure 3-7.
Notice that the shaded area of Figure 3-7, representing ~Xv~Y, is the same as that of Figure 3-4, representing ~(X&Y). The fact
that the same shaded area represents both ~Xv~Y and ~(X&Y) means that the two sentences are true in exactly the same cases and
false in the same cases. In other words, they always have the same truth value. And that is just what we mean by two sentences
being logically equivalent.
Now try to prove the other of De Morgan's laws for yourself using Venn diagrams.
Here are two more laws of logical equivalence:
The Dishibutive Laws: For any three sentences, X, Y, and Z, X&(YvZ) is logically equivalent to (X&Y)v(X&Z). And
Xv(Y&Z) is logically equivalent to (XW&(XvZ).
For example, 'Adam is both bold and either clever or lucky.' comes to the same thing as 'Adam is either both bold and clever or
both bold and lucky.' You should prove these two laws for yourself using Venn diagrams. To do so, you will need a diagram with
three circles, one each representing X, Y, and Z. Again, to make sure that you omit no possible combination of truth values, you
must draw these so that they all overlap, as in Figure 3-8.
Fill in the areas to represent YvZ, and then indicate the area which represents the conjunction of this with X. In a separate diagram,
first fill in the areas representing X&Y and X&Z, and then find the area corresponding to the disjunction of these. If the areas
agree, you will have demonstrated logical equivalence. Do the second of the distributive laws similarly. Also, if you feel you need
more practice with truth tables, prove these laws using truth tables.
 Exercise
3-1. Prove the second of De Morgan's laws and the two distributive laws using Venn diagrams. Do this in the same way that I
proved the first of De Morgan's laws in the text, by drawing a Venn diagram for each proof, labeling the circles in the diagram,
and explaining in a few sentences how the alternate ways of getting the final area give the same result. Use more than one
diagram if you find that helpful in explaining your proof.
2.3.1: Logical Equivalence is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We can't do much with our laws of logical equivalence without using a very simple fact, which our next example illustrates.
Consider
(1) ~~AvB
'~~A' is logically equivalent to 'A'. This makes us think that (1) is logically equivalent to
(2) AvB.
This is right. But it is important to understand why this is right. A compound sentence is made up of component sentences, which
in turn may be made up of further component sentences. How do subsentences (components, or components of components, or the
like) affect the truth value of the original sentence? Only through their truth values. The only way that a subsentence has any effect
on the truth values of a larger sentence is through the subsentence's truth value. (This, again, is just what we mean by saying that
compound sentences are truth -functions.) But if only the truth values matter, then substituting another sentence which always has
the same truth value as the first can't make any difference.
I'll say it again in different words: Suppose that X is a subsentence of some larger sentence. Suppose that Y is logically equivalent
to X, which means that Y and X always have the same truth value. X affects the truth value of the larger sentence only through its
(i.e., X's) truth value. So, if we substitute Y for X, there will be no change in the larger sentence's truth value.
But this last fact is just what we need to show our general point about logical equivalence. The larger sentence will have the same
truth value before and after the substitution; that is, the two versions of the larger sentence will be logically equivalent:
The Law of Substirurion of Logical Equivaknts (SLE): Suppose that X and Y are logically equivalent, and suppose that X
occurs as a subsentence of some larger sentence Z. Let Z* be the new sentence obtained by substituting Y for X in Z. Then Z
is logically equivalent to Z*.
Let's apply these laws to an example. Starting with the sentence
~[(~Av~B)&(~AvB)]
we can apply one of De Morgan's laws. This sentence is the negation of a conjunction, with the conjuncts '~Av~B' and '~AvB'. De
Morgan's law tells us that this first line is logically equivalent to the disjunction of the negation of the two original conjunct&:
~(~Av~B)v~(~AvB) DM
(The 'DM' on the right means that this line was obtained from the previous line by applying one of De Morgan's laws.)
Did you have trouble understanding that one of De Morgan's laws applies to the sentence? If so, try using the idea of the main
connective introduced in chapter 1. Ask yourself: "In building this sentence up from its parts, what is the last thing I do?" You
apply the negation sign to '(~Av-B)&(~AvB)'. So you know the original sentence is a negation. Next, ask yourself, what is the last
thing I do in building '(~Av~B)&(~AvB)' up from its parts? Conjoin '~Av~B' with '~AvB'. So '(~Av~B)&(~AvB)' is a conjunction.
The original sentence, then, is the negation of a conjunction, that is, a sentence of the form ~(X&Y), where, in our example, X is
the sentence '~Av~B' and Y is the sentence '~AvB'. Applying De Morgan's law to ~(X&Y) gives ~Xv~Y; in other words, in our
example, '~(~Av~B)v~(~AvB)'.
Next, we can apply De Morgan's law to each of the components, '~(-Av~B)' and '~(~AvB)', and then use the law of substitution of
logical equivalents to substitute the results back into the full sentence. Doing this, we get
(~~A&~~B)v(~~A&~B) DM, SLE
(As before, 'DM' on the right means that we have used one of De Morgan's laws. 'SLE' means that we have also used the law of
substitution of logical equivalents in getting the last line from the previous one.)
Now we can apply the law of double negation (abbreviated 'DN') to '~~A' and to '~~B' and once more substitute the results into the
larger sentence. This gives
(A&B)v(A&-B) DN, SLE
We have only one more step to do. If you look carefully, you will see that the distributive law (abbreviated 'D') applies to the last
line. So the last line is logically equivalent to
A&(Bv~B) D
This might not be clear at first. As I stated the distributive law, you might think it applies only to show that the very last line is
logically equivalent to the next to last line. But if X is logically equivalent to Y, then Y is logically equivalent to X! Logical
equivalence is a matter of always having the same truth value, so if two sentences are logically equivalent, it does not matter which
one gets stated first. Often students only think to apply a law of logical equivalents in the order in which it happens to be stated.
But the order makes no difference-the relation of logical equivalence is symmetric, as logicians say.
Let's put all the pieces of this problem together. In the following summary, each sentence is logically equivalent to the previous
sentence and the annotations on the right tell you what law or laws give you a line from the previous one.
~[(~Av-B)&(~AvB)]
~(~Av-B)v~(~AvB) DM
(~~A&~~B)v(~~A&~B) DM, SLE
(A&B)v(A&~B) DN, SLE
A&(Bv~B) D
Actually, all I have really proved is that each of the above sentences is logically equivalent to the next. I really want to show that
the first is logically equivalent to the last. Do you see why that must be so? Because being logically equivalent just means having
the same truth value in all possible cases, we trivially have
The Law of Trarrcitivity of Logical Equivaltnce (TLE): For any sentences X, Y, and Z, if X is logically equivalent to Y and Y
is logically equivalent to Z, then X is logically equivalent to Z.
Repeated use of this law allows us to conclude that the first sentence in our list is logically equivalent to the last. Many of you may
find this point obvious. From now on, transitivity of logical equivalence will go without saying, and you do not need explicitly to
mention it in proving logical equivalences.
Here are some more easy, but very important, laws:
The Commutative Law (CM): For any sentences X and Y, X&Y is logically equivalent to Y&X. And XvY is logically
equivalent to YvX.
In other words, order in conjunctions and disjunctions does not make a difference. Note that the commutative law allows us to
apply the distributive law from right to left as well as from left to right. For example, '(A&B)vC' is logically equivalent to
'(AvC)&(BvC)'. You should write out a proof of this fact using the commutative law and the distributive law as I stated it
originally.
Next, the Associate Law tells us that 'A&(B&C)' is logically equivalent to '(A&B)&C'. To check this, try using a Venn diagram,
which in this case gives a particularly quick and clear verification. Or simply note that both of these sentences are true only when
'A', 'B', and 'C" are all true, and are false when one or more of the sentence letters are false. This fact shows that in this special case
we can safely get away with dropping the parentheses and simply writing 'A&B&C', by which we will mean either of the logically
equivalent 'A&(B&C)' or '(A&B)&C'. Better yet, we will extend the way we understand the connective '&'. We will say that '&' can
appear between any number of conjuncts. The resulting conjunction is true just in case all of the conjuncts are true, and the
conjunction is false in all other cases.
The same sort of generalization goes for disjunction. 'Av(BvC)' is logically equivalent to '(AvB)vC'. Both of these are true just in
case one or more of 'A', 'B', and 'C' are true and false only if all three of 'A', 'B', and 'C' are false. (Again, a Venn diagram provides a
particularly swift check.) We extend our definition of 'v' so that it can appear between as many disjuncts as we like. The resulting
disjunction is true just in case at least one of the disjunction is true and the disjunction is false only if all the disjunction are false.
The Associative Law (A): For any sentences X, Y, and Z, X&(Y&Z), (X&Y)&Z, and X&Y&Z are logically equivalent to
each other. And Xv(YvZ), (XvY)vZ, and XvYvZ are logically equivalent to each other. Similarly, conjunctions with four or
more components may be arbitrarily grouped and - similarly for disjunctions with four or more disjuncts.
Here is yet another easy law. Clearly, X&X is logically equivalent to X. Likewise, XvX is logically equivalent to X.
The Law of Redundancy (RD): For any sentence X, X&X is logically equivalent to X. Similarly, XvX is logically equivalent
to X
Let us apply this law in a little example. Again, each line is logically equivalent to the next (RD stands for the law of redundancy):
~(A&B)&(~Av~B)
(~Av~B)&(~Av~B) DM, SLE
~Av~B RD
Before asking you to practice these laws, let me give you a more extended example which illustrates all the laws I have introduced
so far:
~(Av~B)v[(CvB)&(Cv~A)]
~(Av~B)v[CV(B&~A)] D, SLE
~(Av~B)v[(B&~A)vC] CM, SLE
[~(Av~B)v(B&~A)]vC A
[~(Av~B)v(~A&B)]vC CM, SLE
[~(Av~B)v(~A&~~B)]vC DN, SLE
[~(Av~B)v~(Av~B)]vC DM, SLE -
(Av~B)vC RD, SLE.
 Exercise
3-2. Prove the following logical equivalences. Write out your proofs as I did in the text specifying which laws you use in
getting a line from the previous line. You can use the abbreviations for the laws found in the text. Until you feel comfortable
with the easy laws, please include all steps. But when they begin to seem painfully obvious, you may combine the following
laws with other steps and omit mentioning that you have used them: double negation, the associative law, the commutative law,
and the law of substitution of logical equivalents. You must explicitly specify any other law you use.
a) 'Bv~A' is logically equivalent to '~(A&~B)'.
b) '(A&B)vC' is logically equivalent to '(AvC)&(BvC)'. (Show all steps in this problem.)
c) 'A&(~~CVB)' is logically equivalent to '(A&C)v(A&B)'.
d) '~[(A&~B)v(C&~B)]' is logically equivalent to '(~A&~C)vB'.
e) '(AvB)&(CVD)' is logically equivalent to '(A&C)v(B&C)v(A&D)v(B&D)'.
f) '(A&B)v(C&D)' is logically equivalent to '(AvC)&(BvC)&(AvD)&(BvD)'.
g) '(C&A)V(B&C)V[C&~(~B&~A)]' is logically equivalent to 'C&(AvB)'.
h) 'C&~A' is logically equivalent to 'C&[~Av~(~CVA)]'.
i) '~A&B&C' is logically equivalent to C&{~(Av~B)v[B&-(~CVA)])'.
3-3. Give a formal statement of De Morgan's laws in application to negations of conjunctions and disjunctions with three
components. Model your formal statement on the formal statement in the text. It should begin as follows:
De Morgan's Laws: For any sentences X, Y, and Z . . .
2.3.2: Substitution of Logical Equivalents and Some More Laws is shared under a not declared license and was authored, remixed, and/or curated
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Let us look at another interesting example:
A ~A Av~A
t f t
f t f
'Av~A' is true no matter what. Such a sentence is called a Logical Truth.

A sentence of sentence logic is a Logical Truth just in case it is true in all possible cases, that is, just in case it is true for all
assignments of truth values to sentence letters.
Many authors use the word Tautology for a logical truth of sentence logic. I prefer to use the same expression, 'logical truth', for
this idea as it applies in sentence logic and as it applies to predicate logic, which we will study in volume II.
Clearly there will also be sentences which are false no matter what, such as
A ~A A&~A
t f f
f t f
Such a sentence is called a Contradiction.

A sentence of sentence logic is a Contradiction just in case it is false in all possible cases, that is, just in case it is false for all
assignments of truth values to sentence letters.
Later on in the course, logical truths and contradictions will concern us quite a bit. They are interesting here because they provide
several further laws of logical equivalence:
The Law of Logically True Conjunct (LTC): If X is any sentence and Y is any logical truth, then X&Y is logically equivalent
to X
The Law of Contradictory Disjunct (CD): If X is any sentence and Y is any contradiction, then XvY is logically equivalent
to X.
You should be able to show that these laws are true. Furthermore, you should satisfy yourself that a conjunction is always a
contradiction if one of its conjuncts is a contradiction and that a disjunction is always a logical truth if one of its disjuncts is a
logical truth.
3-4. Explain why a disjunction is always a logical truth if one of its disjuncts is a logical truth. Explain why a conjunction is
always a contradiction if one of its conjuncts is a contradiction.
3-5. Further simplify the sentence 'A&(Bv~B)', which was the last line of the first example in section 3-2.
3-6. Prove the following logical equivalences, following the same instructions as in exercise 3-2:
a) 'A&(~AvB)' is logically equivalent to 'A&B'.b) 'AvB' is logically equivalent to 'Av(~A&B)'.
c) 'A' is logically equivalent to '(A&B)v(A&~B)'. (This equivalence is called the Law of Expansion. You may find it
useful in some of the &her problems.)
d) 'A' is logically equivalent to '(AvB)&(Av~B)'.
e) 'A&[Bv(~A&C)I' is logically equivalent to 'A&B'.
f) 'CvB' is logically equivalent to '(C&A)v(B&A)v(C&~A)v(B&~A)'.
g) 'C&B' is logically equivalent to '(CvA)&(Bv~D)&(~AvC)&(DvB)'.
h) '(A&B)v(-A&-B)' is logically equivalent to '(~AvB)&(~BvA)'.
i) '~Av~BvC' is logically equivalent to '~(~AvB)v~AvC'.
3-7. For each of the following sentences, determine whether it is a logical truth, a contradiction, or neither. (Logicians say that
a sentence which is neither a logical truth nor a contradiction is Contingent, that is, a sentence which is true in some cases and
false in others.) Simplify the sentence you are examining, using the laws of logical equivalence, to show that the sentence is
logically equivalent to a sentence you already know to be a logical truth, a contradiction, or neither.
a) (B&A)v(B&~A)
b) B&[)~AvA)&~B)]
c) (A&B)v[(A&~B)&(~A&B)]
d) (AvB)v~(A&C)
e) (AvB)&(Av~B)&(~AVB)&(~Av~B)
f) ~A&~B&(AvB)
g) (AvC)&C&(AvB)
h) (A&B)v~Bv~A
i) (~B&A)v(~A&B)v(~B&~A)
j) (Av~AvB)&(Av~Av~C)
k) (A&C)v~(~BvC)v~(AvB)vC
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Now that we understand logical equivalence, we can use it to put any sentence into a form which shows very clearly what the
sentence says. As usual, we will start by looking at an example. Start with the truth table for Av~B:
Case A B ~B Av~B
1 t t f t
2 t f t t
3 f t f f
4 f f t f
The truth table tells us that 'Av~B' is true in cases 1, 2, and 4. We can easily say what case 1 says using a sentence of sentence
logic. Case 1 just says that 'A' and 'B' are both true, which we can say with 'A&B'. In the same way, case 2 says that 'A' is true and
'B' is false, which we say in sentence logic with 'A&~B'. Finally, '~A&~B' says that case 4 holds, that is, that 'A' is false and 'B' is
false. Of course, none of these things says what 'Av-B' says, which was that either case 1 is true or case 2 is true or case 4 is true.
But, clearly, we can say this in sentence logic by using the disjunction of the three sentences, each one of which describes one of
the cases covered by 'Av-B'. That is
'AV~B' is logically equivalent to '(A&B)v(A&~B)v(~A&~B)'
'(A&B)v(A&~B)v(~A&~B)'. '(A&B)v(A&~B)v(~A&~B)' is said to be in Disjunctive Normal Form, and it says that either 'A' and
'B' are both true or 'A' is true and 'B' is false or 'A' is false and 'B' is false. This disjunction is logically equivalent to 'Av~B' because
the disjunction says just what 'Av-B' says, as shown by its truth table.
Here is a slightly different way of putting the same point. The truth table shows us the possible cases in which the sentence under
study will be true. We can always write a logically equivalent sentence in disjunctive normal form by literally writing out the
information contained in the truth table. For each case in which the original sentence comes out true, write the conjunction of
sentence letters and negated sentence letters which describe that case. Then take the disjunction of these conjunctions. This
sentence will be true in exactly those cases described by the disjuncts (in our example, in the cases described by 'A&B', by 'A&-B',
and by '-A&-B'). But the original sentence is true in just these same cases-that is what its truth table tells us. So the sentence in
disjunctive normal form is true in exactly the same cases as the original, which is to say that the two are logically equivalent.
I want to give you a formal summary description of disjunctive normal form. But first we must deal with three troublesome special
cases. I will neutralize these troublesome cases with what may at first strike you as a very odd trick.
Consider the atomic sentence 'A'. As I have so far defined disjunctions and conjunctions, 'A' is neither a disjunction nor a
conjunction. But because 'A' is logically equivalent to 'AvA', it will do no harm if we extend the meaning of 'disjunction' and say
that, in a trivial way, 'A' will also count as a disjunction-that is, as a degenerate disjunction which has only one disjunct.
We can play the same trick with conjunctions. 'A' is logically equivalent to 'A&A'. So we do no harm if we extend the meaning of
'conjunction' and say that, in a trivial way, 'A' also counts as a conjunction-the degenerate conjunction which has only one conjunct.
Finally, we will want to say the same thing, not just about atomic sentence letters, but about any sentence, X, atomic or compound.
Whatever the form of X, we can always view X as a degenerate disjunction with just one disjunct or as a degenerate conjunction
with just one conjunct.
What is the point of this apparently silly maneuver? I am working toward giving a definition of disjunctive normal form which will
work for any sentence. The idea of disjunctive normal form is that it involves a disjunction of conjunctions. But what should we
say, for example, is the disjunctive normal form of the sentence 'A&B'? If we allow degenerate disjunctions with one disjunct, we
can say that 'A&B' is already in disjunctive normal form-think of it as '(A&B)v(A&B)'. Again, what should we say is the
disjunctive normal form of 'A'? Let's count 'A' as a degenerate conjunction with just one conjunct (think of 'A&A') and let's count
this conjunction as a degenerate disjunction, as in the last example. So we can say that 'A' is already in disjunctive normal form and
still think of disjunctive normal form as a disjunction of conjunctions.
We still have to discuss one more special case. What should we say is the disjunctive normal form of a contradiction, such as
'A&~A'? We will allow repetitions of sentence letters with and without negation signs, so that, again, 'A&~A' will itself already
count as being in disjunctive normal form.
Now we can say very simply:
A sentence is in Disjunctive Normal Form if it is a disjunction, the disjuncts of which are themselves conjunctions of
sentence letters and negated sentence letters. In this characterization we allow as a special case that a disjunction may have
only one disjunct and a conjunction may have only one conjunct.
For any sentence, X, of sentence logic, the disjunctive normal form of X is given by a sentence Y if Y is in disjunctive
normal form and is logically equivalent to X. Except for contradictions, the disjunctive normal form of a sentence is the
sentence's truth table expressed in sentence logic.
The fact that every sentence of sentence logic is logically equivalent to a sentence in disjunctive normal form helps to show
something interesting about the connectives. All our sentences are put together using '&', 'v', and '~'. But are these connectives all
we really need? Could we say new things if we added new connectives? The answer is no, if we limit ourselves to sentences which
can be given in terms of a truth table. Because we can write any truth table in disjunctive normal form, using only '&', 'v' and '~',
anything which we can express using a truth table we can express using just these three connectives. In other words, '&', 'v', and '~'
are enough if we limit ourselves to a logic all the sentences of which are truth functions of atomic sentence letters. We say that '&',
'v', and '~' are, together, Expressively Complete. For given the truth table of any sentence which we might want to write, we can
always write it with a sentence in disjunctive normal form.
Even more interestingly, '&', 'v' and '~' are more than we need. Using De Morgan's laws and double negation, we can always get rid
of a conjunction in favor of a disjunction and some negation signs. And we can always get rid of a disjunction in favor of a
conjunction and some negation signs. (Do you see how to do this?) Thus any sentence which can be represented by a truth table can
be expressed using just '&' and '-'. And any such sentence can be expressed using just 'v' and '~'. So '&' and '~' are expressively
complete, and 'v' and '-' are also expressively complete.
We have just seen that anything that can be represented with truth tables can be expressed with a sentence using just two
connectives. Could we make do with just one connective? Clearly, we can't make do with just '&', with just 'v', or with just '~'. (Can
you see why?) But perhaps we could introduce a new connective which can do everything all by itself. Consider the new
connective 'I*, called the Sheffer Stroke, defined by
X Y X|Y
t t f
t f f
f t f
f f t
Work out the truth table and you will see that X|X is logically equivalent to ~X. Similarly, you can prove that (X|Y)|(X|Y) is
logically equivalent to XvY. With this new fact, we can prove that '|' is expressively complete. We can express any truth function in
disjunctive normal form. Using De Morgan's law and the law of double negation, we can get rid of the '&'s in the disjunctive
normal form. So we can express any truth function using just 'v' and '~'. But now for each negation we can substitute a logically
equivalent expression which uses just '|'. And for each disjunction we can also substitute a logically equivalent expression which
uses just '|'. The final result uses '|' as its only connective. Altogether, the sentence in disjunctive normal form has been transformed
into a logically equivalent sentence using just '1'. And because any truth function can be put in disjunctive normal form, we see that
any truth function, that is, any sentence which could be given a truth table definition, can be expressed using just '|'.
The important idea here is that of expressive completeness:
A connective, or set of connectives, is Expressively Complete for truth functions if and only if every truth function can be
represented using just the connective or connectives.
Actually, the really important idea is that of a truth function. Understanding expressive completeness for truth functions will help to
make sure you have the idea of a truth function clearly in mind.
 Exercise 3-8
Put the following sentences in disjunctive normal form. You can do this most straightforwardly by writing out truth tables for
the sentences and then reading off the disjunctive normal form from the truth tables. Be sure you know how to work the
problems this way. But you might have more fun trying to put a sentence in disjunctive normal form by following this
procedure: First, apply De Morgan's laws to drive all negations inward until negation signs apply only to sentence letters. Then
use other laws to get the sentence in the final disjunctive normal form.
a) ~(A&B)
b) ~[(A&B)v(~A&C)]
3-9. Suppose you are given a sentence in which 'v' occurs. Explain in general how you can write a logically equivalent sentence
in which 'v' does not occur at all. Similarly, explain how a sentence in which '&' occurs can be replaced by a logically
equivalent sentence in which '&' does not occur. (Hint: You will, need to appeal to De Morgan's laws.)
3-10. Define a new connective, '*', as representing the following truth function:
Case X Y X*Y
1 t t f
2 t f t
3 f t t
4 f f t
Show that '*' is expressively complete.

3-11. Show that '&' is not expressively complete. That is, give a truth function and show that this truth function cannot be
expressed by using '&' as the only connective. Similarly, show that 'v' is not expressively complete and show that '~' is not
expressively complete. (You may find this problem hard, but please take a few minutes to try to work it.)
 Chapter Summary Exercise
Once again, you will find below the important terms which I have introduced in this chapter. Make sure you understand all of
them by writing out a short explanation of each. You should refer to the text to make sure that you have correctly explained
each term. Please keep your explanations of these terms in your notebook for reference and review.
a. Logical Equivalence
b. Venn Diagram
c. Law of Double Negation
d. De Morgan's Laws
e. Distributive Laws
f. Law of Substitution of Logical Equivalents
g. Law of Transitivity of Logical Equivalence
h. Commutative Laws
i. Associative Law
j. Law of Redundancy
k. LogicalTruth
l. Contradiction
m. Law of Logically True Conjunct
n. Law of Contradictory Disjunct
If you have read section 3-4, also explain
o. Disjunctive Normal Form
p. Expressively Complete
q. Sheffer Stroke
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SECTION OVERVIEW
2.4: Validity and Conditionals

2.4.1: Validity
2.4.2: Invalidity and Counter Examples
2.4.3: Soundness

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4.1: Validity
Consider the following argument:
AvB Adam loves Eve or Adam loves Bertha.
~A Adam does not love Eve.
B Adam loves Bertha.
If you know, first of all, that either 'A' or 'B' is true, and in addition you know that 'A' itself is false; then clearly, 'B' has to be true.
So from 'AvB' and '~A' we can conclude 'B'. We say that this argument is Valid, by which we mean that, without fail, if the
premises are true, then the conclusion is going to turn out to be true also.
Can we make this idea of validity more precise? Yes, by using some of the ideas we have developed in the last three chapters.
(Indeed one of the main reasons these ideas are important is that they will help us in making the notion of validity very precise.)
Let us write out a truth table for all the sentences appearing in our argument:
Case A B ~A AvB
1 t t f t
2 t f f t
3 f t t t
4 f f t f
We know that cases 1 through 4 constitute all the ways in which any of the sentences in the argument may turn out to be true or
false. This enables us to explain very exactly what we mean by saying that, without fail, if the premises are true, then the
conclusion is going to turn out to be true also. We interpret this to mean that in each possible case (in each of the cases 1 through
4), if the premises are true in that qe, then the conclusion is true in that case. In other words, in all cases in which the premises are
true, the conclusion is also true. In yet other words:
To say that an argument (expressed with sentences of sentence logic) is Valid is to say that any assignment of truth values to
sentence letters which makes all of the premises true also makes the conclusion true.
4.1: Validity is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
4.2: Invalidity and Counter Examples
Let's look at an example of an Invalid argument (an argument which is not valid):
AvB
A
B
Counterexample:
Case A B AvB
1 t t t
2 t f t
3 f t t
4 f f f
I have set up a truth table which shows the argument to be invalid. First I use a '*' to mark each case in which the premises are all
true. In one of these cases (the second) the conclusion is false. This is what can't happen in a valid argument. So the argument is
invalid. I will use the term Counterexample for a case which in this way shows an argument to be invalid. A counterexample to an
argument is a case in which the premises are true and the conclusion is false.
In fact, we can use this idea of a counterexample to reword the definition of validity. To say that an argument is valid is to say that
any assignment of truth values to sentence letters which makes all of the premises true also makes the conclusion true. We reword
this by saying: An argument is valid just in case there is no possible case, no assignment of truth values to sentence letters, in which
all of the premises are true and the conclusion is false. To be valid is to rule out any such possibility. We can break up this way of
explaining validity into two parts:
A Counterexample to a sentence logic argument is an assignment of truth values to sentence letters which makes all of the
premises true and the conclusion false.
An argument is Valid just in case there are no counterexamples to it.
Now let us reexpress an of this using sentences of sentence logic and the idea of logical truth. Let us think of an argument in which
X is the conjunction of an the premises and Y is the conclusion. X and Y might be very complicated sentences. The argument looks
like this:
X
Y
I will express an argument such as this with the words "X. Therefore Y".
A counterexample to such an argument is a case in which X is true and Y is false, that is, a case in which X&~Y is true. So to say
that there are no possible cases in which there is a counterexample is to say that in all possible cases X&-Y is false, or, in all
possible cases ~(X&~Y) is true. But to say this is just to say that ~(X&~Y) is a logical truth. The grand conclusion is that
The argument "X. Therefore Y is valid just in case the sentence ~(X&~Y) is a logical truth.
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2.4.3: Soundness
Logic is largely about validity. So to understand clearly what much of the rest of this book is about, you must clearly distinguish
validity from some other things.
If I give you an argument by asserting to you something of the form "X. Therefore Y", I am doing two different things. First, I am
asserting - the premise or premises, X. Second, I am asserting to you that from these premises the conclusion, Y follows.
To see clearly that two different things are going on here, consider that there are two ways in which I could be mistaken. It could
turn out that I am wrong about the claimed truth of the premises, X. Or I could be wrong about the 'therefore'. That is, I could be
wrong that the conclusion, Y, validly follows from the premises, X To claim that X is true is one thing. It is quite another thing to
make a claim corresponding to the 'therefore', that the argument is valid, that is, that there is no possible case in which the premises
are true and the conclusion is false.
Some further, traditional terminology helps to emphasize this distinction. If I assert that the argument, "X. Therefore Y", is valid, I
assert something about the relation between the premises and the conclusion, that in all lines of the truth table in which the
premises all turn out true, the conclusion turns out true also. In asserting validity, I do not assert that the premises are in fact true.
But of course, I can make this further assertion. To do so is to assert that the argument is not only valid, but Sound:
An argument is Sound just in case, in addition to being valid, all its premises are true.
Logic has no special word for the case of a valid argument with false premises.
To emphasize the fact that an argument can be valid, but not sound, here is an example;
Teller is ten feet tall or Teller has never taught logic. AvB
Teller is not ten feet tall. ~A
Teller has never taught logic. B
Viewed as atomic sentences, 'Teller is ten feet tall.' and 'Teller has never taught logic.' can be assigned truth values in any
combination, so that the truth table for the sentences of this argument looks exactly like the truth table of Section 4-1. The
argument is perfectly valid. Any assignment of truth values to the atomic sentences in which the premises both come out true (only
case 3) is an assignment in which the conclusion comes out true also. But there is something else wrong with the argument of the
present example. In the real world, case 3 does not in fact apply. The argument's first premise is, in fact, false. The argument is
valid, but not sound.
4-1. Give examples, using sentences in English, of arguments of each of the following kind. Use examples in which it is easy
to tell whether the premises and the conclusion are in fact (in real life) true or false.
a) A sound argument
b) A valid but not sound argument with a true conclusion
C) A valid but not sound argument with a false conclusion
d) An argument which is not valid (an invalid argument) all the premises of which are true
e) An invalid argument with one or more false premises
4-2. Use truth tables to determine which of the following arguments are valid. Use the following procedure, showing all your
work: First write out a truth table for all the sentences in the argument. Then use a '*' to mark all the lines of the truth table in
which all of the argument's premises are true. Next look to see whether the conclusion is true in the *ed lines. If you find any
*ed lines in which the conclusion is false, mark these lines with the word 'counterexample'. You know that the argument is
valid if and only if there are no counterexamples, that is, if and only if all the cases in which all the premises are true are cases
in which the conclusion is also true. Write under the truth table whether the argument is valid or invalid (i.e., not valid).
a) ~(A&B) b) ~AvB c) AvB d) AvB e) A
~A A ~BvA ~AvB Bv~C
~B B A A (A&B)v(A&~C)
4-3. Show that X is logically equivalent to Y if and only if the arguments "X. therefore Y and "Y. Therefore X" are both valid.
2.4.3: Soundness is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
In Section 4-2 we saw that the argument, "X. Therefore Y", is intimately related to the truth function ~(X&~Y). This truth function
is so important that we are going to introduce a new connective to represent it. We will define X⊃Y to be the truth function which
is logically equivalent to ~(X&~Y). You should learn its truth table definition:
Truth table Definition of ⊃
X Y X⊃Y
t t t
t f f
f t t
f f t
Again, the connection between X⊃Y and the argument "X Therefore Y" is that X⊃Y is a logical truth just in case the argument "X
Therefore Y" is valid.
Logicians traditionally read a sentence such as 'A⊃B' with the words 'If A, then B', and the practice is to transcribe 'If. . . then . . .'
sentences of English by using '⊃'. So (to use a new example) we would transcribe 'If the cat is on the mat, then the cat is asleep.' as
'A⊃B'.
In many ways, this transcription proves to be problematic. To see why, let us forget '⊃' for a moment and set out afresh to define a
truth functional connective which will serve as a transcription of the English 'If. . . then . . .':
(In the next two paragraphs, think of the example, 'If the cat is on the mat, then the cat is asleep.')
Case A B If A then B
1 t t
2 t f
3 f t
4 f f
That is, by choosing t or f for each of the boxes under 'If A then B' in the truth table, we want to write down a truth function which
says as closely as possible what 'If A then B' says in English.
The only really clear-cut case is case 2, the case in which the cat is on the mat but is not asleep. In this circumstance, the sentence
'If the cat is on the mat, then the cat is asleep.' is most assuredly false. SO we have to put f for case 2 in the column under 'If A then
B'. If the cat is both on the mat and is asleep, that is, if we have case 1, we may plausibly take the conditional sentence to be true.
So let us put t for case 1 under 'If A then B'. But what about cases 3 and 4, the two cases in which A is false? If the cat is not on the
mat, what determines whether or not the conditional, 'If the cat is on the mat, then the cat is asleep.', is true or false?
Anything we put for cases 3 and 4 is going to give us problems. Suppose we put t for case 3. This is to commit ourselves to the
following: When the cat is not on the mat and the cat is asleep somewhereklse, then the conditional, 'If the cat is on the mat, then
the cat is asleep.', is true. But suppose we have sprinkled the mat with catnip, which always makes the cat very lively. Then, if we
are going to assign the conditional a truth value at all, it rather seems that it should count as false. On the other hand, if we put f for
case 3, we will get into trouble if the mat has a cosy place by the fire which always puts the cat to sleep. For then, if we assign a
truth value at all, we will want to say that the conditional is true. Similar examples show that neither t nor f will always work for
case 4.
Our problem has a very simple source: 'If. . . then . . .' in English can be used to say various things, many of which are not truth
functional. Whether or not an 'If. . . then . . .' sentence of English is true or false in these nontruth functional uses depends on more
than just the truth values of the sentences which you put in the blanks. The truth of 'If you are five feet five inches tall, then you
will not be a good basketball player.' depends on more than the truth or falsity of 'You are five feet five inches tall.' and 'You will
not be a good basketball player.' It depends on the fact that there is some factual, nonlogical connection between the truth and
falsity of these two component sentences.
In many cases, the truth or falsity of an English 'If. . . then . . .' sentence depends on a nonlogical connection between the truth and
falsity of the sentences which one puts in the blanks. The connection is often causal, temporal, or both. Consider the claim that 'If
you stub your toe, then it will hurt.' Not only does assertion of this sentence claim that there is some causal connection between
stubbing your toe and its hurting, this assertion also claims that the pain will come after the stubbing. However, sentence logic is
insensitive to such connections. Sentence logic is a theory only of truth functions, of connectives which are defined entirely in
terms of the truth and falsity of the component sentences. So no connective defined in sentence logic can give us a good
transcription of the English 'If. . . then . . .' in all its uses.
What should we do? Thus far, one choice for cases 3 and 4 seems as good (or as bad) as another. But the connection between the
words '. . . therefore . . .' and 'If. . . then . . .' suggests how we should make up our minds. When we use 'If. . . then . . .' to express
some causal, temporal, or other nonlogical connection between things in the world, the project of accurately transcribing into
sentence logic is hopeless. But when we use 'If. . . then . . .' to express what we mean by '. . . therefore . . .' our course should be
clear. To assert "X. Therefore Y", is to advance the argument with X as premise(s) and Y as conclusion. And to advance the
argument, "X Therefore Y, is (in addition to asserting X) to assert that the present case is not a counterexample; that is, it is to
assert that the sentence ~(X&~Y) is true. In particular, if the argument, "X. Therefore Y", is valid, there are no counterexamples,
which, as we saw, comes to the same thing as ~(X&~Y) being a logical truth.
Putting these facts together, we see that when "If X then Y" conveys what the 'therefore' in "X. Therefore Y conveys, we can
transcribe the "If X then Y' as ~(X&~Y), for which we have introduced the new symbol X⊃Y. In short, when 'If. . . then . . .' can
be accurately transcribed into sentence logic at all, we need to choose t for both cases 3 and 4 to give us the truth table for X⊃Y
defined as ~(X&~Y).
Logicians recognize that '⊃' is not a very faithful transcription of 'If . . . then . . .' when 'If . . . then . . .' expresses any sort of
nonlogical connection. But since '⊃ agrees with 'If. . . then . . .' in the clear case 2 and the fairly clear case I, '⊃' is going to be at
least as good a transcription as any alternative. And the connection with arguments at least makes '⊃' the right choice for cases 3
and 4 when there is a right choice, that is, when 'If. . . then . . .' means '. . . therefore . . .'.
We have labored over the introduction of the sentence logic connective '⊃'. Some logic texts just give you its truth table definition
and are done with it. But logicians use the '⊃' so widely to transcribe the English 'If. . . then . . .' that you should appreciate as
clearly as possible the (truth functional) ways in which '⊃' does and the (nontruth functional) ways in which '⊃' does not
correspond to 'If. . . then . . .'.
In these respects, the English 'and' and 'or' seem very different. 'And' and 'or' seem only to have truth functional aspects, so that they
seem to correspond very closely to the truth functionally defined '&' and 'v'. Now that you have been through some consciousness
raising about how English can differ from logic in having nontruth functional aspects, it is time to set the record straight about the
'and' and 'or' of English.
Surely, when I assert, 'Adam exchanged vows with Eve, and they became man and wife.' I do more than assert the truth of the two
sentences 'Adam exchanged vows with Eve.' and 'They became man and wife.' I assert that there is a connection, that they enter
into the state of matrimony as a result of exchanging vows. Similarly, if I yell at you, 'Agree with me or I'll knock your block off!'
I do more than assert that either 'You will agree with me' or 'I will knock your block off is true. I assert that nonagreement will
produce a blow to your head. In these examples 'and' and 'or' convey some causal, intentional, or conventional association which
goes above and beyond the truth functional combination of the truth values of the component sentences. 'And' can likewise clearly
express a temporal relation which goes beyond the truth values of the components. When I say, 'Adam put on his seat belt and
started the car.' I assert not only that 'Adam put Gn his seat belt.' and 'He started the car.' are both true. I also assert that the first
happened before the second.
Although 'and', 'or', and 'If. . . then . . .' all have their nontruth functional aspects, in this respect 'If. . . then . . .' is the most striking.
'⊃' is much weaker than 'If. . . then . . .', inasmuch as '⊃' leaves out all of the nontruth functional causal, temporal, and other
connections often conveyed when we use 'If. . . then . . .'. Students sometimes wonder: If '⊃' (and '&' and 'v') are so much weaker
than their English counterparts, why should we bother with them? The answer is that although truth functional sentence logic will
only serve to say a small fraction of what we can say in English, what we can say with sentence logic we can say with profound
clarity. In particular, this clarity serves as the basis for the beautifully clear exposition of the nature of deductive argument.
When the language of logic was discovered, its clarity so dazzled philosophers and logicians that many hoped it would ultimately
replace English, at least as an all-encompassing exact language of science. Historically, it took decades to realize that the clarity
comes at the price of important expressive power.
But back to '⊃'.
Here are some things you are going to need to know about the connective '⊃':
A sentence of the form X⊃Y is called a Conditional. X is called its Antecedent and Y is called its Consequent.
Look at the truth table definition of X⊃Y and you will see that, unlike conjunctions and disjunctions, conditions are not symmetric.
That is, X⊃Y is not logically equivalent to Y⊃X. So we need names to distinguish between the components. This is why we call
the first component the antecedent and the second component the consequent (not the conclusion conclusion is a sentence in an
argument).
Probably you will most easily remember the truth table definition of the conditional if you focus on the one case in which it is false,
the one case in which the conditional always agrees with English. Just remember that a conditional is false if the antecedent is true
and the consequent is false, and true in all other cases. Another useful way for thinking about the definition is to remember that if
the antecedent of a conditional is false, then the whole conditional is true whatever the truth value of the consequent. And if the
consequent is true, then again the conditional is true, whatever the truth value of the antecedent.
Finally, you should keep in mind some logical equivalences:
The Law of the Conditional (C): X⊃Y is logically equivalent to ~(X&~Y) and (by De Morgan's law) to ~XvY.
The Law of Contraposition (CP): X⊃Y is logically equivalent to ~Y⊃~X.
2.4.4: The Conditional is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We introduce one more connective into sentence logic. Often we will want to study cases which involve a conjunction of the form
(X⊃Y)&(Y⊃X). This truth function of X and Y occurs so often in logic that we give it its own name, the Biconditional, which we
write as X≡Y. Working out the truth table of (X⊃Y)&(Y⊃X) we get as our definition of the biconditional:
Truth table Definition of ≡
X Y X≡Y
t t t
t f f
f t f
f f t
Because a biconditional has a symmetric definition, we don't have different names for its components. We just call them
'components'. You will remember this definition most easily by remembering that a biconditional is true if both components have
the same truth value (both true or both false), and it is false if the two components have different truth values (one true, the other
false). We read the biconditional X≡Y with the words 'X if and only if Y. With the biconditional, we get into much less trouble with
transcriptions between English and sentence logic than we did with the conditional.
Given the way we define '≡', we have the logical equivalence:
The Law of the Biconditiuml (B): X≡Y is logically equivalent to (X⊃Y)&(Y⊃X).
Remember that the conditional, X⊃Y, is a logical truth just in case the corresponding argument, "X. Therefore Y, is valid.
Likewise, there is something interesting we can say about the biconditional, X≡Y, being a logical truth:
X≡Y is a logical truth if and only if X and Y are logically equivalent.
Can you see why this is true? Suppose X≡Y is a logical truth. This means that in every possible case (for every assignment of truth
values to sentence letters) X≡Y is true. But X≡Y is true only when its two components have the same truth value. So in every
possible case, X and Y have the same truth value, which is just what we mean by saying that they are logically equivalent. On the
other hand, suppose that X and Y are logically equivalent. This just means that in every possible case they have the same truth
value. But when X and Y have the same truth value, X≡Y is true. So in every possible case X≡Y is true, which is just what is
meant by saying that X≡Y is a logical truth.
4-4. In section 1-6 1 gave rules of formation and valuation for sentence logic. Now that we have extended sentence logic to
include the connectives '3' and '=', these rules also need to be extended. Write the full rules of formation and valuation for
sentence logic, where sentence logic may now use all of the connectives '-', '&', 'v', '⊃', and '≡'. In your rules, also provide for
three and more place conjunctions and disjunctions as described in section 3-2 in the discussion of the associative law.
4-5. Follow the same instructions as for exercise 4-2.
a) A⊃B b) A⊃~B c) A≡B d) A≡~B
B B Av~B Av~B
A ~A B AvB
e) (AvB)⊃(A&C) f) (AvB)≡(Av-C)
CvA ~BvC
~C AvC
4-6. For each of the following sentences, establish whether it is a logical truth, a contradiction, or neither. Use the laws of
logical equivalence in chapter 3 and sections 4-3 and 4-4, and use the fact that a biconditional is a logical truth if and only if its
components are logically equivalent.
4-7. Discuss how you would transcribe 'unless' into sentence logic. Experiment with some examples, trying out the use of 'v',
'>', and I='. Bear in mind that one connective might work well for one example, another connective for another example. As
you work, pay attention to whether or not the compound English sentences you choose as examples are truth functional. Report
the results of your research by giving the following:
a) Give an example of a compound English sentence using 'unless' which seems to be nontruth functional, explaining
why it is not truth functional.
b) Give an example of a compound English sentence using 'unless' which seems to be truth functional, explaining why it
is truth functional.
c) Give one example each of English sentences using 'unless' which can be fairly well transcribed into sentence logic
using 'v', '⊃', '≡', ' giving the transcriptions into sentence logic.
4-8. Transcribe the following sentences into sentence logic, using the given transcription guide:
A: Adam loves Eve. D: Eve has dark eyes.
B: Adam is blond. E: Eve loves Adam.
C:Eve is clever.
a) If Eve has dark eyes, then Adam does not love her.
b) Adam loves Eve if she has dark eyes.
c) If Adam loves Eve, Eve does not love Adam.
d) Eve loves Adam only if he is not blond.
e) Adam loves Eve if and only if she has dark eyes.
f) Eve loves Adam provided he is blond.
g) Provided she is clever, Adam loves Eve.
h) Adam does not love Eve unless he is blond.
i) Unless Eve is clever, she does not love Adam.
j) If Adam is blond, then he loves Eve only if she has dark eyes.
k) If Adam is not blond, then he loves Eve whether or not she has dark eyes.
l) Adam is blond and in love with Eve if and only if she is clever.
m) Only if Adam is blond is Eve both clever and in love with Adam.
4-9. Consider the following four different kinds of nontruth functional connectives that can occur in English:
a) Connectives indicating connections (causal, intentional, or conventional)
b) Modalities (what must, can, or is likely to happen)
c) So-called "propositional attitudes," having to do with what people know, believe, think, hope, want, and the like
d) Temporal connectives, having to do with what happens earlier, later, or at the same time as something else.
Give as many English connectives as you can in each category. Keep in mind that some connectives will go in more than one
category. ('Since' is such a connective. What two categories does it go into?) To get you started, here are some of these
connectives: 'because', 'after', 'more likely than', 'Adam knows that', 'Eve hopes that'.
 Chapter summary Exercise

Give brief explanations for each of the following. As usual, check your explanations against the text to make sure you get them
right, and keep them in your notebook for reference and review.
a) Valid
b) Invalid
c) Counterexample
d) Sound
e) Conditional
f) Biconditional
g) Law of the Conditional
h) Law of Contraposition
i) Law of the Biconditional
2.4.5: The Biconditional is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
2.5: Natural Deduction for Sentence Logic - Fundamentals


2.5: Natural Deduction for Sentence Logic - Fundamentals is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
In chapter 4 you learned that saying an argument is valid means that any case which makes all of the argument's premises true also
makes its conclusion true. And you learned how to test for validity by using truth tables, by exhaustively checking all the relevant
cases, that is, all the lines of the truth table. But truth tables are horribly awkward. It would be nice to have a way to check validity
which looked more like the forms of argument we know from everyday life.
Natural deduction does just that. When we speak informally, we use many kinds of valid arguments. (I'll give some examples in a
moment.) Natural deduction makes these familiar forms of argument exact. It also organizes them in a system of valid arguments in
which we can represent absolutely any valid argument.
Let's look at some simple and, I hope, familiar forms of argument. Suppose I know (say, because I know Adam's character) that if
Adam loves Eve, then he will ask Eve to many him. I then find out from Adam's best friend that Adam does indeed love Eve.
Being a bright fellow, I immediately conclude that a proposal is in the offing. In so doing I have used the form of argument
traditionally called 'modus ponens', but which I am going to call Conditional Elimination.
Conditional Elimination
X⊃Y
X ⊃E
Y
Logicians call such an argument form a Rule of Inference. If, in the course of an argument, you are given as premises (or you have
already concluded) a sentence of the form X⊃Y and the sentence X, you may draw as a conclusion the sentence Y. This is because,
as you can check with a truth table, in any case in which sentences of the form X⊃Y and X are both true, the sentence Y will be
true also. You may notice that I have stated these facts, not for some particular sentences 'A⊃B', 'A', and 'B', but for sentence forms
expressed with boldfaced 'X' and 'Y'. This is to emphasize the fact that this form of argument is valid no matter what specific
sentences might occur in the place of 'X' and 'Y'.
Here is another example of a very simple and common argument form, or rule of inference:
Disjunction Elimination
XvY
~X vE
Y
If I know that either Eve will marry Adam or she will marry no one, and I then somehow establish that she will not marry Adam
(perhaps Adam has promised himself to another), I can conclude that Eve will marry no one. (Sorry, even in a logic text not all love
stories end happily!) Once again, as a truth table will show, this form of argument is valid no matter what sentences occur in the
place of 'X' and in the place of 'Y'.
Though you may never have stopped explicitly to formulate such rules of argument, all of us use rules like these. When we argue
we also do more complicated things. We often give longer chains of argument which start from some premises and then repeatedly
use rules in a series of steps. We apply a rule to premises to get an intermediate conclusion. And then, having established the
intermediate conclusion, we can use it (often together with some of the other original premises) to draw further conclusions.
Let's look at an example to illustrate how this process works. Suppose you are given the sentences 'A⊃B', 'B⊃C', and 'A' as
premises. You are asked to show that from these premises the conclusion 'C' follows. How can you do this?
It's not too hard. From the premises 'A⊃B' and 'A', the rule of conditional elimination immediately allows you to infer 'B':
A⊃B
A ⊃E
C
But now you have 'B' available in addition to the original premise 'B⊃C'. From these two sentences, the rule of conditional
elimination allows you to infer the desired conclusion 'C':
B⊃C
B ⊃E
C
I hope this example is easy to follow. But if I tried to write out an example with seven steps in this format, things would get
impossibly confusing. We need a streamlined way of writing chains of argument.
The basic idea is very simple. We begin by writing all our premises and then drawing a line to separate them from the conclusions
which follow. But now we allow ourselves to write any number of conclusions below the line, as long as the conclusions follow
from the premises. With some further details, which I'll explain in a minute, the last example looks like this:
1 | A⊃B P
2 | B⊃C P
3|AP
4 | B 1,3,⊃E
5 | C 2,4,⊃E
Lines 1 through 5 constitute a Derivation of conclusions 4 and 5 from premises 1, 2, and 3. In thinking about such a derivation, you
should keep most clearly in mind the idea that the conclusions are supposed to follow from the premises, in the following sense:
Any assignment of truth values to sentence letters which makes the premises all true will also make all of the conclusions true.
In a derivation, every sentence below the horizontal line follows from the premises above the line. But sentences below the line
may follow directly or indirectly. A sentence follows directly from the premises if a rule of inference applies directly to premises to
allow you to draw the sentence as a conclusion. This is the way I obtained line 4. A sentence follows indirectly from the premises if
a rule of inference applies to some conclusion already obtained (and possibly also to an original premise) to allow you to draw the
sentence as a conclusion. The notation on the right tells you that the first three sentences are premises. It tells you that line 4 is
Licensed (i.e., permitted) by applying the rule of conditional elimination to the sentence of lines 1 and 3. And the notation for line 5
tells you that line 5 is licensed by applying the rule of conditional elimination to the sentences of lines 2 and 4.
For the moment don't worry too much about the vertical line on the left. It's called a Scope Line. Roughly speaking, the scope line
shows what hangs together as one extended chain of argument. You will see why scope lines are useful when we introduce a new
idea in the next section. You should be sure you understand why it is legitimate to draw conclusions indirectly from premises, by
appealing to previous conclusions. Again, what we want to guarantee is that any case (i.e., any assignment of truth values to
sentence letters) which makes the premises true will also make each of the conclusions true. We design the rules of inference so
that whenever they apply to sentences and these sentences happen to be true, then the conclusion licensed by the rule will be true
also. For short, we say that the rules are Truth Preserving.
Suppose we have a case in which all of the premises are true. We apply a rule to some of the premises, and because the rule is truth
preserving, the conclusion it licenses will, in our present case, also be true. (Line 4 in the last example illustrates this.) But if we
again apply a rule, this time to our first conclusion (and possibly some premise), we are again applying a rule to sentences which
are, in the present case, all true. So the further conclusion licensed by the rule will be true too. (As an illustration, look at line 5 in
the last example.) In this way, we see that if we start with a case in which all the premises are true and use only truth preserving
rules, all the sentences which follow in this manner will be true also.
To practice, let's try another example. We'll need a new rule:
Disjunction Introduction -
X vl
XvY
which says that if X is true, then so is XvY. If you recall the truth table definition of 'v', you will see that disjunction introduction is
a correct, truth preserving rule of inference. The truth of even one of the disjuncts in a disjunction is enough to make the whole
disjunction true. So if X is true, then so is XvY, whatever the truth value of Y.
Let's apply this new rule, together with our two previous rules, to show that from the premises 'A⊃~B', 'BvC', and 'A', we can draw
the conclusion 'CvD'. But first close the book and see if you can do it for yourself.
The derivation looks like this:
1 | A⊃~B P
2 | B⊃C P
3|AP
4 | ~B 1, 3, ⊃E
5 | C 2, 4, vE
6 | CvD 5, vl
The sentence of line 4 (I'll just say "line 4" for short) is licensed by applying conditional elimination to lines 1 and 3. Line 5 is
licensed by applying disjunction elimination to lines 2 and 4. Finally, I license line 6 by applying disjunction introduction to line 5.
 Exercise
5-1. For each of the following arguments, provide a derivation which shows the argument to be valid. That is, for each
argument construct a derivation which uses as premises the argument's premises and which has as final conclusion the
conclusion of the argument. Be sure to number and annotate each step as I have done with the examples in the text. That is, for
each conclusion, list the rule which licenses drawing the conclusion and the line numbers of the sentences to which the rule
applies.
a) ~P⊃~D b) ~C⊃~D c) Fv~G d) A⊃B e) Lv~M
~D⊃~F ~C ~F A ~L
~P ~DvE GvK B⊃~C MvD
~F K CvD D⊃H
DvE H
f) C g) (Kv~D)⊃F h) D
C⊃(HvA) K (DvB)⊃~G
~H FvD (~Gv~H)⊃(GvQ)
Av~K Qv~A
i) (Mv~T)⊃(AvJ)
~A
BvM
~A⊃~B
JvD
2.5.1: The Idea of Natural Deduction is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Many of you have probably been thinking: So far, we have an "introduction" and an "elimination" rule for disjunction and just an
"elimination" rule for the conditional. I bet that by the time we're done we will have exactly one introduction and one elimination
rule for each connective. That's exactly right. Our next job is to present the introduction rule for the conditional, which involves a
new idea.
How can we license a conclusion of the form X⊃Y? Although we could do this in many ways, we want to stick closely to
argument forms from everyday life. And most commonly we establish a conclusion of the form X⊃Y by presenting an argument
with X as the premise and Y as the conclusion. For example, I might be trying to convince you that if Adam loves Eve, then Adam
will marry Eve. I could do this by starting from the assumption that Adam loves Eve and arguing, on that assumption, that
matrimony will ensue. Altogether, I will not have shown that Adam and Eve will get married, because in my argument I used the
unargued assumption that Adam loves Eve. But I will have shown that if Adam loves Eve, then Adam will marry Eve.
Let's fill out this example a bit. Suppose that you are willing to grant, as premises, that if Adam loves Eve, Adam will propose to
Eve ('A⊃B'), and that if Adam proposes, marriage will ensue ('B⊃C'). But neither you nor I have any idea whether or not Adam
does love Eve (whether 'A' is true). For the sake of argument, let's add to our premises the temporary assumption, 'A', which says
that Adam loves Eve, and see what follows. Assuming 'A', that Adam loves Eve, we can conclude 'B' which says that Adam will
propose (by conditional elimination, since we have as a premise 'A⊃B', that if Adam loves Eve, he will propose). And from the
conclusion 'B', that Adam will propose, we can further conclude 'C', that marriage will ensue (again by conditional elimination, this
time appealing to the premise 'B⊃C', that proposal will be followed by marriage). So, on the temporary assumption 'A', that Adam
loves Eve, we can conclude 'C', that marriage will ensue. But the assumption was only temporary. We are not at all sure that it is
true, and we just wanted to see what would follow from it. So we need to discharge the temporary assumption, that is, restate what
we can conclude from our permanent premises without making the temporary assumption. What is this? Simply 'A⊃C', that if
Adam loves Eve, marriage will ensue.
Presenting this example in English takes a lot of words, but the idea is in fact quite simple. Again, we badly need a streamlined
means of representing what is going on. In outline, we have shown that we can establish a conditional of the form X⊃Y not on the
basis of some premises (or not from premises alone), but on the strength of an argument. We need to write down the argument we
used, and, after the whole argument, write down the sentence which the argument establishes. We do it like this:
3||A
4 | | A⊃B
5 | | B⊃C
6 | | B 3, 4, ⊃E
7 | | V 5, 6, ⊃E
||
8 | A⊃C 3-7, Conditional Introduction (31)
For right now, don't worry about where lines 4 and 5 came from. Focus on the idea that lines 3 through 7 constitute an entire
argument, which we call a Subderivation, and the conclusion on line 8 follows from the fact that we have validly derived 'C' from
'A'. A subderivation is always an integral part of a larger, or Outer Derivation. Now you can see why I have been using the vertical
scope lines. We must keep outer derivations and subderivations separated. A continuous line to the left of a series of sentences
indicates to you what pieces hang together as a derivation. A derivation may have premises, conclusions, and subderivations,
which are fullfledged derivations in their own right.
A subderivation can provide the justification for a new line in the outer derivation. For the other rules we have learned, a new line
was justified by applying a rule to one or two prior lines. Our new rule, conditional introduction (>I), justifies a new line, 8 in our
example, by appealing to a whole subderivation, 3-7 in our example. When a rule applies to two prior lines, we list the line
numbers separated by commas-in the example line 6 is licensed by applying ⊃E to lines 3 and 4. But when we justify a new line (8
in our example) by applying a rule (here, ⊃I) to a whole subderivation, we cite the whole subderivation by writing down its
inclusive lines numbers (3-7 in our example).
Now, where did lines 4 and 5 come from in the example, and why did I start numbering lines with 3? I am trying to represent the
informal example about Adam and Eve, which started with the real premises that if Adam loves Eve, Adam will propose (A⊃B),
and that if Adam proposes, they will marry (B⊃C). These are premises in the original, outer derivation, and I am free to use them
anywhere in the following argument, including in any subderivation which forms part of the main argument. Thus the whole
derivation looks like this:
1 | A⊃B P
2 | B⊃C P
3 | | A Assumption (A)
4 | | A⊃B 1, Reiteration (R)
5 | | B⊃C 2, Reiteration (R)
6 | | B 3, 4, ⊃E
7 | | C 5, 6, ⊃E
8 | A⊃C 3-7, Conditional Introduction (⊃1)
I am licensed to enter lines 4 and 5 in the subderivation by the rule:
Reiteration: If a sentence occurs, either as a premise or as a conclusion in a derivation, that sentence may be copied
(reiterated) in any of that derivation's lower subderivations, or lower down in the same derivation.
In the present example, 'A⊃B' and 'B⊃C' are assumed as premises of the whole argument, which means that everything that is
supposed to follow is shown to be true only on the assumption that these original premises are true. Thus we are free to assume the
truth of the original premises anywhere in our total argument. Furthermore, if we have already shown that something follows from
our original premises, this conclusion will be true whenever the original premises are true. Thus, in any following subderivation,
we are free to use any conclusions already drawn.
At last I can give you the full statement of what got us started on this long example: the rule of Conditional Introduction. We have
been looking only at a very special example. The same line of thought applies whatever the details of the subderivation. In the
following schematic presentation, what you see in the box is what you must have in order to apply the rule of conditional
introduction. You are licensed to apply the rule when you see something which has the form of what is in the box. What you see in
the circle is the conclusion which the rule licenses you to draw.
Conditional Introduction
In words: If you have, as part of an outer derivation, a subderivation with assumption X and final conclusion Y, then X⊃Y
may be entered below the subderivation as a further conclusion of the outer derivation. The subderivation may use any
previous premise or conclusion of the outer derivation, entering these with the reiteration rule.
You will have noticed that the initial sentences being assumed in an outer, or main, derivation get called "premises," while the
initially assumed sentence in a subderivation gets called an "assumption." This is because the point of introducing premises and
assumptions is slightly different. While we are arguing, we appeal to premises and assumptions in exactly the same way. But
premise? always stay there. The final conclusion of the outer derivation is guaranteed to be true only in those cases in which the
premises are true. But an assumption introduced in a subderivation gets Discharged.
This is just a new word for what we have been illustrating. The point of the subderivation, beginning with assumption X and
ending with final conclusion Y, is to establish X⊃Y as part of the outer derivation. Once the conclusion, X⊃Y, has been
established and the subderivation has been ended, we say that the assumption, X, has been discharged. In other words, the scope
line which marks the subderivation signals that we may use the subderivation's special assumption only within that subderivation.
Once we have ended the subderivation (indicated with the small stroke at the bottom of the vertical line), we are not, in the outer
derivation, subject to the restriction that X is assumed to be true. If the premises of the original derivation are true, X⊃Y will be
true whether X is true or not.
It's very important that you understand why this last statement is correct, for understanding this amounts to understanding why the
rule for conditional introduction works. Before reading on, see if you can state for yourself why, if the premises of the original
derivation are true, and there is a subderivation from X as assumption to Y as conclusion, X⊃Y will be true whether or not X is
true.
The key is the truth table definition of X⊃Y. If X is false, X⊃Y is, by definition, true, whatever the truth value of Y. So we only
have to worry about cases in which X is true. If X is true, then for X⊃Y to be true, we need Y to be true also. But this is just what
the subderivation shows: that for cases in which X is true, Y is also true. Of course, if the subderivation used premises from the
outer derivation or used conclusions that followed from those premises, the subderivation only shows that in all cases in which X
and the original premises are true, Y will also be true. But then we have shown that X⊃Y is true, not in absolutely all cases, but in
at least those cases in which the original premises are true. But that's just right, since we are entering X⊃Y as a conclusion of the
outer derivation, subject to the truth of the original premises.
 Exercise
5-2. Again, for each of the following arguments, provide a derivation which shows the argument to be valid. Be sure to number
and annotate each step to show its justification. All of these exercises will 68 Natural Lkdvcfia for Smtmce Logic require you
to use conditional introduction and possibly other of the rules you have already learned. You may find the use of conditional
introduction difficult until 'you get accustomed to it. If so, don't be alarmed, we're going to work on it a lot. For these problems
you will find the following strategy very helpful: If the final conclusion which you are trying to derive (the "target conclusion")
is a conditional, set up a subderivation which has as its assumption the antecedent of the target conclusion. That is, start your
outer derivation by listing the initial premises. Then start a subderivation with the target conclusion's antecedent as its
assumption. Then reiterate your original premises in the subderivation and use them, together with the subderivation's
assumptions, to derive the consequent of the target conclusion. If you succeed in doing this, the rule of conditional introduction
licenses drawing the target conclusion as your final conclusion of the outer derivation.
a) A⊃B b) NvP c) B d) ~B e) K⊃~D
B⊃C ~N⊃P A⊃B (BvC)⊃C DvH
C⊃D K⊃H
A⊃D
f) A⊃B g) F⊃(CvM) h) (DvB)⊃J i) A⊃K
A⊃(BvC) ~C D⊃J (KvP)⊃L
F⊃M A⊃L
j) Q⊃~S k) P
Q⊃(SvF) (~DvK)⊃B
Q⊃F P⊃(Kv~F)
(Fv~D)⊃(Bv~P)
2.5.2: Subderivations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We now have in place all the basic ideas of natural deduction. We need I only to complete the rules. So &at you will have them all
in one place for easy reference, I will simply state them all in abbreviated form and then comment on the new ones. Also, I will
now state all of the rules using the same format. For each rule I will show a schematic derivation with one part in a box and another
part in a circle. In the box you will find, depending on the rule, either one or two sentence forms or a subderivation form. In the
circle you will find a sentence form. To apply a given rule in an actual derivation, you proceed as follows: You look to see whether
the derivation has something with the same form as what's in the box. If so, the rule licenses you to write down, as a new
conclusion, a sentence with the form of what's in the circle.
Reiterntion: If a sentence occurs, either as a premise or as a conclusion in a derivation, that sentence may be copied
(reiterated) in any of that derivation's lower subderivations, or lower down in the same derivation.
In interpreting these schematic statements of the rules, you must remember the following: When a rule applies to two sentences, as
in the case of conjunction introduction, the two sentences can occur in either order, and they may be separated by other sentences.
The sentences to which a rule applies may be premises, an assumption, or prior conclusions, always of the same derivation, that is,
lying along the same scope line. Also, the sentence which a rule licenses you to draw may be written anywhere below the licensing
sentences or derivation, but as part of the same derivation, again, along the same scope line.
Conjunction introduction and elimination are so simple we rarely bother to mention them when we argue informally. But to be
rigorous and complete, our system must state and use them explicitly. Conjunction introduction states that when two sentences, X
and Y, appear in a derivation, in either order and whether or not separated by other sentences, we may conclude their conjunction,
X&Y, anywhere below the two conjunct~. Conjunction elimination just tells us that if a conjunction of the form X&Y appears on a
derivation, we may write either conjunct (or both, on different lines) anywhere lower down on the derivation. We have already
discussed the rules for disjunction and the conditional. Here we need only add that in the elimination rules, the sentences to which
the rules apply may occur in either order and may be separated by other sentences. For example, when applying disjunction
elimination, the rule applies to sentences of the form XvY and ~X, in whatever order those sentences occur and whether or not
other sentences appear between them.
Biconditional introduction and elimination really just express the fact that a biconditional of the form X≡Y is logically equivalent
to the conjunction of sentences of the form X⊃Y and Y⊃X. If the two conditionals appear on a derivation, whatever the order, and
whether or not separated by other sentences, we may write the biconditional lower down as a conclusion. Conversely, if a
biconditional of the form X≡Y appears, one may write lower down, as a conclusion, X⊃Y, Y⊃X, or both (on separate lines).
Note that negation elimination licenses dropping a double negation, and is justified by the fact that X is always logically equivalent
to ~~X. Negation introduction requires some comment. Once again, natural deduction seeks to capture and make precise
conventional forms of informal argument. This time we express what traditionally goes under the name of "reductio ad absurdum,"
or "reduction to the absurd." Here the idea is that if we begin with an assumption from which we can deduce a contradiction, the
original assumption must be false. Natural deduction employs this strategy as follows: Begin a subderivation with an assumption,
X. If one succeeds in deriving both a sentence of the form Y and its negation, ~Y, write the sentence of the form ~X as a
conclusion of the outer derivation anywhere below the subderivation.
As with the other rules, you should be sure you understand why this rule works. Suppose in a subderivation we have drawn the
conclusions Y and ~Y from the assumption X. This is (by the rules for conjunction) equivalent to deriving the contradiction Y&~Y
from X. Now, X must be either true or false. If it is true, and we have drawn from it the conclusion that Y&~Y, we have a valid
argument from a true premise to a false conclusion. But that can't happen--our rules for derivations won't let that happen. So X
must have been false, in which case ~X must be true and can be entered as a conclusion in the outer derivation. Finally, if the
subderivation has used premises or conclusions of the outer derivation, we can reason in exactly the same way, but subject to the
restriction that we consider only cases in which the original premises were true.
In annotating negation introduction, keep in mind the same consideration which applied in annotating conditional introduction. The
new line is justified by appeal, not to any one or two lines, but to a whole argument, represented by a subderivation. Consequently,
the justification for the new line appeals to the whole subderivation. Indicate this fact by writing down the inclusive line numbers
of the subderivation (the first and last of its line numbers separated by a dash).
In applying these rules, be sure to keep the following in mind: To apply the rules for conditional and negation introduction, you
must always have a completed subderivation of the form shown. It's the presence of the subderivation of the right form which
licenses the introduction of a conditional or a negated sentence. To apply any of the other rules, you must have the input sentence
or sentences (the sentence or sentences in the box in the rule's schematic statement) to be licensed to write the output sentence of
the rule (the sentence in the circle in the schematic presentation). But an input sentence can itself be either a prior conclusion in the
derivation or an original premise or assumption.
Incidentally, you might have been puzzled by the rule for negation introduction. The rule for negation elimination has the form
"~~X. Therefore X". Why not, you might wonder, use the rule "X. Therefore ~~X for negation introduction? That's a good
question. The rule "X. Therefore ~~X" is a correct rule in the sense that it is truth preserving. It will never get you a false sentence
out of true ones. But the rule is not strong enough. For example, given the other rules, if you restrict yourself to the rule "X.
Therefore ~~X for negation introduction, you will never be able to construct a derivation that shows the argument
~A
~(A&B)
to be valid. We want our system of natural deduction not only to be Sound, which means that every derivation represents a valid
argument. We also want it to be Complete, which means that every valid argument is represented by a derivation. If we use the rule
"X. Therefore ~~X" for negation introduction, our system of natural deduction will not be complete. The rules will not be strong
enough to provide a correct derivation for every valid argument.
 Exercise
5-3. Below you find some correct derivations without the annotations which tell you, for each line, which rule was used in
writing the line and to which previous line or lines the rule appeals. Copy the derivations and add the annotations. That is, for
each line, write the line number of the previous line or lines and the rule which, applying to those previous lines, justifies the
line you are annotating.
a) 1 | B&(B⊃~A) P c) 1 | A⊃~B P
2 | B 2 | BvC P
3 | B⊃~A 3 | |
4 | ~A | | A
b) 1 | ~C≡(AvB) P 4 | | A⊃~B
2 | A P 5 | | ~B
3 | (AvB) ⊃ ~C 6 | | BvC
4 | AvB 7 | | C
5 | ~C 8 | A ⊃ C
d) 1 | D P e) 1 | AvB P
2 | (D&A)⊃C P 2 | | ~A&~B
3 | | A 3 | | ~A
4 | | D 4 | | ~B
5 | | D&A 5 | | AvB
6 | | (D&A)⊃C 6 | | B
7 | | C 7 | ~(~A&~B)
8 | A⊃C
f) 1 | A&B P g) 1 | ~A⊃B P
2 | | A 2 | ~A⊃~B P
3 | | A&B 3 | | ~A
4 | | B 4 | | ~A⊃B
5 | A⊃B 5 | | B
6 | | B 6 | | ~A⊃~B
7 | | A&B 7 | | ~B
8 | | A 8 | ~~A
9 | B⊃A 9 | A
10 | A≡B
5-4. For each of the following arguments, provide a derivation which shows the argument to be valid. Follow the same
directions as you did for exercises 5-1 and 5-2.
a) C&~H b) JvD c) A&B d) A⊃~D
~H ~~~D B&A ~~A
J ~D
e) G⊃D f) A≡~B g) M h) A&(B&C)
G⊃~D ~B⊃A Rv~H (A&B)&C
~G M&(Rv~H)
i) ~C⊃D j) A≡~B k) ~C⊃~~~A l) K⊃~B
D⊃~C ~B ~C B&F
D≡~C A ~A ~K
m) ~P n) (N⊃K)&(N⊃L) o) D⊃(AvF) p) H≡J
~Q N⊃(K&L) D⊃~F H≡K
~(PvQ) ~A J≡K
~D&~A
5-5. In chapter 3 we defined triple conjunctions and disjunctions, that is, sentences of the form X&Y&Z and XvYvZ. Write
introduction and elimination rules for such triple conjunctions and disjunctions.
5-6. Suppose we have a valid argument and an assignment of truth values to sentence letters which makes one or more of the
premises false. What, then, can we say about the truth value of the conclusions which follow validly from the premises? Do
they have to be false? Can they be true? Prove what you say by providing illustrations of your answers.
 chapter summary Exercise
Give brief explanations for each of the following, referring back to the text to make sure your explanations are correct and
saving your answers in your notebook for reference and review.
a) Derivation
b) Subderivation
c) Outer Derivation
d) Scope Line
e) Premise
f) Assumption
g) Rule of Inference
h) License (to draw a conclusion)
i) Truth Reserving Rule
j) Discharging an Assumption
k) Conjunction Introduction
l) Conjunction Elimination
m) Disjunction Introduction
n) Disjunction Elimination
o) Conditional Introduction
p) Conditional Elimination
q) Biconditional Introduction
r) Biconditional Elimination
s) Negation Introduction
t) Negation Elimination
u) Reiteration
2.5.3: The Complete Rules of Inference is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
2.6: Natural Deduction for Sentence Logic - Strategies


2.6: Natural Deduction for Sentence Logic - Strategies is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
Knowing the rules for constructing derivations is one thing. Being able to apply the rules successfully is another. There are no
simple mechanical guidelines to tell you which rule to apply next, so constructing derivations is a matter of skill and ingenuity.
Long derivations can be extremely difficult. (It's not hard to come up with problems which will stump your instructorl) At first,
most students feel they don't even know how to get started. But with a bit of practice and experience, you will begin to develop
some intuitive skill in knowing how to organize a derivation. To get you started, here are some examples and practical strategies.
Usually you will be setting a problem in the following form: You will be given some premises and a conclusion. You will be told to
prove that the conclusion follows validly from the premises by constructing a derivation which begins with the given premises and
which terminates with the given conclusion. So you already know how your derivation will begin and end.
Your job is to fill in the intermediate steps so that each line follows from previous lines by one of the rules. In filling in, you should
look at both the beginning and the end of the derivation.
Let's illustrate this kind of thinking with a simple example. Suppose you are asked to derive 'B&C' from the premises 'A⊃B',
'A⊃C', and 'A'. Right off, you know that the derivation will take the form
1 | A⊃B P
2 | A⊃C P
3|AP
|?
|?
|?
| B&C
where you still have to figure out what replaces the question marks.
First, look at the conclusion. It is a conjunction, which can most straightforwardly be introduced with the rule for &I. (From now
on, I'm going to use the shorthand names of the rules.) What do you need to apply that rule? You need 'B' and you need 'C'. So if
you can derive 'B' and 'C', you can apply &I to get 'B&C'. Can you derive 'B' and 'C'? Look at the premises. Can you get 'B' out of
them? Yes, by applying ⊃E to lines 1 and 3. Similarly, you can derive 'C' from lines 2 and 3. Altogether, the derivation will look
like this:
1 | A⊃B P
2 | A⊃C P
3|AP
4 | B 1, 3, ⊃E
5 | C 2, 3, ⊃E
6 | B&C 4, 5, &I
Let's try a slightly harder example. Suppose you are asked to derive 'C~A' from the premises 'AvB' and 'C⊃~B'. Your target
conclusion is a conditional. Well, what rule allows you to conclude a conditional? ⊃1. So you will try to set things up so that you
can apply ⊃I. This will involve starting a subderivation with 'C' as its assumption, in which you will try to perive 'A'. In outline, the
derivation you are hoping to construct can be expected to look like this:
1 | A⊃B P
2 | C⊃~B P
3||CA
||?
||?
||A
| C⊃A
(Your derivation won't have to look like this. In every case there is more than one correct derivation of a conclusion which follows
from a given set of premises. But in this case, this is the obvious thing to try, and it provides the simplest correct derivation.)
To complete the derivation, you must fill in the steps in the subderivation to show that (given the premises of the outer derivation)
'A' follows from 'C'.
How will you do this? Let's study what you have available to use. In the subderivation you are allowed to use the subderivation's
assumption and also any previous premise or conclusion in the outer derivation. Notice that from 'C' and the premise 'C⊃~B' you
can get '~B' by ⊃E. Is that going to do any good? Yes, for you can then apply vE to '~B' and the premise 'AvB' to get the desired
'A'. All this is going to take place in the subderivation, so you will have to reiterate the premises. The completed derivation looks
like this:
1 | AvB P
2 | C⊃~B P
3||CA
4 | | C⊃~B 2, R
5 | | ~B 3, 4, ⊃E
6 | | AvB 1, R
7 | | A 5, 6, vE
8 | C⊃A 3-7, ⊃l
If you are still feeling a little lost and bewildered, reread the text from the beginning of this section.
When you have understood the examples given so far, you are ready for something new. Let's try to derive 'A⊃~B' from 'B⊃~A'.
As in the second example, our first effort to derive a conditional should be by using ⊃I. So we want a subderivation with 'A' as
assumption and '~B' as final conclusion:
1 | B⊃~A P
2||AA
||?
||?
| | ~B
| A⊃~B ⊃l
But how can we get '~B' from the assumption of 'A', using the premise of the outer derivation?
'~B' is the negation of the sentence 'B'. Unless there is some really obvious simple alternative, one naturally tries to use ~I. ~I
works by starting a subderivation with the sentence to be negated as assumption and then deriving some sentence and its negation.
In the present situation this involves something that might not have occurred to you, namely, creating a subderivation of a
subderivation. But that's fine. All the rules for working on a derivation apply to subderivations also, including the creation of
subderivations. The only difference between a subderivation and a derivation is that a subderivation ends when we discharge its
assumption, returning to its outer derivation; and that in a subderivation we may reiterate prior premises or conclusions from an
outer derivation (or from any outer-outer derivation, as you will see in a moment). This is because in a subderivation we are
working under the assumption that all outer assumptions and premises are true.
Will this strategy work? Before writing anything down, let me illustrate the informal thinking you should go through to see whether
a strategy promises to be successful. Look back at the outline we have already written of how we hope the derivation will look. We
are proposing to start a sub-sub-derivation with the new assumption 'B'. That sub-sub-derivation can use the original premise
'B⊃~A', which, together with the assumption 'B', will give '~A' by ⊃E. But the sub-sub-derivation is also within its outer
derivation beginning with the assumption of 'A'. So 'A' is also being assumed in the sub-sub-derivation, which we express by
reiterating 'A' in the sub-sub-derivation. The sub-sub-derivation now has both 'A' and '~A', which constitutes the contradiction we
needed:
1 | B⊃~A P
2 | |A A
3|||BA
4 | | | B⊃~A 1, R
5 | | | ~A 3, 4, ⊃E
6 | | | A 2, R
7 | | ~B 3-6, ~I
8 | A⊃~B 2-7, ⊃I
How are you doing? If you have had trouble following, rest for a moment, review to be sure you have gotten everything up to this
point, and then we'll try something one step more involved.
Let's try deriving 'A≡~B' from 'AvB' and '~(A&B)'. The conclusion is a biconditional, and one derives a biconditional most easily
by using ≡I. Think of a biconditional as the conjunction of two conditionals, the two conditionals we need to derive the
biconditional using ≡I. So you should aim to develop a derivation along these lines:
1 | AvB P
2 | ~(AvB) P
3|?
4|?
5 | ~B⊃A
6|?
7|?
8 | A⊃~B
9 | A≡~B ≡I
We have reduced the complicated problem of deducing 'A≡~B' to the simpler problems of deducing '~B⊃A' and 'A⊃~B'.
In constructing derivations, you should learn to think in this kind of pattern. Try to resolve the problem of deriving the final
conclusion (your target conclusion) by breaking it down into simpler problems of deriving . simpler sentences (your new target
conclusions). You may actually need to resolve your simpler problems intostill more simple problems. You continue working
backward toward the middle until you can see how to derive your simple sentences from the premises. At this point you start
working from the premises forward and fill everything in.
How, in this example, can we derive our simplified new target conclusions? They are both conditionals, and as we saw in the
second example, the straightforward way to derive conditionals uses ⊃I. This involves starting one subderivation for each of the
conditionals to be derived:
1 | AvB P
2 | ~(AvB) P
| |~B A
||?
||?
||A
| ~B⊃A ⊃I
||AA
||?
||?
| | ~B
| A⊃~B ⊃I
| A≡~B ≡I
We have now resolved our task into the problem of filling in the two subderivations.
Can you see how to complete the subderivations by working with the premises of the outer derivation? The first subderivation is
easy: '~B' and 'AvB' give 'A' by vE. The second subderivation presents more of a challenge. But we can complete it by using the
same tactics illustrated in the previous example. We've assumed 'A' and want to get '~B'. To get '~B', we can try ~I (unless a really
simple alternative suggests itself). ~I will require us to start a subsub-derivation with 'B' as assumption. In this sub subderivation
we can reiterate anything which is above in an outer derivation of the derivation on which we are working. So we can reiterate 'A',
which, with 'B' , will give us 'A&B'; and we can reiterate the original premise '~(A&B)', thus giving us our contradiction. (Note that
the contradiction can be in the form of any sentence and its negation. Neither sentence has to be atomic.) Since from 'B' we have
derived a sentence and its negation, we can use ~I to discharge the assumption 'B', giving the conclusion '~B' at the end of the
subderivation which began with 'A'. This is just what we needed.
If you find this tangle of thinking hard to unravel, read it over again, following each step in the completed derivation below to see
how it all fits together.
1 | AvB P
2 | ~(AvB) P
3 | | ~B A
4 | | AvB 1, R
5 | | A 3, 4, vE
6 | ~B⊃A 3-5, ⊃I
7||AA
8|||BA
9 | | | A 7, R
10 | | | A&B 8, 9, &I
11 | | | ~(A&B) 2, R
12 | | ~B 8-11, ~I
13 | A⊃~B 7-12, ⊃I
14 | A≡~B 6, 13, ≡I
Now let's tackle something different. You are asked to derive 'C' from 'A&B' and '~C⊃~B'. What can you do? If you are stuck, you
can at least write down the premises and conclusion so that you can start to see how the derivation will look:
1 | A&B P
2 | ~C⊃~B P
|?
|?
|C
No rule applies immediately to the premises to give 'C'. Because 'C' is atomic, no introduction rule for a connective will give 'C'.
What on earth can you do?
Sometimes when you are stuck, you can succeed by arranging to use ~-I in what I am going to call the Reductio Ad Absurdum
strategy. This strategy proceeds by assuming the negation of what you want and then from this assumption (and prior premises and
conclusions) deriving a contradiction. As you will see in the example, you will then be able to apply ~I to derive the double
negation of what you want, followed by ~E to get rid of the double negation. In outline, the reductio absurdum strategy, applied to
this problem, will look like this:
1 | A&B P
2 | ~C⊃~B P
3 | | ~C A
| | X ('X' here stands for some specific sentence, but I don't yet know what it will be.)
| | ~X
| ~~C ~I
| C ~E
Will this strategy work in this case? If you assume '~C', you will be able to use that assumption with the premise '~C⊃~B' to get
'~B'. But '~B' will contradict the 'B' in the premise 'A&B', and you can dig 'B' out of 'A&B' with &E. In sum, from '~C' and the
premises you will be able to derive both 'B' and '~B'. ~I then allows you to conclude '~~C' (the negation of the assumption which
led to the contradiction). ~E finally gives 'C':
1 | AvB P
2 | ~C⊃~B P
3 | | ~C A
4 | | ~C⊃~B 2, R
5 | | ~B 3, 4, ⊃E
6 | | A&B 1, R
7 | | B 6, &E
8 | ~~C 3-7, ~I
9 | C 9, ~E
The first time you see an example like this it may seem tricky. But you will soon get the hang of it.
You do need to be a little cautious in grasping at the reductio strategy when you are stuck. Often, when students have no idea what
to do, they assume the opposite of what they want to conclude and then start blindly applying rules. This almost never works. To
use the reductio strategy successfully, you need to have a more specific plan. Ask yourself: "Can I, by assuming the opposite of
what I want to derive, get a contradiction (a sentence and its negation) out of the assumption?" If you can see how to do this, you
are all set, and you can start writing down your derivation. If you think you see a way which might work, it may be worth starting
to write to clarify your ideas. But if you have no idea of how you are going to get a contradiction out of your assumption, go slow.
Spend a little time brainstorming about how to get a contradiction. Then, if you find you are getting nowhere, you may well need to
try an entirely different approach to the problem.
I should also comment on the connection between what I have called the reductio ad absurdum strategy and the rule for ~I. They
really come to pretty much the same thing. If you need to derive a sentence of the form ~X, consider assuming X, trying to derive a
contradiction, and applying ~I to get ~X. To derive a sentence of the form X, assume ~X, and derive ~~X by ~I. Then eliminate the
double negation with ~E.
 Exercise 6-1.
For each of the following arguments, provide a derivation, complete with annotations, which shows the argument to be valid. If
you find you are having difficulty with these problems, go over the examples earlier in this chapter and then try again.
a) Kv~I b) ~C⊃A c) ~D⊃~K d) ~F⊃G
~(~K&I) B⊃~A K G⊃~E
B ~KvH E⊃F
C D&H
e) A≡~B f) AvB
~A⊃B B⊃C
~CvD
~D⊃A
2.6.1: Constructing Correct Derivations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Suppose you are asked to provide a derivation which shows the following argument to be valid:
(1) (A⊃B)&[C≡(A⊃B)]
(2) C
The premise is a mess! How do you determine which rule applies to it? After your labor with some of the exercises in the last
chapter, you probably can see that the key lies in recognizing the main connective. Even if you got all of those exercises right, the
issue is so important that it's worth going over from the beginning.
Let's present the issue more generally. When 1 stated the rules of inference, I expressed them in general terms, using boldface
capital letters, 'X' and 'Y'. For example, the rule for &E is
(3) | X&Y and | X&Y
|X|Y
The idea is that whenever one encounters a sentence of the form X&Y in a derivation, one is licensed to write either the sentence X
or the sentence Y (or both on separate lines) further down. Focus on the fact that this is so whatever sentences X and Y might be.
This is the point of using boldface capital letters in the presentation. 'X' and 'Y' don't stand for any particular sentences. Rather, the
idea is that if you write any sentence you want for 'X' and any sentence you want for 'Y', you will have a correct instance of the rule
for &E. This is what I mean by saying that I have expressed the rule "in general terms" and by talking about a sentence "of the form
X&Y".
How will these facts help you to deal with sentence (I)? Here's the technique you should use if a sentence such as (1) confuses you.
Ask yourself: "How do I build this sentence up from its parts?" You will be particularly interested in the very last step in putting (1)
together from its parts. In this last step you take the sentence
(4) A⊃B which you can think of as X
and the sentence
(5) C=(A⊃B) which you can think of as Y
and put them on either side of an '&' to get the sentence
(A⊃B)&[C≡(A⊃B)] which has the form X&Y
You have just established that (1) has the form of X&Y; that is, it is a conjunction with sentences (4) and (5) as its conjuncts.
Consequently, you know that the rule for &E, (3), applies to sentence (l), so that if (1) appears in a derivation you are licensed to
write sentence (4) or (5) (or both) below.
Similarly, if in a derivation you are faced with the sentence
(6) C≡(A⊃B)
ask yourself "What is the last thing I do to build this sentence up from its parts?" You take the sentence
(7) C which you can think of as X
and the sentence
(8) A⊃B which you can think of as Y
and you put them on either side of a biconditional, '≡', giving
(9) C≡(A⊃B) which thus has the form X≡Y
Consequently, if you find sentence (6), you can apply the rule of inference for ≡E:
| X≡Y and | X≡Y
| X⊃Y | Y⊃X
which, when we put in sentences (7) and (8) for X and Y, look like
| C≡(A⊃B) | C≡(A⊃B)
| C⊃(A⊃B) ≡E and | (A⊃B)⊃C ≡E
Thus ≡E applies to sentence (6), licensing us to follow (6) on a derivation either with the sentence 'C⊃(A⊃B)' , or the sentence
'(A⊃B)⊃C', or both on separate lines.
In a moment we will apply what we have just done to provide a derivation which shows how (2) follows from (1). But we will need
to see how to treat one more compound sentence. This time, try to figure it out for yourself. What is the form of the sentence
'(A⊃B)⊃C'?
The last thing you do in putting this one together from its parts is to put '(A⊃B)' and 'C' on either side of a conditional, '⊃'. So the
sentence has the form X⊃Y, with 'A⊃B' as X and 'C' as Y. If we have 'A⊃B' as well as '(A⊃B)⊃C' in a derivation, we can apply
⊃E to the two to derive 'C'.
Perhaps you can now see how we can write a derivation which shows (2) to follow from (1). In this case, because the desired
objective, 'C', is atomic, we can't get it by an introduction rule. So it is useless to try to work backward from the end. The reductio
ad absurdum strategy could be made to work, but only by doing more indirectly what we're going to have to do anyway. In this
case the best strategy is to use elimination rules to take the premise apart into simpler pieces.
When we think through what these pieces will look like, we will see that they provide just what we need to derive 'C'. &E applies
to the premise. '(A⊃B)&[C≡(A⊃B)]' to give us 'A⊃B' and 'C≡(A⊃B)'. In turn, ≡E applies to 'C≡(A⊃B)' to give us '(A⊃B)⊃C',
which, together with the 'A⊃B', immediately gives us 'C' by ⊃E. (≡E applied to 'C=(A⊃B)' also gives 'C⊃(A⊃B)'. But we have
no use for 'C⊃(A⊃B)', so although licensed to write it down as a conclusion, we don't bother.) Altogether, the completed
derivation looks like this:
1| (A⊃B)&[C≡(A⊃B)] P
2| A⊃B 1, &E
3 | C≡(A⊃B) 1, &E
4 | (A⊃B)⊃C 3, ≡E
5 | C 2, 4, ⊃E
The, key idea you need to untangle a sentence such as (1) is that of a main connective:
The Main Connective in a sentence is the connective which was used last in building up the sentence from its component or
components.
(A negated sentence, such as '~(Av~B)', has just one component, 'Av~B' in this case. All other connectives use two components in
forming a sentence.) Once you see the main connective, you will immediately spot the component or components to which it has
been applied (so to speak, the X and the Y), and then you can easily determine which rules of inference apply to the sentence in
question.
Let's practice with a few examples:
SENTENCE MAIN CONNECTIVE COMPONENT OR COMPONENTS
(AvB)⊃~(B&D) ⊃ AvB and ~(B&D)
[(A⊃B)v~D)v(~A≡D) v (A⊃B)v~D and ~A≡D
~[(A⊃B)⊃~(B⊃~A)] ~ (A⊃B)⊃~(B⊃~A)
The second and third examples illustrate another fact to which you must pay attention. In the second example, the main connective
is a 'v'. But which occurrence of 'v'? Notice that the sentence uses two 'v's, and not both occurrences count as the main connective!
Clearly, it is the second occurrence, the one used to put together the components '(A⊃B)v~D' and '~A≡D', to which we must pay
attention. Strictly speaking, it is an occurrence of a connective which counts as the main connective. It is the occurrence last used in
putting the sentence together from its parts. In the third example, '~' occurs three times! Which occurrence counts as the main
connective? The very first.
In the following exercises you will practice picking out the main connective of a sentence and determining which rule of inference
applies. But let's pause first to say more generally how this works:
The elimination rule for '&' applies to a sentence only when an '&' occurs as the sentence's main connective. The same thing
goes for 'v', '⊃', and ≡'. The components used with the main connective are the components to which the elimination rule
makes reference.
The elimination rule for '~' applies only to a doubly negated sentence, ~~X; that is, only when '~' is the sentence's main
connective, and the '~' is applied to a component, ~X, which itself has a '~' as its main connective.
The introduction rule for '&' licenses you to write as a conclusion a sentence, the main connective of which is '&'. The same
thing goes for 'v', '⊃', '≡', and '~'
 Exercise 6-2.
Give derivations which establish the validity of the following arguments:

a) (AvB)&(AvB)⊃CJ b) A
C (AvB)≡[(A⊃K)&(B⊃K)]
K
c) [A⊃(Dv~B)]&{[A⊃(Dv~B)]⊃(B⊃A)}
B⊃A
2.6.2: Recognizing the Main Connective is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
This chapter and chapter 5 have described, explained, and illustrated derivations. Let's pull these thoughts together with some
explicit definitions and further terminology.
A Rule of Inference tells when you are allowed, or Licensed, to draw a conclusion from one or more sentences or from a
whole argument (as represented by a subderivation).
A Derivation is a list of which each member is either a sentence or another derivation. If a first derivation has a second
derivation as one of the first derivation's parts, the second derivation is called a Subderivation of the first and the first is
called the Outer Derivation, of the second. Each sentence in a derivation is a premise or assumption, or a reiteration of a
previous sentence from the same derivation or an outer derivation, or a sentence which follows by one of the rules of
inference from previous sentences or subderivations of the derivation.
In practice, we always list the premises or assumptions of a derivation at its beginning, and use a horizontal line to separate them
from the further sentences which follow as conclusions. What's the difference between premises and assumptions? From a formal
point of view, there is no difference at all, in the sense that the rules of inference treat premises and assumptions in exactly the
same way. In practice, when an unargued sentence is assumed at the beginning of the outermost deduction, we call it a premise.
When an unargued sentence is assumed at the beginning of a subderivation, we call it an assumption. The point is that we always
terminate subderivations before the end of a main derivation, and when we terminate a subderivation, in some sense we have gotten
rid of, or Discharged, the subderivation's assumptions.
To make these ideas clearer and more precise, we have to think through what the vertical lines, or Scope Lines, are doing for us?
A Scope Line tells us what sentences and subderivations hang together as a single derivation. Given a vertical scope line, the
derivation it marks begins where the line begins and ends where the line ends. The derivation marked by a scope tine
includes all and only the sentences and subderivations immediately to the right of the scope tine.
To help sort out these definitions, here is a schematic example:
|Q
|R
|S
|.
|.
|.
||T
||.
||.
||.
| 2| U
|.
|.
|.
||V
||.
||.
||.
|||W
|||.
|||.
|||.
| | 4| X
|3|Y
|.
|.
|.
1|Z
Notice that at the bottom of each scope line I have written a number to help us in talking about the different component derivations.
Consider first the main derivation, derivation 1, marked with the leftmost scope line numbered '1' at its bottom. Derivation 1
includes premises Q and R and has a first conclusion S, other conclusions not explicitly shown, indicated by . . . , and the final
conclusion Z. Derivation 1 also includes two subderivations, derivations 2 and 3. Derivation 2 has assumption T, various
conclusions not explicitly indicated (again signaled by . . .), and final conclusion U. Derivation 3 starts with assumption V, has final
conclusion Y, and includes a subderivation of its own, derivation 4.
This organization serves the purpose of keeping track of what follows from what. In the outermost derivation 1, all the conclusions
of the derivation (S . . . Z) follow from the derivation's premises, Q and R. This means that every assignment of truth values to
sentence letters which makes the premises Q and R true will make all the conclusions of derivation 1 true. But the conclusions of a
subderivation hold only under the subderivation's additional assumption. For example, the conclusion U of subderivation 2 is
subject to the assumption T as well as the premises Q and R. This means that we are only guaranteed that any assignment of truth
values to sentence letters which makes Q, R, and T all true will make U true also. In other words, when we start a subderivation,
we add an additional assumption which is assumed in effect just in the subderivation. Any premises or assumptions from outer
derivations also apply in the subderivation, since they and their consequences can be reiterated into the subderivation.
You should particularly notice that when a subderivation has ended, its special assumption is no longer assumed. It is not assumed
in any conclusions drawn as part of the outer derivation, nor is it assumed as part of a new subderivation which starts with a
different assumption. Thus the truth of T is not being assumed anywhere in derivation 1, 3, or 4. This is what we mean by saying
that the assumption of a subderivation has been discharged when the subderivation is terminated.
These facts are encoded in the reiteration rule which we can now spell out more clearly than before. The reiteration rule spells out
the fact that a subderivation assumes the truth, not only of its own assumption, but of the prior assumptions, premises, and
conclusions of any outer derivation. Thus, in subderivation 2, reiteration permits us to write, as part of 2, Q, R, S, or any other
conclusion of 1 which appears before 2 starts. This is because inside 2, we assume that the premises of outer derivation 1 are true.
Andrbecause whenever the premises are true, conclusions which follow from them are true, we may also use the truth of any such
conclusions which have already followed from these premises.
But we cannot reiterate a sentence of 2 in, for example, 1. This is because when we end subderivation 2 we have discharged its
premise. That is, we are no longer arguing under the assumption that the assumption of 2 is true. So, for example, it would be a
mistake to reiterate U as part of 1. U has been proved only subject to the additional assumption T. In 1, T is not being assumed. In
the same way, we cannot reiterate U as part of 3 or 4. When we get to 3, subderivation 2 has been ended. Its special assumption, T,
has been discharged, which is to say that we no longer are arguing under the assumption of T.
Students very commonly make the mistake of copying a conclusion of a subderivation, such as U, as a conclusion of an outer
derivation-' m our schematic example, listing U as a conclusion in derivation 1 as well as in subderivation 2. I'll call this mistake
the mistake of hopping scope lines. Don't hop scope lines!
We can, however, reiterate Q, R, S, or any prior conclusion in 1 within sub-sub-derivation 4. Why? Because 4 is operating under its
special assumption, W, as well as all the assumptions and premises of all derivations which are outer to 4. Inside 4 we are
operating under all the assumptions which are operative in 3, which include not only the assumption of 3 but all the premises of the
derivation of which 3 is a part, namely, 1. All this can be expressed formally with the reiteration rule, as follows: To get a premise
or prior conclusion of 1 into 4, first reiterate the sentence in question as part of 3. Now that sentence, officially part of 3, can be
reiterated again in 4. But we can dispense with the intermediate step.
Incidentally, once you clearly understand the reiteration rule, you may find it very tiresome always to have to explicitly copy the
reiterated sentences you need in subderivations. Why, you may wonder, should you not be allowed, when you apply other rules,
simply to appeal to prior sentences in outer derivations, that is, to the sentences which the reiteration rule allows you to reiterate? If
you fully understand the reiteration rule, you will do no harm in thus streamlining your derivations. I will not use this abbreviation,
because I want to be sure that all of my readers understand as clearly as possible how reiteration works. You also should not
abbreviate your derivations in this way unless your instructor gives you explicit permission to do so.
Scope lines also indicate the sentences to which we can apply a rule in deriving a conclusion in a derivation or subderivation. Let
us first focus on rules which apply only to sentences, that is, rules such as vE or ⊃E, which have nothing to do with subderivations.
The crucial feature of such a rule is that, if the sentences to which we apply it are true, the conclusion will be true also. Suppose,
now, we apply such a rule to the premises Q and R of derivation 1. Then, if the premises are true, so will the rule's conclusion, so
that we can write any such conclusion as part of derivation 1. In further application of such rules in reaching conclusions of
derivation 1, we may appeal to 1's prior conclusions as well as its premises, since if the premises are true, so will the prior
conclusions. In this way we are still guaranteed that if the premises are true, so will the new conclusion.
But we can't apply such a rule to assumptions or conclusions of a subderivation in drawing conclusions to be made part of
derivation 1. For example, we can't apply a rule to sentences S and U in drawing a conclusion which will be entered as a part of
derivation 1. Why not? Because we want all the conclusions of 1 to be guaranteed to be true if 1's premises are true. But
assumptions or conclusions of a subderivation, say, 2, are only sure to be true if 1's premises and 2's special assumption are true.
In sum, when applying a rule of inference which provides a conclusion when applied to sentences ("input sentences"), the input
sentences must already appear before the rule is applied, and all input sentences as well as the conclusion must appear in the same
derivation. Violating this instruction constitutes a second form of the mistake of hopping scope lines.
What about ⊃I and ~I, which don't have sentences as input? Both these rules have the form: If a subderivation of such and such a
form appears in a derivation, you may conclude so and so. It is important to appreciate that these two rules do not appeal to the
sentences which appear in the subderivation. They appeal to the subderivation as a whole. They appeal not to any particular
sentences, but to the fact that from one sentence we have derived certain other sentences. That is why when we annotate these rules
we cite the whole subderivation to which the rule applies, by indicating the inclusive line numbers of the subderivation.
Consider ⊃I. Suppose that from T we have derived U, perhaps using the premises and prior conclusions of our outer derivation.
Given this fact, any assignment of truth values to sentence letters which makes the outer derivation's premises true will also make
the conditional T⊃U true. (1 explained why in the last chapter.) Thus, given a subderivation like 2 from T to U, we can write the
conclusion T⊃U as part of the outer derivation 1. If 1's premises are true, T⊃U will surely be true also.
The key point to remember here is that when ⊃I and ~I apply to a subderivation, the conclusion licensed appears in the same
derivation in which the input subderivation appeared as a part. Subderivation 2 licenses the conclusion T⊃Y as a conclusion of 1,
by ⊃I; and ⊃I, similarly applied to derivation 4, licenses concluding W⊃X as part of 3, but not as part of 1.
By this time you may be feeling buried under a pile of details and mistakes to watch out for. Natural deduction may not yet seem all
that natural. But, as you practice, you will find that the bits come to hang together in a very natural way. With experience, all these
details will become second nature so that you can focus on the challenging task of devising a good way of squeezing a given
conclusion out of the premises you are allowed to use.
For each of the following arguments, provide a derivation which shows the argument to be valid. If you get stuck on one
problem, try another. If you get good and stuck, read over the examples , in this chapter, and then try again.
a) R b) (~A&B)vC c) ~(Hv~D) d) F⊃(O⊃M)
(RvD)&(RvK) AvD F⊃H (F&O)⊃M
~C⊃D ~F
e) P&~Q f) (K&G)⊃S g) ~(A&~F) h) ~(N⊃I)
~R⊃~[P⊃(RvQ)] K⊃(G⊃S) D⊃A ~I⊃C
D⊃F C
i) ~M~L j) QvF k) ~(S&T) l) ~C⊃(AvB)
~L⊃~K Q⊃A SvT ~D⊃(Cv~B)
K⊃M F⊃A ~S≡T ~(cvD)
A (AvB)&(Cv~B)
m) G≡~H n) P≡Q o) (N⊃S)&(G⊃D)
~GH ~(P≡~Q) (SvD)⊃{[F⊃(FvK)]⊃(N&G)}
N≡G
p) (S&B)⊃K q) CvB r) H⊃(D⊃K)
(G⊃P))&(GvP) ~(C&~B) (K&M)⊃P
~B≡(~P&G) ~(~C&B) J⊃~(M⊃P)
S⊃K C&(Bv~C) H⊃(D⊃~I)
6-4. Write a rule of inference for the Sheffer stroke, defined in section 3-5.
This chapter has focused on improving your understanding of material introduced in chapter 5, so there are only a few new
ideas. Complete short explanations in your notebook for the following terms. But also go back to your explanations for the
terms in the chapter summary exercises for chapter 5 and see if you can now explain any of these terms more accurately and
clearly.
a. Reductio Ad Absurdum Strategy
b. Main Connective
c. Discharging an Assumption
d. Hopping Scope Lines.
2.6.3: Derivations- Overview, Definitions, and Points to Watch out for is shared under a not declared license and was authored, remixed, and/or
curated by LibreTexts.
SECTION OVERVIEW
2.7: Natural Deduction for Sentence Logic - Derived Rules and Derivations without
Premises
2.7.1: Derived Rules

2.7: Natural Deduction for Sentence Logic - Derived Rules and Derivations without Premises is shared under a not declared license and was
authored, remixed, and/or curated by LibreTexts.
7.1: Derived Rules
This section begins with a somewhat strange example. We will first follow our noses in putting together a derivation using the
strategies I have recommended. When we are done, we will notice that some of the steps, although perfectly corn, do no real work
for us. We will then find interesting ways to avoid the superfluous steps and to formulate in a general way some methods for
making derivations easier.
Let's derive 'A⊃(B⊃C)' from '(A⊃B)⊃C'. Our derivation must begin like this:
1 | (A⊃B)⊃C P
|?
|?
| A⊃(B⊃C)
We will pursue the obvious strategy of getting the conclusion by constructing a subderivation from the assumption of 'A' to 'B⊃C'
as conclusion:
1 | (A⊃B)⊃C P
2||AA
||?
||?
| | B⊃C
| A⊃(B⊃C) ⊃I
We have reduced our task to that of deriving 'B⊃C' from 'A', w can use the outer derivation's premise. But how are we going to do
that?
The target conclusion we now need to derive, 'B⊃C', is itself a conditional. So let's try setting up a sub-subderivation with 'B' as
assumption from which we are going to try to derive 'C'. We are shooting for a derivation which looks like this:
1 | (A⊃B)⊃C P
2||AA
|||BA
|||?
|||?
|||C
| | B⊃C ⊃I
| A⊃(B⊃C) ⊃I
How are we going to squeeze 'C' out of 'B'? We have not yet used our premise, and we notice that the consequence of the premise is
just the needed sentence 'C'. If only we could also get the antecedent of the premise, 'A⊃B', in the sub-sub-derivation, we could use
that and the premise to get 'C' by ⊃E.
It might look rough for getting 'A⊃B' into the sub-sub-derivation, but once you see how to do it, it's not hard. What we want is
'A⊃B', which, again, is a conditional. So we will have to start a sub-sub-sub-derivation with 'A' as assumption where we will try to
get 'B' as a conclusion. But that's easy because this sub-sub-sub-derivation is a subderivation of the derivation with 'B' as its
assumption. So all we have to do is reiterate 'B' in our sub-sub-sub-derivation.
If this is a little confusing, follow it in the completed derivation below, rereading the text if necessary to see clearly the thought
process which leads me along step by step:
1 | (A⊃B)⊃C P
2||AA
3|||BA
4 | | | (A⊃B)⊃C 1, R
5||||AA
6 | | | 4| B 3, R
7 | | | A⊃B 5-6, ⊃I
8 | | 3 | C 4, 7, ⊃E
9 | 2 | B⊃C 3-8, ⊃I
10 | A⊃(B⊃C) 2-9 , ⊃I
I've carefully gone through this example for you because I wanted to illustrate again our strategies for constructing derivations. In
this case, though, we have produced a derivation which, while perfectly correct, has an odd feature. Notice that I got 'B' in step 6 by
just reiterating it. I never used :he assumption, 'A'! In just the same way, I never used the assumption of 'A' on line 2 in deriving
'B>C' in line 9. The fact that 'A' was assumed (twice no less!), but never used, in no way shows anything technically wrong with
the derivation. Any derivation correctly formed, as this one is, following the rules, counts as correct even if it turns out that parts
were not used. No one ever said that a line, either an assumption or a conclusion, has to be used.
I should refine what I just said: The assumptions of 'A', both in line 2 and in line 5, were not used in deriving the target conclusions
of the subderivations in which 'A' was assumed. But we had to assume 'A' in both cases to permit us to apply ⊃I to introduce a
conditional with 'A' as - the antecedent. However, if in subderivation 2 the assumption 'A' was never used in deriving 'B⊃C', you
would suspect that we could derive not just 'A⊃(B⊃C)' but the stronger conclusion 'B⊃C' from the original premise. And, indeed,
we can do just that:
1 | (A⊃B)⊃C P
2||BA
3 | | | (A⊃B)⊃C 1, R
4||||AA
5 | | | | B 2, R
6 | | A⊃B 4, 5, ⊃I
7 | | C 3, 6, ⊃E
8 | B⊃C 2-7, ⊃I
Now we notice that we could have worked the original problem in a different way. We could have first derived 'B⊃C', as I have
just done. Then we could have modified this derivaton by inserting the subderivation beginning with 'A', the subderivation 2 in the
previous derivation, and then applying ⊃I. In other words, if we can derive 'B⊃C', we can always derive 'A⊃(B⊃C)' by simply
assuming 'A', deriving 'BX' by whatever means we used before, and then applying >I. In fact, we can organize things most clearly
by separating the two steps. First derive 'B⊃C', then create a subderivation with assumption 'A' and conclusion 'B⊃C' obtained by
reiterating 'B⊃C' from above. Then apply ⊃I. The relevant parts of the total derivation, beginning with the previously derived
conclusion, 'B⊃C', will look like this:
|.
|.
|.
| B⊃C
||AA
| | B⊃C R
| A⊃(B⊃C) ⊃I
We have just discovered something extremely interesting: Nothing in the above line of thought turns on the two sentences involved
being 'B⊃C' and 'A'. This procedure will work for any old sentences X and Y. For any sentences X and Y, if we can derive Y, we
can always extend the derivation to a new derivation with conclusion X⊃Y. If
|.
|.
|.
|Y
stands for the part of the derivation in which we derive Y, the new derivation will look like this:
|.
|.
|.
|Y
||XA
||YR
| X⊃Y ⊃I
Because X and Y can be any sentences at all, we can express the fact we have just discovered as a "new" rule of inference:
|Y
| X⊃ Y Weakening (W)
In other words, if any sentence, Y, appears in a derivation, you are licensed to write X⊃Y anywhere below, using any sentence you
like for X. This rule of inference is not really new (that's why a moment ago I put quotes around the word "new"). If we want to, we
can always dispense with the weakening rule and use our original rules instead. Wherever we have used the weakening rule, we can
always fill in the steps as shown above by assuming X, reiterating Y, and applying ⊃I.
Dispensable, shortcut rules like weakening will prove to be extraordinarily useful. We call them Derived Rules.
A Derived Rule is a rule of inference which can always be replaced by some combination of applications of the original rules
of inference. The original rules are called the Primitive Rules of inference.
A proof of a derived rule is a demonstration which shows how the derived rule may be systematically replaced by application
of the primitive rules of inference.
The weakening rule is proved in the schematic derivation which you saw immediately above.
By using the derived weakening rule, we can immensely simplify the derivation we have been studying in this section. For we can
use weakening instead of both of the subderivations which start from 'A' as an assumption. In addition to the use of weakening
which we have already i seen, we can use it in the subderivation which begins by assuming 'B'. Given 'B' as an assumption,
weakening immediately licenses us to write 1 'A⊃B', which we need for applying ⊃E.
1 | (A⊃B)⊃C P
2||BA
3 | | | (A⊃B)⊃C 1, R
4 | | | A⊃B 2, W
5 | | | C 3, 4, ⊃E
6 | B⊃C 2-5, ⊃I
7 | A⊃(B⊃C) 6, W
Isn't that easy!
7.1: Derived Rules is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Once we see how much work derived rules can save us, we will want others. Indeed, many derivations which are prohibitively
complicated if we use only primitive rules become quite manageable when we can use derived rules also. Here is another example
of a derived rule:
Argument by Cases
| XvY
| X⊃Z
| Y⊃Z
| Z AC
Here is a proof of this derived rule:
.
.
.
1 | XvY (Input for the derived rule)
2 | X⊃Z (Input for the derived rule)
3 | Y⊃Z (Input for the derived rule)
4 | | ~Z A
5|||XA
6 | | | X⊃Z 2, R
7 | | | Z 5, 6, ⊃E
8 | | 3 | ~Z 4, R
9 | | ~X 5-8, ~I
10 | | XvY 1, R
11 | | Y 9, 10, vE
12 | | Y⊃Z 3, R
13 | 2 | Z 11, 12, ⊃E
14 | ~~Z 4-13, ~I
15 3| Z 14, ~E
Again, the point of this proof is this: Suppose you have just used the derived rule Argument by Cases. Then, if you really want to,
you can go back and rewrite your derivation using only the primitive rules. This proof shows you how to do it. Whatever sentence
you have used for X and whatever sentence you have used for Y, just substitute the above in your derivation, after having written in
your sentences for X and Y. Of course, you will have to redo the line numberings to fit with those in your full derivation. 1 have
put in line numbers above to help in talking about the proof.
(A small point to notice in this example: In line 14 I have appealed to subderivation 2, lines 4-13, to use negation introduction. But
where in the subderivation did I conclude both a sentence and its negation? The point is that the assumption can count as one of
these sentences. Why? Because any sentence can be derived from itself, by reiteration. Thus, in derivation 2, I can fairly enough
count both ~Z and Z as following from ~Z.)
Here is another derived rule, which we will also call Argument by Cases because it differs only superficially from the last:
Argument by Cases (second form)
| XvY
|.
|.
|.
||XA
||.
||.
||.
||Z
|.
|.
|.
||YA
||.
||.
||.
||Z
| Z AC
In words, if in a derivation you already have a sentence of the form XvY, a subderivation from X as assumption to Z
as conclusion, and a second subderivation from Y as assumption to Z as conclusion, you are licensed to write Z as a
conclusion anywhere below.
This second form of argument by cases is actually the form you will use most often.
The proof of the second form of argument by cases goes like this:
1 | XvY (Input for the derived rule)
|.
|.
||XA
| | . (Input for the derived rule)
||.
||Z
2 | X⊃Z ⊃I
|.
|.
||YA
| | . (Input for the derived rule)
||.
||Z
3 | Y⊃Z ⊃I
| Z From lines 1, 2, 3 and the first form of Argument by Cases
Note that in this proof I have used a previously proved derived rule (the first form of argument by cases) in proving a new derived
rule (the second form of argument by cases). Why should I be allowed to do that, given that a proof of a derived rule must show
that one can dispense with the derived rule and use primitive rules instead? Can you see the answer to this question?
Suppose we want to rewrite a derivation which uses a new derived rule so that, after rewriting, no derived rules are used. First
rewrite the derivation dispensing with the new derived rule, following the proof of the new derived rule. The resulting derivation
may use previously proved derived rules. But now we can rewrite some more, using the proofs of the previously proved derived
rules to get rid of them. We continue in this way until we arrive at a (probably horrendously long) derivation which uses only
primitive rules.
Argument by cases is an especially important derived rule, so much so that in a moment I'm going to give you a batch of exercises
designed exclusively to develop your skill in applying it. Its importance stems not only from the fact that it can greatly shorten your
work. It also represents a common and important form of reasoning which gets used both in everyday life and in mathematics. In
fact, many texts use argument by cases as a primitive rule, where I use what I have called disjunction elimination. In fact, given the
other rules, what I have called argument by cases and disjunction elimination are equivalent. I prefer to start with my form of
disjunction elimination because I think that students learn it more easily than argument by cases. But now that your skills have
developed, you should learn to use both rules effectively.
 Exercise
7-1. Use argument by cases as well as any of the primitive rules to construct derivations to establish the validity of the
following arguments:
a) AvB b) Av(BvC) c) (AvB)&(B>C) d) (A&B)v(A&C)
BvA (AvB)vC AvC A&(BvC)
e) A&(BvC) f) Av(B&C) g) (AvB)&(AvC) h) KvL
(A&B)v(A&C) (AvB)&(AvC) Av(B&C) K≡L
K&L
i) (D⊃G)v(D⊃I) j) ~CvK k) ~HvM l) (S&J)v(~S&~J)
D⊃(GvI) A⊃D ~M⊃~C S≡J
(AvC)⊃(KvD) (HvC)⊃M
m) K⊃(Fvc)
J⊃(CvD)
~C
~(FvD)⊃~(KvJ)
7-2. Show that in the presence of the other primitive rules, vE is equivalent to AC. (Half of this problem is already done in the
text. Figure out which half and then do the other half I) , Biconditional Introduction Biconditional Elimination and Disjunction
Elimination Denying the Consequent Reductio Ad Absurdum Vraditionally called "Modus Tolens")
2.7.2: Argument by Cases is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Here are some further derived rules. In naming these rules I have used the same name for similar or closely connected primitive
rules.
Biconditional Introduction Bioconditional Elimination
| | X | X≡Y
||.|X
| | . | Y ≡E
| | Y and
| | X≡Y
||Y
| | X ≡E
||Y
||.
||.
||X
| X≡Y ≡I
Disjunction Elimination
| ~XvY | Xv~Y
| X and | Y
| Y vE | X vE
Denying the Consquent Reductio Ad Absurdum
(Traditionally called "Modus Tolems")
| X⊃Y | X⊃~Y | ~X⊃Y | | ~X
| ~Y | Y | ~Y | | .
| ~X DC | ~X DC | X DC | | .
||Y
| | ~Y
| X RD
The reductio ad absurdum rule looks like a negation elimination rule. Actually, as you will see when you do the exercises, it uses
both ~I and ~E.
We can get further derived rules from the laws of logical equivalence (chapters 3 and 4). For example, any sentence of the form ~
(XvY) is logicalley equivalent to ~X&~Y. Because two logically equivalent sentences have the same truth value in any assignment
of truth values to sentence letters, if one of these sentences validly follows from some premises and assumptions, so does the other.
We can use these facts to augment our stock of derived rules:
| ~(XvY) | ~X&~Y
| ~X&~Y DM | ~(XvY) DM
Similarly, one can use other logical equivalences to provide derived rules. Here is a list of such rules you may use, given in a
shortened notation. You should understand the first line as the pair of derived de Morgan rules immediately above. Understand the
following lines similarly.
DE MORGAN'S RULES
~(XvY) and ~X&~Y are mutually derivable (DM).
~(X&Y) and ~Xv~Y are mutually derivable (DM).
CONTRAPOSITION
X⊃Y and ~Y⊃~X are mutually derivable (CP).
~X⊃Y and ~Y⊃X are mutually derivable (CP).
X⊃~Y and Y⊃~X are mutually derivable (CP).
CONDITIONAL RULES
X⊃Y and ~XvY are mutually derivable (C).
~(X⊃Y) and X&~Y are mutually derivable (C).
The letters in parentheses are the abbreviations you use to annotate your use of these rules.
We could add further rules of mutual derivability based on the distributive, associative, commutative, and other laws of logical
equivalence. But in practice the above rules are the ones which prove most useful.
It is not hard to prove that these rules of mutual derivability follow from the primitive rules-in fact, you will give these proofs in the
exercises.
Many texts use rules of logical equivalence to augment the rules for derivations in a much more radical way. These strong rules of
replacement, as they are called, allow you to substitute one logical equivalent for a subsentence inside a larger sentence. Thus, if
you have derived '(AvB)⊃C', these strong rules allow you to write down as a further conclusion '(Av~~B)⊃C', where you have
substituted '~~B' for the logically equivalent 'B'.
By the law of substitution of logical equivalents, we know that such rules of replacement must be correct, in the sense that they will
always take us from true premises to true conclusions. But it is not so easy to prove these replacement rules as derived rules. That
is, it is hard to prove that one can always systematically rewrite a derivation using one of these replacement rules as a longer
derivation which uses only primitive rules. For this reason I won't be using these replacement rules. Your instructor may, however,
explain the replacement rules in greater detail and allow you to use them in your derivations. Your instructor may also choose to
augment the list of logical equivalents you may use in forming such rules.
 Exercise
7-3. Prove all the derived rules given in the text but not proved there. In each of your proofs use only those derived rules
already proved.
7-4. Provide derivations which establish the validity of the following arguments. You may use any of the derived rules. In fact,
many of the problems are prohibitively complex if you don't use derived rules!
a) M&(~BvC) b) M⊃(D⊃P) c) (K⊃S)⊃(S⊃H)
B⊃C M⊃S S
M⊃P K⊃H
d) F⊃O e) ~[(F&H)v~F] f) B⊃(HvR)
L⊃J ~H B⊃(~H⊃R)
(FvL)⊃(OvJ)
g) ~(~M&D) h) (A⊃B)&(D⊃~B) i) P⊃(DvM)
F⊃~M (CvD)&(C⊃~B) (P⊃D)v(P⊃M)
Fv~D ~A
~D
j) (G&~M)⊃(~M&K) k) AvB l) S=J
K⊃~G ~B≡(CvD) (S&J)v(~S&~J)
G⊃M (D&E)v[D&(F⊃G)]
A
m) ~C⊃[Fv~(DvN)] n) (GvA)⊃(H⊃B) o) F⊃(KvB)
~N⊃D [H⊃(H&B)]⊃K (~FvG)&(~Gv~K)
~F⊃C G⊃K F⊃B
p) Dv(M⊃J) q) Q-~(A&F) r) (J&~T)⊃P
[M⊃(M&J)]⊃(PvK) ~(MvA)⊃~H ~A⊃~T
(P⊃D)&(K⊃F) ~(Q&A)vF ~TvC
DvF Q⊃(H⊃M) C⊃D
~P⊃[I⊃(D&A)]
s) B⊃(NvM)
N⊃(C&K)
C⊃(K⊃P)
~(P&B)
B⊃M
2.7.3: Further Derived Rules is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
When we discovered the derived weakening rule, we stumbled across the fact that a derivation (or a subderivation) does not have to
use all, or even any, of its premises or assumptions. This fact is about to become important in another way. To fix ideas, let me
illustrate with the simplest possible example:
0|BP
1||AA
2 | | A 1, R
3 | A⊃A 1-2, ⊃I
The premise, 'B', never got used in this derivation. But then, as I put it before, who ever said that all, or even any, premises have to
be used?
Once you see this last derivation, the following question might occur to you: If the premise, 'B', never gets used, do we have to
have it? Could we just drop it and have a derivation with no premises? Indeed, who ever said that a derivation has to have any
premises?
A derivation with no premises, which satisfies all the other rules I have given for forming derivations, will count as a perfectly
good derivation. Stripped of its unused premise, the last derivation becomes:
1||AA
2 | | A 1, R
3 | A⊃A 1-2, ⊃I
(You might now wonder: Can subderivations have no assumptions? We could say yes, except that an assumptionless subderivation
would never do any work, for a subderivation helps us only when its assumption gets discharged. So I will insist that a
subderivation always have exactly one assumption.)
All right - a derivation may have no premises. But what does a premiseless derivation mean?
Remember that the significance of a derivation with one or more premises lies in this: Any case, that is, any assignment of truth
values to sentence letters, which makes all the premises true also makes all of the derivation's conclusions true. How can we
construe this idea when there are no premises?
To approach this question, go back to the first derivation in this section, the one beginning with the premise 'B'. Since the premise
never got used, we could cross it out and replace it by any other sentence we wanted. Let us indicate this fact symbolically by
writing X for the' premise, thereby indicating that we can write in any sentence we want where the 'X' occurs
0|XP
1||AA
2 | | A 1, R
3 | A⊃A 1-2, ⊃I
For example, for X we could put the logical truth, 'AV~A'. Because the result is a correct derivation, any assignment of truth values
to sentence letters which makes the premise true must also make all conclusions true. But 'Av~A' is true for all cases. Thus the
conclusion, 'A⊃A', must be true in all cases also. That is, 'A⊃A' is a logical truth. I can make the same point another way. I want to
convince you that 'A⊃A' is true in all cases. So I'll let you pick any case you want. Given your case, I'll choose a sentence for X
which is true in that case. Then the above derivation shows 'A⊃A' to be true in that case also.
Now, starting with any derivation with no premises, we can go through the same line of reasoning. Adding an arbitrary, unused
premise shows us that such a derivation proves all its conclusions to be logical truths. Since we can always modify a premiseless
derivation in this way, a premiseless derivation always proves its conclusions to be logical truths:
A derivation with no premises shows all its conclusion to be logical truth
Armed with this fact, we can now use derivations to demonstrate that a given sentence is a logical truth. For example, here is a
derivation which shows 'Av~A' to be a logical truth:
1 | | ~(Av~A) A
2 | | ~A&~~A 1, DM
3 | | ~A 2, &E
4 | | ~~A 3, &E
5 | Av~A 1-4, RD
I devised this derivation by using the reductio strategy. I assumed the negation of what I wanted to prove. I then applied the derived
De Morgan and reductio rules. Without these derived rules the derivation would have been a lot of work.
Let's try something a bit more challenging. Let's show that
{[A⊃(B&~C)]&(~BvD)}⊃(A⊃D)
is a logical truth. This is not nearly as bad as it seems if you keep your wits about you and look for the main connective. What is the
main connective? The second occurrence of '⊃', just after the ')'. Since we want to derive a conditional with '[A⊃(B&~C)]&
(~BvD)' as antecedent and 'A⊃D' as consequence, we want a subderivation with the first sentence as assumption and the second as
final conclusion:
1 | | [A⊃(B&~C)]&(~BvD) A
||
||?
||?
| | A⊃D
| {[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) ⊃I
What do we do next? Work in from both ends of the subderivation. The conclusion we want is the conditional with 'A' as
antecedent. So probably we will want to start a sub-sub-derivation with 'A' as assumption. At the top, our assumption has an '&' as
its main connective. So &E will apply - to give us two simpler conjuncts which we may be able to use. The first of these conjuncts
is a conditional with 'A' as antecedent. We are going to be assuming 'A' as a new assumption any way, so most likely we will be
able to apply ⊃E. Let's write down what we have so far:
1 | | [A⊃(B&~C)]&(~BvD) A
2|||AA
3 | | | [A⊃(B&~C)]&(~BvD) 1, R
4 | | | A⊃(B&~C) 3, &E
5 | | | ~BvD 3, &E
6|||
|||?
|||D
| | A⊃D ⊃I
| {[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) ⊃I
To complete the derivation, we note that from lines 2 and 4 we can get the conjunction 'B&~C' by ⊃E. We can then extract 'B' from
'B&~C by &E and apply the derived form of vE to 'B' and '~BvD' to get 'D' as we needed:
1 | | [A⊃(B&~C)]&(~BvD) A
2|||AA
3 | | | [A⊃(B&~C)]&(~BvD) 1, R
4 | | | A⊃(B&~C) 3, &E
5 | | | ~BvD 3, &E
6 | | | B&~C 2, 4, ⊃E
7 | | | B 6, &E
8 | | | D 5, 7, vE
9 | | A⊃D 2-8, ⊃I
10 | {[A⊃(B&~C)]&(~BvD)}⊃(A⊃D) 1-9, ⊃I
You might be entertained to know how I dreamed up this horriblelooking example. Note that if, in the last derivation, we
eliminated line 10 and the outermost scope line, line 1 would become the premise of a derivation with 'A⊃D' as its final
conclusion. In other words, I would have a derivation that in outline looked like this:
|XP
|.
|.
|.
| Y final conclusion
But starting with such a derivation I can obviously do the reverse. I get back to the former derivation if I add back the extra outer
scope line, call what was the premise the assumption of the subderivation, and add as a last step an application of ⊃I. In outline, I
have
||XA
||.
||.
||.
||Y
| X⊃Y ⊃I
Looking at the last two schematic diagrams you can see that whenever you have a derivation in the form of one, you can easily
rewrite it to make it look like the other. This corresponds to something logicians call the Deduction Theorem.
Here is one last application. Recall from chapter 3 that a contradiction is a sentence which is false for every assignment of truth
values to sentence letters. We can also use derivations to establish that a sentence is a contradiction. Before reading on, see if you
can figure out how to do this.
A sentence is a contradiction if and only if it is false in every case. But a sentence is false in every case if and only if its negation is
true in every case. So all we have to do is to show the negation of our sentence to be a logical truth:
To demonstrate a sentence, X, to be a contradiction, demonstrate its negation, ~X, to be a logical truth. That is, construct a
derivation with no premises, with ~X as the final conclusion.
7-5. Demonstrate the correctness of the following alternative test for contradictions:
A derivation with a sentence, X as its only premise and two sentences, Y and ~Y, as conclusions shows X to be a
contradiction.
7-6. Provide derivations which establish that the following sentences are logical truths. Use derived as well as primitive rules.
a) (AvB)⊃~(B⊃A)
b) Mv~(M&N)
c) [H⊃(O⊃N)]⊃[(H&O)⊃N]
d) (D⊃B)⊃{D⊃T)⊃[D⊃(B&T)]}
e) (K⊃F)⊃[~F⊃~(K&P)]
f) [(FvG)⊃(P&Q)]⊃(~Q⊃~F)
g) [L⊃(M⊃N)]⊃[(L⊃M)⊃(L⊃N)]
h) [(SvT)⊃F]⊃{(FvG)⊃H]⊃(S⊃H)}
i) (I&~J)v[(J&K)v~(K&I)]
j) {[C&(AvD)]v~(C&F)}v~(A&~G)
7-7. Provide derivations which establish that the following sentences are contradictions:
a) A&~A
b) (Hv~B)&[(~B⊃H)&~H)]
c) [(H&F)⊃C]&~[H⊃(F⊃C)]
d) [~(GvQ)&(K⊃G)]&~(Pv~K)]
e) [K⊃(D⊃P)&[(~KvD)&~(K⊃P)]
f) ~[~(Nv~R)⊃(N≡~R)]
g) (FvG)≡(~F&~G)
h) [~(FvG)v(P&Q)]&~(~Q⊃~F)
i) (A⊃D)&[(A&~B)v(A&~C)]&[(B&~D)v(B&C)]
(Exercise i is unreasonably long unless you use a derived rule for the distributive law. You have really done the work for
proving this law in problem 7-ld.
j) (A≡B)=(~A≡B)
7-8. Consider the definition
A set of sentence logic sentences is Inconsistent if and only if there is no assignment of truth values to sentence
letters which makes all of the sentences in the set true.
a) Explain the connection between inconsistency as just defined and what it is for a sentence to be a contradiction.
b) Devise a way of using derivations to show that a set of sentences is inconsistent.
c) Use your test to establish the inconsistency of the following sets of sentences:
c1) C≡G, G≡~C
c2) FvT, (FvT)⊃(~F&~T)
c3) JvK, ~Jv~K, J≡K
c4) (GvK)⊃A, (AvH)⊃G, G&~A
c5) D=(~P&~M), P≡(I&~F), ~Fv~D, D&J
7-9. Devise a way of using derivations which will apply to two logically equivalent sentences to show that they are logically
equivalent. Explain why your method works. Try your method out on some logical equivalences taken from the text and
problems of chapter 3. I
Provide short explanations for each of the following. Check against the text to make sure your explanations are correct, and
save your answers for reference and review.
a) Main Connective
b) Primitive Rule
c) Derived Rule
d) Weakening Rule
e) Contraposition Rule
f) De Morgan's Rules
g) Conditional Rules
h) Reductio Ad Absurdurn Rule
i) Derivations without Premises
j) Tests for Logical Truths and Contradictions
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SECTION OVERVIEW
2.8: Truth Trees for Sentence Logic - Fundamentals

2.8.1: Proving Validity with Truth Trees
2.8.4: The Complete Rules for the Connectives

2.8: Truth Trees for Sentence Logic - Fundamentals is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
8.1: Proving Validity with Truth Trees
You know, from the exercises of chapter 4, that you can use truth tables to check the validity of any argument of sentence logic. But
such truth table checks for validity are extremely tedious. If you read chapters 5, 6, and 7, you also know how to establish validity
by using derivations. Many logicians like this natural deduction technique because (for those with a taste for logic) derivations are
challenging and fun, and because derivations express and make precise informal modes of deductive reasoning.
In this and the next chapter I will present a third means for determining the validity of sentence logic arguments-the truth tree
method. This method is more nearly mechanical than is natural deduction. This fact makes truth trees less fun, because they
provide less of a challenge, but also less aggravating, because they are easier to do. Truth trees also have the advantage of making
the content of sentence logic sentences dear, in a way which helps in proving general facts about systems of logic, as you will see if
you study part II of Volume II.
As a basis for the truth tree method we need to remember two fundamental facts from sections 4-1 and 4-2. We know that an
argument is valid if and only if every possible case which makes all its premises true makes its conclusion true. And we know that
this comes to the same thing 114 Truth Trees fm Sentence Logic Fundamentals 8-1. Proving Validity with Truth Trees 115 as an
argument having no counterexamples, that is, no cases which make the premises true and the conclusion false.
The truth tree method proceeds by looking for counterexamples in.an organized way. The method has been cleverly designed so
that it is guaranteed to turn up at least one counterexample to an argument if there are any counterexamples. If the method finds a
counterexample, we know the argument is invalid. If the method runs to completion without turning up a counterexample, we know
there are no counterexamples, so we know that the argument is valid. Finally, the method considers whole blocks of possible cases
together, so in most cases it works much more efficiently than truth tables.
We will introduce the truth tree method with a specific example:
AvB
~BvC
AvC
We are trying to find a counterexample to this argument. That is, we are looking for an assignment of truth values to sentence
letters which makes the premises true and the conclusion false. Now, if we try to make some sentences true and another sentence
false, things are going to get very confusing. It would be much more straightforward if we could follow a procedure which tried to
make all of the sentences considered true.
Can we do that and still be looking for a counterexample to our argument? Yes - if we replace the conclusion of the argument with
its negation. So we begin the truth tree method for looking for a counterexample by listing the premises of the argument and the
negation of the conclusion:
1 AvB P(Premise)
2 ~BvC P (Premise)
3 ~(AvC) ~C (Negation of the Conclusion)
Our method now proceeds by trying to make lines 1, 2, and 3 true. "Make true" just means finding an assignment of truth values to
sentence letters for which the sentences are true. If we succeed, we will have a case in which the argument's premises, lines 1 and 2,
are true. And this same case will be one in which the conclusion is false, because the negation of the conclusion, line 3, will be true.
So if we can make lines 1, 2, and 3 true, we will have our counterexample.
Note that I have numbered the lines, and written some information off to the right of each line which tells you why the line got put
there. You should always number and annotate your lines in this way so that we can talk about them easily.
Now to work. Let's begin by making line 1 true. We see that there are two alternative ways of making 'AvB' true. First, we can
make it true just by making 'A' true. If we make 'A' true, we have made 'AvB' true, whatever eventually happens to the truth value
of 'B'. But the same is true of 'B'. If we make 'B' true, 'AvB' will be true whatever eventually happens to the truth value of 'A'. So
making 'A' true is a first and making 'B' true is a second, alternative way of making 'AvB' true. We need the truth tree method to be
guaranteed to find a way of making all initial lines true if there is a way. So we must be sure to write down all ways in which a line
might be true, and if there are alternative ways, we must list them independently.
We mark this fact by extending the tree as follows:
We have split the original path into two paths or branches. Each branch will represent one way of making true all the sentences
which appear along it. The left branch will make line 1 true by making 'A' true. The right branch will make line 1 true by making
'B' true. Since the paths have branched, they represent alternative ways of making line 1 true. What happens along one path will
have no effect on what happens along the other path below the point at which they branch.
I have added some notation to the tree. The '1, v' at the right of line 4 means that I got line 4 from line 1 by working on a
disjunction. I have checked line 1 to remind myself that I have now done everything which needs doing to make it true. I won't
have to work on that line anymore.
Now let's work on line 2. I have to make it true, and I will do this exactly as I did for line 1. I will make each of the two disjuncts
true; that is, I will make '~B' true and I will independently make 'C' true along a separate branch. But I have to "add on" the ways of
making line 2 true to each of the ways of making line 1 true. Remember, we are looking for some way of making all the original
lines true together. So I must first write each of the alternative ways of making line 2 true as an extension of the first (left branch)
way of making line 1 true. That is, I must extend the left branch with two new branches each of which represents one of the two
ways of making line 2 true. So the left branch will itself split and look like this:
The same reasoning obviously applies to the right branch. I must add on both of the ways of making line 2 true to the second (right
branch) way of making line 1 true. We represent this by splitting the right branch so it looks like this:
Putting these pieces together, after working on line 2, the tree looks like this:
Each branch represents one of the ways of making line 1 true combined with one of the ways of making line 2 true. The leftmost,
or first, branch makes line 1 true by making 'A' true and makes line 2 true by making '-B' true. The second branch makes line 1 true
by making 'A' true and makes line 2 true by making 'C' true. (Do you see how we combine the alternative ways of making the first
two lines true?)
The third branch makes line 1 true by making 'B' true and makes line 2 true by making '~B' true. Whoops! Surely something has
gone wrong! Surely we can't make line 1 true by making 'B' true and at the same time make line 2 true by making '~B' true. If we
were to make '~B' true, this would be to make 'B' false, and we just said that along the third branch 'B' would be made true to make
line 1 true. We can't make 'B' both true and false. What is going on?
What has happened is that the third branch represents an inconsistent way of making lines 1 and 2 true. Focus on the fact that each
branch represents one of the four ways of trying to make lines 1 and 2 true together. It turns out that the third of these ways won't
work. One way of making line 1 true is by making 'B' true. One way of making line 2 true is by making '~B' true. But we can't
combine these ways of making the two lines true into one way of making the lines true at the same time, because doing this would
require making 'B' both true and false. We ' mark this fact by writing an 'x' at the bottom of the branch on which both 'B' and '~B'
appear. The 'x' tells us that we cannot consistently make true all of the sentences which appear along the branch. We say that the
branch is Closed. We never have to work on a closed branch again:
We say that a branch is Closed when a sentence, X, and its negation, ~X, both appear on the branch. We mark a branch as
closed by writing an ' X ' at the bottom of the branch. Do not write anything further on a branch once it has been marked as
closed.
The sentence X may be an atomic sentence, such as 'A', or a compound sentence, such as 'Bv~(C&~A)'. Also note that the sentence
and its negation which cause a branch to close must both appear as entire sentences at points along a branch. If one or both appear
as part of some larger compounds, that does not count. To illustrate this point, look at the tree drawn up to line 4, as presented on
page 115. On the right-hand branch you see 'B' as the entire sentence at the end of the branch, and '~B' as part of a compound on
line 2. The branch does not close at this point because there is no conflict between line 2 and the right branch at line 4. It is
perfectly possible for 'B' and '~BvC' both to be true.
It is the fact that branches can close which gives the truth tree method its simplifying power. Here is how simplification occurs. We
still have to make line 3 true, and we have to combine the ways of making line 3 true with the ways of making lines 1 and 2 true.
But we can forget about one of these ways of (trying to) make lines 1 and 2 true because it turned out to be inconsistent. This
corresponds to ruling out certain lines of the truth table before we have completely worked them out. Because a truth tree avoids
working out what corresponds to some of the lines of the corresponding truth table, truth trees often save work.
Note how I annotated the tree in the last step: I put '2, v' to the right of line 5 to indicate that I got line 5 from line 2 by working on
a disjunction. And I checked line 2 to remind myself that I have done everything needed to make line 2 true. I won't need to worry
about line 2 anymore.
Let's complete the tree by working on line 3. What has to be done to make line 3 true? Line 3 is the negation of a disjunction. The
negation of a disjunction is true just in case the disjunction itself is false. So, to make '~(AvC)' true, I have to make 'AvC' false.
How do I make 'AvC' false? According to the truth table definition of 'v', I can do this only by making both 'A' false and 'C' false.
So the result of making line 3 true will not be a branch of two alternative ways of making a sentence true. There is only one way to
make line 3 true: a stack which first makes 'A' false and then makes 'C' false. But to keep things clearly organized I only write
down true sentences. So what I have to write is a stack which makes '~A' true and makes '~C' true.:
~A
~C
By making '~A' true followed by '~C' true, I have made 'A' and 'C' both false, which makes 'AvC' false, in the only way that this can
be done. Finally, making 'AvC' false makes '~(AvC)' true, which was what we wanted to accomplish.
Where do I write this stack? I must combine the results of making line 3 true with all the ways I already have for making lines I and
2 true. At first thought, this would seem to indicate that I should write my '~A', '~C' stack at the bottom of every branch. This is
almost right. But I won't have to write the results of working on line 3 at the bottom of the closed (third) branch. Remember that the
closed branch represents a way of trying to make previous sentences true that can't work because it is inconsistent. So I only have
to write my stack at the bottom of every Open Branch, that is, every branch which does not have an 'x' at its bottom:
A branch which is not closed is Open.
Adding the stack to the open branches gives a tree that looks like this:
The '3, ~v' means that I got the lines from line 3 by working on a negated disjunction. I also checked line 3.
We have one more thing to do to this tree. Note that now the first, second, and fourth branches all have the problem that the third
branch had. To make all the sentences along the first branch true, I would have to make 'A' true and '~A' true. That is, I would have
to make 'A' both true and false. I can't do that, so I put an 'x' at the bottom of the first branch. I do exactly the same to tk second
branch for exactly the same reason. The fourth branch has both 'C' and '~C', so I must close it with an ' x ' also. The final tree looks
like this:
Here is the final result: All branches have closed. That means that every possible way of trying to make all the initial sentences true
results in a conflict. Each way of trying to assign truth values to sentence letters requires that some sentence be made both true and
false, and the rules of the game do not permit that. We agreed in chapter 1 that in a given problem a sentence letter would be true or
false, but not both. Since there is no way of making all initial sentences true, there is no way of making the argument's premises
true and its conclusion false. That is, there is no counterexample to the argument. So the argument is valid.
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The example we have just completed contains all the ideas of the truth tree method. But explaining two more points will help in
understanding the rest of the rules.
First, why did I write a stack when I worked on line 3 and a two-legged branch when I worked on lines 1 and 2? Here is the
guiding idea: If there are two alternative ways of making a sentence true I must list the two ways separately. Only in this way will
the method nose out all the possible different combinations of ways of making the original sentences true. I can make line I true by
making 'A' true and, separately, by making 'B' true. I just list these two alternative ways separately, that is, on separate branches.
This will enable me to combine the alternatives separately with all possible combinations of ways of making the other lines true.
But when I tried to make line 3 true there was only one way to do it. I can make ' ~(AvC)' true only by making both '~A' and '~C'
true. Because there is only one way of making line 3 true, line 3 does not generate two branches. It generates only one thing: the
extension of all existing open branches with the stack composed of '~A' followed by '~C'.
I have just explained how I decide whether to write a branch or a stack when working on a sentence. (I'll call the sentence we are
working on the "target" sentence.) But how do I decide what new sentence to write on the resulting branches or the resulting stack?
Since each path represents a way of making all the sentences along it true, I should put enough sentences to ensure the truth of the
target sentence along each branched or stacked path. But I also want to be sure that I don't miss any of the possible ways of making
the target sentence true.
It turns out that I can achieve this objective most efficiently by writing the least amount on a branch which gives one way of
making the target sentence true. I must then write, along separate branches, all the different ways in which I can thus make my
target sentence true with as few components as possible. I will express this idea by saying that, when working on a sentence, I must
write, along separate branches, each minimally sufficient sentence or stack of sentences which ensures the truth of my target
sentence. This will make sure that no avoidable inconsistency will arise. And in this way the method will be sure to find a way of
making all the initial sentences true if there is a way.
In sum, in working on a target sentence, I do the following. I figure out all the minimally sufficient ways of making my target
sentence true. If there is only one such way, I write it at the bottom of every open path on which the target sentence appears. If there
is more than one, I write each separate minimally sufficient way on a separate branch at the bottom of every open path on which
the target sentence appears.
This reasoning works to explain all the further rules.
 Exercise
8-1. Use the truth tree method to show that the following arguments are valid. Show your trees, following the models in the
text, complete with annotations to the right indicating at each stage which rule you have used and what line you have worked
on.
a) D b) ~(Mv~N) c) ~(FvP) d) Hvl
J N ~Fv~P ~IvJ
DvJ ~JvK
Hvk
8-2. If you ran into a conjunction on a tree, what would you do? If you don't see right away, try to figure this out on the basis of
the description in this section of how the rules work and the two rules you have seen. If you succeed in stating the rule for a
conjunction, try to state the rules for negated conjunctions, conditionals, and negated conditionals. If you succeed again, try
stating the rules for biconditionals and negated biconditionals, which are a little harder.
2.8.2: The General Strategy for the Rules is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Let us now look at an example of an invalid argument:
AvB
A&B
Keep in mind that this argument is invalid just in case there is a counterexample to it, and that the truth tree method works by
searching for such counterexamples.
We begin by listing the premise and the negation of the conclusion.
1 AvB P
2 ~(A&B) ~C
An assignment of truth values which makes both of these sentences true will constitute a counterexample, demonstrating the
invalidity of the original argument.
We already know what to do with line 1. But when we get to line 2 we will have a new kind of sentence to work on, a negated
conjunction. To make the negated conjunction '~(A&B)' true, I must make the conjunction '(A&B)' false. Making 'A' false is
minimally sufficient for making '(A&B)' false, and so for making '~(A&B)' true. So for line 2 I must produce one branch which
makes 'A' false, which I do by writing a branch which makes '~A' true. Making 'B' false is likewise minimally sufficient for making
'A&B' false. So for line 2 I also need a second, separate branch with '~B' on it. Altogether, the result of working on line 2 will look
like this:
I will write this at the bottom of every open path on which line 2 appears. Note that this rule is very different from the rule for the
negated disjunction '~(AvB)'. In working on the negated disjunction, '~(AvB)', I had to write a stack of the negated disjuncts ('~A'
followed by '~B') at the bottom of every open branch. This is because only the stack, '~A' followed by '-B', is minimally sufficient
for making the negated disjunction true. In working on the negated conjunction, '~(A&B)', I must write a branch with '~A' on one
leg and '-B' on the other leg. This is because each of the negated conjuncts ('-A' or '-B') is by itself minimally sufficient for making
the negated conjunction true.
We now know everything we need to test our new argument for validity:
I first worked on line 1, producing a branch with 'A' on one leg and 'B' on the other, representing the two minimally sufficient ways
of making line 1 true. I checked line 1 to remind myself that I have done everything needed to make it true. I then worked on line 2.
At the bottom of each open path I put a new branch, one leg representing one minimally sufficient way of making line 2 true, and
the second leg representing the second minimally sufficient way of making line 2 true. The annotation on the right of line 4
indicates that I got line 4 by applying my rule for a negated conjunction to line 2. And I checked line 2 to remind myself that I have
done all that is needed to make it true. Next I inspected each path to see if it contained both a sentence and the negation of the same
sentence. The first path has both 'A' and '~A'. The fourth path has both 'B' and '~B'. So I marked both these paths closed. At this
point I observed that there are no unchecked compound sentences which I could make true by making some simpler sentences true.
How does this tree show the argument to be invalid? You see that this completed tree has two open paths. What do they mean?
Look, for example, at the first open path. Reading up from the bottom, suppose we make '~B' hue and make 'A' true. This
constitutes an assignment of truth values to sentence letters ('B' false and 'A' true) which will make the original sentences 1 and 2
true. This is because we made '~F true as one way of making line 2 true. And we made 'A' true as one way of making line 1 true. So
by assigning truth values f to 'B' and t to 'A', we make all sentences along the first open path true.
This sort of thinking works generally for open paths. Keep in mind how we constructed the paths. Each time we found a sentence
with an '&' or a 'v' in it, we wrote a shorter sentence farther down the path. We did this in a way which guaranteed that if we made
the shorter sentence true, the longer sentence from which it came would be true also. So if we start at the bottom of a path and work
up, making each sentence letter and each negated sentence letter true, that will be enough to make all sentences along the path true.
Any sentence longer than a sentence letter or negated sentence letter will be made true by one or more sentences farther down the
path.
Even if we start with very complicated sentences at the top, each sentence is made true by shorter sentences below, and the shorter
sentences are made true by still shorter sentences farther down, until we finally get to the shortest sentences possible, sentence
letters and negated sentence letters. Then we can work backward. Making the sentence letters and negated sentence letters along
the path true will, step by step, also make true all longer sentences higher up on the path.
We have seen that the assignment of f to 'B' and t to 'A' makes all sentences along the first open path true. In particular, it makes the
original first two sentences true. These sentences were our argument's premise and the negation of our argument's conclusion. So
making the initial two sentences true makes our argument's premise true and its conclusion false. In short, 'B' false and 'A' true
constitutes a counterexample to our argument. Thus the argument is invalid. One counterexample is enough to show an argument to
be invalid, but you should note that the second open path gives us a second counterexample. Reasoning in exactly the same way as
we did for the first open path, we see that making '~A' true and 'B' true makes every sentence along the second open path true. So
making 'A' false and 'B' true constitutes a second counterexample to the argument.
Our last step is to record all this information next to the completed tree. We write 'invalid' next to the tree and write the
counterexamples which show the argument to be invalid. We can do this most easily with the sentences of sentence logic which
describe the counterexamples. The sentence '~B&A' describes the counterexample given by the first open path and the sentence
'~A&B' describes the counterexample given by the second open path.
A last detail will conclude this section: The order of the cases in the description of the counterexample obviously does not matter. I
described the first counterexample with the sentence '~B&A' because I read the counterexample off the path by reading from the
bottom up. As a matter of practice, I recommend that you do the same. But in principle the equivalent description 'A&~B' describes
the same counterexample just as well.
8-3. Use the truth tree method to show that the following arguments are invalid. Show your trees, being careful to show which
branches are closed. In each problem give any counterexamples which show the argument being tested to be invalid.
a) F b) ~(~S&T) C) KvH d) ~(I&P)
F&K S ~K PvF
H&D IvF
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8.4: The Complete Rules for the Connectives
All we have left to do is to state the rules for the remaining connectives. And this further development involves nothing very
different from what we have done already. The reasoning that enables us to figure out the new rules is just like the reasoning we
went over in explaining the rules for disjunction, negated disjunction, and negated conjunction.
To see how to state the rules generally, let us start with one we have already seen in action:
Rule V: If a disjunction of the form XvY appears as the entire sentence at a point on a tree, write the branch
at the bottom of every open path on which XvY appears.

Notice that I have stated the general rule using bold face 'X' and 'Y' instead of sentence letters 'A' and 'B'. This is because I want to
be able to apply the rule to more complicated sentences, which result if we substitute compound sentences for the 'X' and the 'Y'.
Don't worry about this for now. We will work on this complication in the next chapter.
We continue with a statement of the other two rules which we have already seen in action:
Rule ~v: If a negated disjunction of the form ~(XvY) appears as the entire sentence at a point on a tree, write the stack
~X
~Y
at the bottom of every open path on which ~(XvY) appears.
Rule -&: If a negated conjunction of the form ~(X&Y) appears as the entire sentence at a point on a tree, write the branch
at the bottom of every open path on which ~(X&Y) appears.

Did you try your hand at exercise 8-2? If not, or if you did and had trouble, try again, now that you have seen how I have given a
general statement of the rules we have already worked with. Try to figure out the new rules for the other connectives by thinking
through what, for example, you would need to do to a tree to make true a conjunction which appears on a tree, or a conditional, or a
negated conditional.
In order to make a conjunction true on a path we have to make both. conjuncts true. This is the only minimally sufficient way of
making the conjunction true. So the rule for conjunction is
Rule &: If a conjunction of the form X&Y appears as the entire sentence at a point on a tree, write the stack
X
Y
at the bottom of every open path on which X&Y occurs.
Now let's try the conditional. How can we make a conditional true? If we make the antecedent false, that does it right there. Making
the antecedent false is minimally sufficient for making the conditional true. Similarly, making just the consequent true is minimally
sufficient for making the conditional true. Clearly, we need a branch, each fork of which has one of the minimally sufficient ways
of making the conditional true. One fork of the branch will have the negation of the antecedent, and the other fork will have the
consequent:
Rule⊃: If a sentence of the form X⊃Y appears as the entire sentence at a point on a tree, write the branch
at the bottom of every open path on which X⊃Y occurs.
What about negated conditionals? A negated conditional produces a stack as do conjunctions and negated disjunctions. A negated
conditional can be made true only by making its antecedent true and its consequent false at the same time. So our rule for negated
conditionals reads ,
Rule ~⊃: If a sentence of the form ~(X⊃Y) appears as the entire sentence at a point on a tree, write the stack
X
~Y
at the bottom of every open path on which ~(X⊃Y) appears.
The rule for the biconditional is just a bit more complicated. You can't make the biconditional X≡Y true just by making one
component true or false. You have to assign a truth value to two components to make it true. So far, the rule for X≡Y is looking like
that for X&Y. But there are two independent ways of making X≡Y true. We can make it true by making both components true, and
we can make it true by making both components false. Finally, these are all the minimally sufficient ways of making X≡Y true. So
we will have to make a branch, each fork of which will have two sentences:
Rule ≡: If a biconditional of the form X≡Y appears as the entire sentence at a point on a tree, write the branch
at the bottom of every open path on which X≡Y appears.

Note that the rule ≡ looks different from all the previous rules. Each previous rule instructed you to branch or stack, but not both.
The rule ≡ requires both branching and stacking. We need a branch with two forks, and each fork has a stack of two sentences.
Only thus do we get all the minimally sufficient ways of making a biconditional true.
The reasoning behind the rule for the negated biconditional looks very similar to that for the conditional. A negated biconditional is
true just in case the biconditional itself is false. Under what conditions is a biconditional false? It's false when the components have
different truth values. This can happen in two ways: The biconditional is false when the first component is true and the second
component is false, and it is false when the first component is false and the second component is true. This gives us the rule Rule
Rule ~≡: If a negated biconditional of the form ~(X≡Y) appears as the entire sentence at a point on a tree, write the branch
at the bottom of every open path on which ~(X≡Y) appears.

As with the conditional, we need two branches, each with a stack of two sentences. Only in this way will we get all the minimally
sufficient ways of making a negated biconditional true.
We need one last truth tree rule. Consider the following argument and the tree for testing its validity:
How did I get the branch on the right to close? We can look at this in two equally correct ways. We can note that '~~B' is the
negation of '~B'. Thus we have an inconsistent pair of sentences on the same branch. This is what I hope occurred to you when you
met double negations in the previous exercises in this chapter. I can no more make '~B' and '~~B' both true than I can make 'B' and
'~B' both true. So the branch closes. Also, we can observe that '~~B' is logically equivalent to 'B'. Clearly, the one will be true if
and only if the other is true. So we can make '~~B' true by making 'B' true. We could rewrite the right branch on the above tree as
follows:
~~B 1, ~&
B 3, ~~
A 2, ~⊃
~B 2, ~⊃
x
We will formalize this move in a further rule:
Rule ~~: If the sentence ~~X appears as the entire sentence at a point on a tree, write X at the bottom of every open path on
which ~~X appears.
These nine rules for the connectives tell you what to do when working on a sentence which appears as the entire sentence at a point
on a tree. The examples should give you a pretty good idea of how to go about applying these rules. But let's review the basic ideas:
A truth tree works by looking for an assignment of truth values to sentence letters which makes all of the tree's original
sentences true.
We will see in the next chapter that a truth tree can be used for other things in addition to checking arguments for validity. But so
far we have studied only validity checking:
To use the truth tree method to test an argument for validity, list, as the initial sentences of a tree, the argument's premises
and the negation of its conclusion.
You then apply the rules in this way:
1) Starting with a tree's initial sentences, you may apply the rules in any order.
2) After applying a rule to a sentence, check the sentence, to remind yourself that you do not need to work on that sentence
again.
3) After working on a sentence, look to see if any paths have closed. Mark any closed paths with an ' x '.
4) Continue working on unchecked sentences until the tree has been completed, which happens when either
a) All paths have closed, or
b) No unchecked sentences are left to which you can apply a rule.
Now we need to review how you should interpret a tree once it has been completed:
An open path provides an assignment of truth values to sentence letters as follows: Assign the truth value t to each sentence
letter which occurs as the entire sentence at some point along the open path. Assign the truth value f to each sentence letter
whose negation occurs as the entire sentence along the open path.
In a completed tree the assignment of truth values to sentence letters provided by an open path makes all sentences along the
path true. If all paths have closed, there is no assignment of truth values to sentence letters which makes all the initial
sentences in the tree true.
When we use a tree to test an argument for validity, we apply this last general fact about trees, as follows:
Suppose we have a completed tree whose initial sentences are an argument's premises and the negation of the argument's
conclusion. An open path in this tree provides an assignment of truth values to sentence letters which makes all of the initial
sentences true, and so makes the argument's premises true and its conclusion false. Such an assignment is a counterexample
to the argument, showing the argument to be invalid. If all of the tree's paths have closed, there is no such assignment, hence
no counterexample, and the argument is valid.
We will also always use the following practical procedure:
When using the truth tree method to test an argument for validity, write next to your completed tree the word 'valid' or
'invalid' to record what the tree shows about the argument. Also, if the tree shows the argument to be invalid, write down all
the counterexamples which the tree provides to the argument.
Of course, one counterexample is enough to show that an argument is invalid. But you should practice your skill at constructing
trees, and as part of this practice it is useful to read off all the counterexamples a tree provides.
8.4: The Complete Rules for the Connectives is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
The summary statement tells you that you may be apply the rules in any order you like. Indeed, the order does not matter in getting
the right answer. But the order can make a practical difference as to how quickly you get the right answer. In all the examples you
have seen so far, I started with the first sentence on the tree and worked downward. To show you that order can make a practical
difference I am going to redo the first example and work on line 3 first.
Compare this tree with the first way I worked the problem, and you will see that this one is a good bit shorter. I worked our the
problem the longer way the first time because I wanted to illustrate how branched get stacked on top of several other branches. And
I wanted to show you how working on th sentences in a different order can make a difference as to how quickly you can do the
problem.
But how can you tell what the shortest way will be? I have no surefire formula which will work for you all the time. In general, you
have to "look ahead: and try to see how the problem will work our to decide on an order in which to work the lines. Your objective
is to get as many branches to close as quickly as possible. If you don't make the best choice, the worst will happen is that your
problem will take a little longer to do.
There are several practical rules of thumb to help you out. First,
Practical guide: Work on lines that produce stacks before lines that produce branches.
Branches make trees messy. Stacks line up more sentences along a path and improve your chances of getting a path to close. In the
problem which I redid I knew I was likely to get a shorter tree by working on line 3 before working on lines 1 and 2. I knew this
because line 3 produces a stack while both 1 and 2 produce branches.
A second suggestion:
Practical guide: When you have as sentences on which to work only ones which are going to produce branches, work first on
one which will produce at least some closed paths.
If the first. two suggestions don't apply, here is a final piece of advice:
Practical guide: If the first two guides don't apply, then work on the longest sentence.
If you put off working on a long sentence, you may have to copy the results of working on it at the bottom of many open branches.
By working on a long sentence early on, you may get away with its long pieces at the bottom of relatively few branches, making
your tree a little less complicated.
Now you should sharpen your understanding of the rules by working the following problems. The easiest way to remember the
rules is to understand how they work. Once you understand the reasoning which led us to the rules, you will very quickly be able to
reconstruct any rule you forget. But you may also refer to the rule summary on the inside back cover. You should understand the
boxes and circles in this way: Whenever you find a sentence with the form of what you see in a box occurring as the entire sentence
at a point along a tree, you should write what you see in the circle at the bottom of every open path on which the first sentence
occu'rred. Then check the first sentence. I have written the abbreviated name of the rule above each rule.
 Exercise
8-4. Use the truth tree method to determine whether or not the following arguments are valid. In each case show your tree,
indicating which paths are closed. Say whether the argument is valid or invalid, and if invalid give all the counterexamples
provided by your tree.
a) A b) ~C⊃H c) K⊃F d) P⊃M e) FvG f) J⊃D
B ~H⊃C F M⊃B ~GvH K⊃D
A&B K P⊃B F&H J&K
D
g) (KvS) h) J⊃D i) M≡N j) T≡I k) ~(F&~L)
~(K&H) K⊃D M=~N J&A ~(L&~C)
K⊃H JvK ~Tv~A F≡L
D
l) F≡G m) C⊃N n) ~(Q⊃D) o) R≡~A p) O≡~F
G≡H ~(I⊃C) ~(Q&B) ~(R=T) ~(F≡K)
F≡H ~(Nv~I) ~(B&~R) R&~S O⊃K
~N⊃~I Dv~Q
8-5. The rules, as I have stated them, specify an order for the branches and an order for sentences in a stack. For example, the
rule for a negated conditional, ~(X⊃Y) instructs you to write a stack
X
~Y
But could you just as well write the stack
~Y
X
at the bottom of every open path on which ~(X⊃Y) appears? If so, why? If not, why not? Likewise, the rule for the
conditional, X⊃Y, instructs you to write the branches
But could you just as well write the branches in the other order, writing
at the bottom of every open path on which X⊃Y appears? If so, why? If not, why not? Comment on the order of branches and
the order within stacks in the other rules as well. I
8-6. If we allow ourselves to use certain logical equivalences the truth tree method needs fewer rules. For example, we know
from 1 chapter 4 that, for any sentences X and Y, X⊃Y is logically equivalent to ~XvY. Now suppose we find a sentence of
the form X⊃Y on a tree. We reason as follows: Our objective is to make this sentence true by making other (in general shorter)
sentences true. But since ~XvY is logically equivalent to X⊃Y, we can make X⊃Y true by making ~XvY true. So I will write
~-XvY at the bottom of every open branch on which X⊃Y appears, check X⊃Y, and then apply the rule for disjunctions to
~XvY. In this way we can avoid the need for a special rule for conditional sentences.
Apply this kind of reasoning to show that, by appealing to de Morgan's rules, we can do without the rules for negated
conjunctions and negated disjunctions, using the rules for disjunctions and conjunctions in their place. Also show that we could
equally well do without the rules for conjunctions and disjunctions, using the rules for negated disjunctions and negated
conjunctions in their place. 8-7. In chapter 3 I extended the definition of conjunctionsand disjunctions to include sentences with
three or more conjuncts and sentences with three or more disjuncts. But we have not yet stated truth tree rules for such
sentences.
a) State truth tree rules for conjunctions of the form X&Y&Z and for disjunctions of the form XvYvZ.
b) State truth tree rules for conjunctions and disjunctions of arbitrary length.
8-8. Write a truth tree rule for the Sheffer stroke, defined in section 3-5.
Here are this chapter's important new ideas. Write a short explanation for each in your notebook.
a) Truth Tree
b) Counterexample
c) Branching Rule
d) Nonbranching Rule
e) Closed Branch (or Path)
f) Open Branch (or Path)
g) Rule ~~
h) Rule &
i) Rule ~&
j) Rule v
k) Rule ~v
1) Rule ⊃
m) Rule -⊃
n) Rule ≡
0) Rule -≡
2.8.5: In Which Order Should You Work on the Sentences in a Tree? is shared under a not declared license and was authored, remixed, and/or
SECTION OVERVIEW
2.9: Truth Trees for Sentence Logic - Applications


2.9: Truth Trees for Sentence Logic - Applications is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
This is going to be a short chapter. You really have all the facts about truth trees. It only remains to see how to apply these facts in
some new ways.
In the last chapter I was careful to give you only problems in which the sentences were very simple. But now that you have the
hang of the rules, let's see how to apply them in testing the validity of an argument like this:
Following the suggestion of working first on nonbranching lines, I began with line 2. But what, then, should I do with line l? Line 1
is a disjunction of conjunctions. Which rule applies? And how? Keep in mind what the rules are supposed to do. Our objective is to
make line 1 true, in all the minimally sufficient ways that this can be done. Because line 1 is a disjunction, we must do this by
making each of the disjuncts true along a separate leg of a branch. That is, we must make the full subsentence 'A&B' true along one
branch and the full subsentence '~A&C' true along a second branch. The subsentences 'A&B' and '~A&C' are themselves
compound sentences to which we must apply the rule for conjunction, which I have done in lines 6 and 7.
How can you tell which rule applies to line l? Ask yourself: What do 1 have to do to the whole sentence appearing on line 1 which
will guarantee that the whole sentence is true? The answer will direct you to the components which must be written at points farther
down on the tree.
To see more clearly how this works, let us look at some more examples. If the sentence is
(A&B)⊃(CvD)
I say to myself: This sentence is a conditional, the antecedent of which is the conjunction 'A&B' and the consequent of which is the
disjunction 'CvD'. A conditional can be made true by making the antecedent false. And it can alternatively be made true by making
the consequent true. So at the bottom of every open path on which '(A&B)⊃(CvD)' appears, I must write the branch
What should I do with

-[(BvC)⊃A]?
This sentence is a negated conditional. To make it true I must simultaneously make the conditional's antecedent true and its
consequent false. So at the bottom of every open path on which it appears, I must write the stack
BvC
~A
We will look at some nastier examples in a moment. But first let's discuss an explicit prescription for applying the rules to
sentences no matter how complex. Suppose you are faced with a very complex sentence and you are not sure which rule applies.
Ask yourself the following question: In building this sentence up from smaller parts, what was the last step? What was the very last
connective that got used in putting together this compound sentence from its components?
I hope that by this time you are recognizing the idea of a sentence's main connective from chapter 1. (If you did the chapters on
natural deduction, you may find this section repetitious. Please bear with those who are learning the tree method without having
learned natural deduction first.)
The Main Connective in a compound sentence is the connective which was used last in building up the sentence from its
component or components.
To determine which rule applies to a sentence, first determine the sentence's main connective. If the main connective is not '~' all
you have to do is to apply the rule for the main connective. If the main connective is '~', determine the main connective of the
negated sentence. Then apply the corresponding rule ~v, ~&, ~⊃, ~≡, or ~~. Let us see how the rules apply to a few more
examples. Consider
(A⊃B)⊃[(CvA)⊃B)]
What is the main connective? It is '⊃'. But which occurrence of '⊃'? It's the second occurrence. The parentheses tell you that the
very last step in building up this sentence is to take 'A⊃B' and '(CvA)⊃B' and to make the first the antecedent and the second the
consequent of a conditional.
Here is another example:
~{[(A≡B~)≡C]≡[C⊃(~A≡B)]}
This is a negated biconditional, and the occurrence of '≡' to which you have to apply the rule is the third. In building the sentence
up from its parts, the very last thing that was done was to apply the outermost negation sign. The step before that was to form a
biconditional from the components '(A≡~B)≡C' and 'C⊃(~A≡B)'. So the rule for negated biconditional applies, using '(A≡~B)≡C'
and 'C⊃(~A≡B)' as the components. At the bottom of every open branch, we write
Before turning you loose on some exercises, I should mention a small side point. When you worked problem Wg, one of the open
branches displayed '~H' and 'K' but neither 'S' nor '~S'. So what counterexamples does this branch represent? What happened in this
case is that making 'H' false and 'K' true is already enough to make everything on the branch true. If 'H' is false and 'K' is true, the
initial sentences are true whether 'S' is true or false. But, strictly speaking, an assignment of truth values to sentence letters for a
sentence must assign a truth value to each sentence letter in the sentence. So, strictly speaking, a counterexample for this problem
must specify a truth value for 'S'. Thus we really should say that the assignment of truth values which you read off the open branch,
'H' false and 'K' true, is an abbreviation for the pair of counterexamples, ' ~H&K&S' and '~H&K&~S'.
However, having said this, we will record counterexamples by reading off the truth values for the sentence letters and negated
sentence letters which occur on an open branch. If some sentence letters have been left out, we know that our list is an abbreviation
for all the truth value assignments which result by arbitrarily assigning truth values for the neglected sentence letters.
 Exercise
9-1. Determine the main connective in the following sentences:

a) A&[Bv(C⊃D)]
b) ~[(Hv~K)≡F]⊃)~Rv~F)
c) ~{[Iv~P0≡M]⊃(~Iv~G)}
d) {[F⊃(Bv~N)]⊃)~J⊃N)}⊃{Nv[F⊃~(BvJ)]}
9-2. Test the following arguments for validity. Show your trees, showing which paths are closed. Say whether the argument is
valid or invalid, and if invalid give the counterexamples provided by the finished tree.
Before beginning these problems, you should review the practical guides at the end of chapter 8. Also, try to stay clear of the
following pitfalls that often catch students: A tree is not completed until either all branches have closed or until all sentences
have been checked. Sometimes you can see before completing a tree that there will surely be at least one counterexample.
(This can happen, for example, when completing the left branch produced by an original sentence, before the right branch is
complete.) But, both for safety and to make sure you get plenty of practice, please don't quit on a tree until all compound
sentences have been checked.
Sometimes students try to take shortcuts, writing down the results of working on two rules at once. Often this produces
mistakes, and makes it terribly hard for anyone to correct your papers. Finally, pay constant attention to the main connective.
Only by correctly identifying the main connective in a compound sentence will you correctly apply the rules.
a) B b) DvF c) N⊃(D⊃P) d) (H&P)v(S&~J)
(BvC)&(BvD) K⊃~F N⊃D J⊃~(H&D)
~F⊃(DvK) N⊃P (J⊃~D)
e) (RvL)⊃(CvA) f) (~I&~D)v~D g) (G⊃B)&(~G⊃N)
R⊃~G ~[(I&~J)v(D&~J)] Bv~N
~(B&~R) Nv~B
~BvA
h) (D⊃H)⊃P i) I⊃(J⊃K) j) F&[Pv~(D⊃F)]
D⊃~(FvG) (I⊃J)⊃K Dv~(FvK)
FvH K⊃D
D⊃P
k) (T&G)v(G&~M) l) (H≡~Q)≡(H≡~M) m) Q≡M
T&~M H⊃(Qv~(~Q&M) (H≡~Q)≡(H≡~M)
n) J⊃[D⊃(BvP)] o) ~[~K&~(~A&~B)] p) (GvA)⊃(H⊃B)
DvP (A⊃K)&(K⊃B) [H⊃(H&B)]⊃K
H⊃P G⊃K
q) Dv(M⊃J) r) [(F&~B)vQ]⊃[A&(SvT)] s) F⊃(KvB)
[M⊃(M&J)]⊃(PvK) F&~(S&A) (~FvG)&(~Gv~K)
(P⊃D)&(K⊃F) SvA F⊃B
DvF B⊃(A⊃S)
t) F≡[~Jv(C&T)] u) Q≡~(A&F) v) (I&~T)⊃P
Av(C≡O) ~(MvA)⊃~H ~A⊃~T
~[F⊃(K⊃~O)] ~(Q&A)vF ~TvC
F≡(T&A) Q⊃(H⊃M) C⊃D
~P⊃[I⊃(D&A)]
2.9.1: Application of the Rules to Complex Sentences is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
So far we have used truth trees exclusively to test the validity of arguments. But now that we understand how truth trees work we
can easily apply them to do all sorts of other jobs. For example, suppose I hand you the sentence '~(AvB)&(~A⊃B)' and ask you
whether it is a contradiction. You could always work this problem with truth tables. If in all lines of the truth table the sentence is
false, it is, by definition, a contradiction. If the sentence is true in one or more cases it is not a contradiction. But a truth tree will get
the job done faster. If we make a sentence the initial line on a truth tree, we know the truth tree will find us a case in which the
sentence is true, if there is such a case. If there is no such case, all paths on the tree will close. Can you see how the truth tree will
tell us whether the sentence is a contradiction?
To explain this more thoroughly, we can again use the idea of a counterexample. A contradiction is false in all cases. So a case in
which a sentence is true constitutes a Counterexample to its being a contradiction. The truth tree method applies immediately to
look for counterexamples to a sentence being a contradiction. We make the sentence to be tested the first line of a tree. If there are
one or more counterexamples, that is, cases in which the sentence is true, the tree method is guaranteed to find them. If the tree
method does not turn up a counterexample, that is, if all paths close, we know there are no cases in which the sentence is true. But
if there are no cases in which it is true, the sentence is false in all cases; in other words, it is a contradiction. We can summarize this
test by saying
To test a sentence for being a contradiction, make the sentence the first line of a truth tree. If there is an open path in the tree,
this path provides a counterexample to the sentence being a contradiction. If all paths close, the sentence is a contradiction.
Applying this test to our example we get
Is the sentence 'A≡(~AvC)' a contradiction?
Note that I have written down the result of my test, that the sentence to be tested is not a contradiction. And note how I also put
down the counterexample which shows this.
A final small point about this example. In the annotation for line 4, 1 have listed two rules. This is because I applied the rule v to
the disjunction on the left branch of line 3, and I separately applied the rule ~V to the separate sentence of line 3 on the right
branch.
Can we use truth trees to determine whether a given sentence is a logical truth? Sometimes students propose the following test for a
sentence being a logical truth: List the sentence and see if all branches remain open. Close, but no cigari The proposed test
mistakenly tells us that 'A' is a logical truth. If we make 'A' the initial line of a tree, there is nothing to do, and all branches are
open. But 'A' is not a logical truth. We also get the wrong answer if we apply the proposed test to (Av~A)v(A&~A):
One branch of this tree does close. But the initial sentence is a logical truth, as you can tell from the fact that one of its disjuncts is a
logical truth.
However, there is a very simple way to use the tree method to test for a logical truth. Just use the test for contradictions? How? A
sentence is a logical truth just in case it is true in all cases. But a sentence is true in all cases just in case its negation is false in all
cases. So a sentence is a logical truth if and only if its negation is a contradiction. Suppose, now, that I ask you whether a sentence
is a logical truth, for example, the sentence of the very last example. Take the negation of the sentence. Determine whether this
negation is a contradiction. If the negation is a contradiction, then the original sentence was a logical truth. If the negation is not a
contradiction, then the original sentence was not a logical truth:
Notice that this last tree is an example of a completed tree in which not all compound sentences have been checked. I never worked
on line 3 because all branches closed before I got to line 3. Once all the branches close, the tree is finished. There is no way to
make all the initial sentences true. If any sentences have not been worked when all branches close, continuing and working them
would give us no new information.
Let us similarly test '(A&B)v~A' to see whether it is a logical truth:
How should we understand the counterexample here? The case A&~B ('A' true and 'B' false) is a case in which '~[(A&B)v~A]', the
sentence tested for being a contradiction, is true. But '~[(A&B)v~A]' is true in a case if and only if '(A&B)v~A', the sentence tested
for being a logical truth, is false in the case. A case in which a sentence is f&e proves that the sentence is not a logical truth. Such a
case constitutes a counterexample to the sentence being a logical truth. So the case A&~B is a counterexample to '(A&B)v~A'
being a logical truth. Clearly, this will hold generally:
For any sentence X, any case which is a counterexample to ~X being a contradiction will also be a counterexample to X
being a logical truth.
To summarize the test for logical truth:
To test a sentence for being a logical truth, make the negation of the sentence the first line of a truth tree. If d the paths close,
the sentence is a logical truth. An open path gives a counterexample to the original sentence being a logical truth.
This summary makes no mention of the intermediate role of a test for contradiction. If you do not understand the test as just
summarized, go back over the last two examples and make sure you can see how they fit in with the summary I have just given.
We can do yet more with the truth tree method. Recall that two sentences are logically equivalent just in case their biconditional is
a logical truth. Thus we can use the test we have just devised for logical truth to determine whether two sentences are logically
equivalent:
To test whether X and Y are logically equivalent, test X≡Y for being a logical truth. If X≡Y is a logical truth, X and Y are
logically equivalent. If X≡Y is not a logical truth, X and Y are not logically equivalent. A counterexample to X≡Y being a
logical truth is also a counterexample to the logic equivalence of X and Y. That is, it is a case in which one of the two
sentences, X and Y, is true and the other is false.
Can you see why a counterexample to the logical truth of X≡Y is also a case in which one of the two sentences, X and Y, is true
and the other is false? A counterexample to X≡Y being a logical truth is a case in which X≡Y is false. But a biconditional is false if
and only if one of its components is true and the other is false, that is, if X is true and Y is false or the other way around. Finally,
you can see why we would call such a case a counterexample to the logical equivalence of X and Y. X and Y are logically
equivalent if and only if in all cases they have the same truth value. So if we have come up with a case in which one sentence is
true and the other is false, we have a case which proves by counterexample that they are not logically equivalent.
To illustrate this method of determining logical equivalence, I will use it to verify one of De Morgan's rules. To show: '~(A&B)' is
logically equivalent to '~Av~B'.
This section will introduce you to one more notion:

A set of one or more sentence logic sentences is consistent if and only if there is at least one assignment of truth values to
sentence letters which makes all of the sentences true.
The truth tree method applies immediately to test a set of sentences for consistency. Before reading on, see if you can figure out
this test for yourself.
The tree method works by finding a case in which all initial sentences on a tree are true, if there is such a case. So, to determine
whether the sentences in a given set are consistent, list the sentences as the initial part of a tree. If there is a case in which all of
these sentences are true together, the tree method will find it. Such a case constitutes what logicians call a Model. which shows the
initial sentences to constitute a consistent set. If the tree closes, the set of sentences has no model and is Inconsistent:
A Model of a set of sentence logic sentences is an assignment of truth values to sentence letters which makes all of the
sentences in the set true.
To test a finite set of sentences for consistency, make the sentence or sentences in the set the initial sentences of a tree. If the
tree closes, there is no assignment of truth values to sentence letters which makes all the sentences true (there is no model),
and the set is inconsistent. An open branch gives a model and shows the set to be consistent.
To make the statement of the test correct, I have had to use the notion of a Finite Set of sentences, that is, a set or collection of a
limited, as opposed to an unlimited, number of sentences. You do not need to understand the distinction between a finite as opposed
to an infinite set to understand anything in Volume I and Volume 11, Part I of this text. But you may explore the idea and its
implications for consistency by working through exercise 9-7.
It's time to practice these new applications of truth trees.
 Exercise
9-3. Use the truth tree method to determine which of the following sentences are logical truths. Show your completed trees. If a
sentence is not a logical truth, give the counterexample or counterexamples which show this.
a) ~(~F&G)v~(F&G)
b) [H⊃(O⊃N)]⊃[(H&O)⊃N]
c) [~S&(GvK)]v[~G⊃(SvK)]
d) [(FvG)⊃(P&Q)]⊃(~Q⊃~F)
e) [Lv(M&N)]⊃(LvM)&N]
f) [(K&M)&P]⊃[(~K&P)v(Kv~M)]
g) [L⊃(M⊃N)]⊃[(L⊃M)⊃(L⊃N)]
h) [(SvT)⊃F]⊃{[(FvG)⊃H]⊃(S⊃H)}
i) (I&~J)v[(J&K)v~(K&I)]
j) {[C&(AvD)]v~(C&F)}v~(A&~G)
9-4. Use the truth tree method to determine which of the following sentences are contradictions. Show your completed trees,
and for each sentence you determine not to be a contradiction, give the counterexample or counterexamples which show this.
a) (A&B)&(~Av~B)
b) (FvG)&(~Fv~G)
c) (FvG)≡(~F&~G)
d) [Iv(J&K)]⊃[(IvJ)&K]
e) [(H&F)⊃C]&~[H⊃(F⊃C)]
f) ~{[B&(Mv~P)]⊃[P⊃(~Mv~B)]}
g) (A≡B)≡(~A≡B)
h) [(~FvQ)v(P&Q)&~(~Q⊃~F)
i) [K⊃(G⊃P)&[(~KvD)&~(K⊃P)]
j) [(~GvQ)&(K⊃G)]&~(Pv~K)
9-5. Use the information suggested in exercise 4-3 to state a new truth tree test for logical equivalence. Comment on the
relation between this test and the test given in the text.
9-6. Use the truth tree method to determine which of the following pairs of sentences are logically equivalent. You may use
either the test given in the text or the closely related test you discovered in exercise 9-4. Show your completed trees, and when
you find a pair which is not logically equivalent, give the counterexample or counterexamples which show this.
a) A⊃~A and ~A
b) ~(lvJ) and ~I&~J
c) Mv~H and ~M⊃H
d) ~(F&P) and ~F&~P
e) (D&N)⊃J and D⊃(N⊃J)
f) (I⊃Q)⊃D and l⊃(Q⊃D)
g) L&(SvT) and (L&S)v(L&T)
h) Hv~(~Pv~Q) and (HvP)&(HvQ)
9-7. Consider the following definition:
(C1) A sentence of sentence logic is consistent if and only if it is not a contradiction.
a) Are all logical truths consistent according to definition (C1)? Explain why or why not.
b) Show that a sentence logic sentence is consistent according to definition (C1) if and only if there is at least one assignment
of truth values to sentence letters which makes the sentence true.
If you go on in your study of logic, the more general notion of consistency already given in the text will turn out to be very
important:
(C2) A set of one or more sentence logic sentences is consistent if and only if there is at least one assignment of truth
values to sentence letters which makes all of the sentences true.
c) Show that a set of sentences is consistent according to definition (C2) if and only if the conjunction of all the sentences in
the set is consistent according to definition (C1).
This last problem seems to show that the two definitions of consistency come to the same thing. Why, then, did I say that the
second definition is more general? Actually, there is something not quite right about exercise (c). To see what this is, you need
to understand the difference between a finite and an infinite set of sentences. A finite set of sentences has some definite number
of sentences in it, such as 2, or 47, or 1,007,859. In an infinite set of sentences the list of sentences goes on without end. There
is no integer which gives you the number of sentences in an infinite set.
d) Here is an example of an infinite set of sentences:
~A, ~~A, ~~~A, ~~~~A, ~~~~~A, . . .
The first sentence is '~A'. The second sentence is '~~A'. The third, fourth, fifth, and further sentences are 'A' preceded by 3, 4,
5, and further negation signs, so that the list goes on forever. Question: Is this set consistent?
The difficulty with exercise (c) is that it makes sense only if you assume that the set of sentences is finite. For if the set is
infinite, there is no sentence which is the conjunction of all its members. This is because in sentence logic, all sentences are
finite in length.
Now you can see why definition (C2) is more general than (C1). (C2) gives sense to the consistency of an infinite set of
sentences. The two definitions really come to the same thing, but only for finite sets of sentences.
e) Describe a consistent infinite set of sentences.
f) Use the truth tree method to test the following sets of sentences for consistency. In each case, show your tree. Write next to
your tree whether the set is consistent or inconsistent, and when consistent, give all the models for the set which the truth tree
provides.
f1) PvS, P⊃S
f2) (~F⊃S)>F, ~F, S
f3) K&[(~K&H)vH]
f4) NvB, Nv~B, -NvB
From a logical point of view, you should really think of the truth tree method as a method for testing a set of sentences for
consistency. This general method then has more specific applications, such as testing an argument for validity. This is because
An argument is valid if and only if the set comprised by the argument's premises and the negation of its conclusion is
inconsistent.
We say the same thing in other words by saying that an argument is valid if and only if the negation of the conclusion is
inconsistent with (the set of) the argument's premises.
g) Explain why the last offset statement is correct.
Some textbooks first present the truth tree method as a test of consistency and then apply it to argument validity. I introduced
trees as a test for argument validity because I wanted to motivate the introduction of trees with something you already know
about, namely, arguments. It is initially hard for many students to understand the interest in consistency and inconsistency, but
these notions will become very important in Volume 11, Part I1 of the text.
Here are the important terms and ideas from this chapter. Write your explanations for them as usual. This list repeats some
terms from previous chapters.
a) Main Connective
b) Logical Truth
c) Truth Tree Test for Logical Truths
d) Contradiction
e) Truth Tree Test for Contradictions
f) Logical Equivalence
g) Truth Tree Test for Logical Equivalence
h) Consistency
i) Model
j) Infinite Set of Sentences
k) Truth Tree Test for Consistency of a Finite Set of Sentences
2.9.2: Other Uses for Truth Trees is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
3: Volume II- Predicate Logic
The first part of Volume II lays out predicate logic, including identity, functions, and definite descriptions. The second part
introduces metatheory, including mathematical induction, soundness, and completeness. A predicate is commonly understood to be
a Boolean-valued function P: X→ {true, false}, called the predicate on X. However, predicates have many different uses and
interpretations in mathematics and logic, and their precise definition, meaning and use will vary from theory to theory.

3: Volume II- Predicate Logic is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
1
SECTION OVERVIEW
3.1: Predicate Logic - Syntax

3.1.2: Quantifiers and Variables

3.1: Predicate Logic - Syntax is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
In Volume I you gained a firm foundation in sentence logic. But there must be more to logic, as you can see from the next
examples. Consider the following two English arguments and their transcriptions into sentence logic:
(1 ) Everyone lwes Adam. A (2) Eve loves Adam. B
Eve loves Adam. B Someone loves Adam. C
In sentence logic, we can only transcribe the sentences in these arguments as atomic sentence letters. But represented with sentence
letters, both natural deduction and truth trees tell us that these arguments are invalid. No derivation will allow us to derive 'B' from
'A' or 'C' from 'B'. A&~B is a counterexample to the first argument, and B&~C is a counter-example to the second. An argument is
valid only if it has no counterexamples.
Something has gone terribly wrong. Clearly, if everyone loves Adam, then so does Eve. If the premise is true, without fail the
conclusion will be true also. In the same way, if Eve loves Adam, then someone loves Adam. Once again, there is no way in which
the premise could be true and the conclusion false. But to say that if the premises are true, then without fail the conclusion will be
true also is just what we intend when we say that an argument is valid. Since sentence logic describes these arguments as invalid, it
looks like something has to be wrong with sentence logic.
Sentence logic is fine as far as it goes. The trouble is that it does not go far enough. These two arguments owe their validity to the
internal logical structure of the sentences appearing in the arguments, and sentence logic does not describe this internal logical
structure. To deal with this shortcoming, we must extend sentence logic in a way which will display the needed logical structure
and show how to use this structure in testing arguments for validity. We will keep the sentence logic we have learned in Volume I.
But we will extend it to what logicians call Predicate Logic (also sometimes called Quantificational Logic).
Predicate logic deals with sentences which say something about someone or something. Consider the sentence 'Adam is blond.'
This sentence attributes the property of being blond to the person named 'Adam'. The sentence does this by applying the predicate
(the word) 'blond' to the name 'Adam'. A sentence of predicate logic does the same thing but in a simplified way.
We will put capital letters to a new use. Let us use the capital letter 'B', not now as a sentence letter, but to transcribe the English
word 'blond'. And let us use 'a' to transcribe the name 'Adam'. For 'Adam is blond.', predicate logic simply writes 'Ba', which you
should understand as the predicate 'B' being applied to the name 'a'. This, in turn, you should understand as stating that the person
named by 'a' (namely, Adam) has the property indicated by 'B' (namely, the property of being blond).
Of course, on a different occasion, we could use 'B' to transcribe a different English predicate, such as 'bachelor', 'short', or 'funny'.
And we could use 'a' as a name for different people or things. It is only important to stick to the same transcription use throughout
one problem or example.
Predicate logic can also express relations which hold between things or people. Let's consider the simple statement that Eve loves
Adam. This tells us that there is something holding true of Eve and Adam together, namely, that the first loves the second. To
express this in predicate logic we will again use our name for Adam, 'a'. We will use a name for Eve, say, the letter 'e'. And we will
need a capital letter to stand for the relation of loving, say, the letter 'L'. Predicate logic writes the sentence 'Eve loves Adam.' as
'Lea'. This is to be read as saying that the relation indicated by 'L' holds between the two things named by the lowercase letters 'e'
and 'a'. Once again, in a different example or problem, 'L', 'a', and 'e' could be used for different relations, people, or things.
You might be a little surprised by the order in which the letters occur in 'Lea'. But don't let that bother you. It's just the convention
most often used in logic: To write a sentence which says that a relation holds between' two things, first write the letter which
indicates the relation and then write the names of the things between which the relation is supposed to hold. Some logicians write
'Lea' as 'L(e,a)', but we will not use this notation. Note, also, the order in which the names 'e' and 'a' appear in 'Lea'. 'Lea' is a
different sentence from 'Lae'. 'Lea' says that Eve loves Adam. 'Lae' says that Adam loves Eve. One of these sentences might be true
while the other one is false! Think of 'L' as expressing the relation, which holds just in case the first thing named loves the second
thing named.
Here is a nasty piece of terminology which I have to give you because it is traditional and you will run into it if you continue your
study of logic. Logicians use the word Argument for a letter which occurs after a predicate or a relation symbol. The letter 'a' in 'Ba'
is the argument of the predicate 'B'. The letters 'e' and 'a' in 'Lea' are the arguments of the relation symbol 'L'. This use of the word
'argument' has nothing to do with the use in which we talk about an argument from premises to a conclusion.
At this point you might be perplexed by the following question. I have now used capital letters for three different things. I have
used them to indicate atomic sentences. I have used them as predicates. And I have used them as relation symbols. Suppose you
encounter a capital letter in a sentence of predicate logic. How are you supposed to know whether it is an atomic sentence letter, a
predicate, or a relation symbol?
Easy. If the capital letter is followed by two lowercase letters, as in 'Lea', you know the capital letter is a relation symbol. If the
capital letter is followed by one lowercase letter, as in 'Ba', you know the capital letter is a predicate. And if the capital letter is
followed by no lowercase letters at all, as in 'A', you know it is an atomic sentence letter.
There is an advantage to listing the arguments of a relation symbol after the relation symbol, as in 'Lea'. We can see that there is
something important in common between relation symbols and predicates. To attribute a relation as holding between two things is
to say that something is true about the two things taken together and in the order specified. To attribute a property as holding of one
thing is to say that something is true about that one thing. In the one case we attribute something to one thing, and in the other we
attribute something to two things.
We can call attention to this similarity between predicates and relations in a way which also makes our terminology a bit smoother.
We can indicate the connection by calling a relation symbol a Two Place Predicate, that is, a symbol which is very like an ordinary
predicate except that it has two argument places instead of one. In fact, we may sometimes want to talk about three place
predicates (equally well called 'three place relation symbols'). For example, to transcribe 'Eve is between Adam and Cid', I
introduce 'c' as a name for Cid and the three place predicate 'K' to indicate the three place relation of being between. My
transcription is 'Keac', which you can think of as saying that the three place relation of being between holds among Eve, Adam, and
Cid, with the first being between the second and the third.
This is why our new logic is called 'predicate logic': It involves predicates of one place, two places, three places, or indeed, any
number of places. As I mentioned, logicians also refer to these symbols as one place, two place, or many place relation symbols.
But logicians never call the resulting system of logic 'relation logic'. I have no idea why not.
Our familiar sentence logic built up all sentences from atomic sentence letters. Predicate logic likewise builds up compound
sentences from atomic sentences. But we have expanded our list of what counts as an atomic sentence. In addition to atomic
sentence letters, we will include sentences such as 'Ba' and 'Lea'. Indeed, any one place predicate followed by one name, any two
place predicate followed by two names, and so on, will now also count as an atomic sentence. We can use our expanded stock of
atomic sentences to build up compound sentences with the help of the connectives, just as before.
How would you say, for example, 'Either Eve loves Adam or Adam is not blond.'? 'Lea v ~Ba'. Try 'Adam loves himself and if he is
blond then he loves Eve too.': 'Laa & (Ba ⊃ Lae)'.
In summarizing this section, we say
In predicate logic, a capital letter without a following lowercase letter is (as in sentence logic) an atomic sentence. Predicate
logic also includes predicates applied to names among its atomic sentences. A capital letter followed by one name is a One
applied to one name. A capital letter followed by two names is a Two Place Predicate
Place Predicate
applied to two names, where the order of the names is important. Predicates with three or more
places are used similarly.
 Exercise
In the following exercises, use this transcription guide:
a: Adam
e: Eve
c: Cid
Bx: x is blond
Cx: x is a cat
Lxy: x loves y
Txy: x is taller than y
1-1. Transcribe the following predicate logic sentences into English:
a. Tce
b. Lce
c. ~Tcc
d. Bc
e. Tce ⊃ Lce
f. Lce v Lcc
g. ~(Lce & Lca)
h. Bc ≡ (Lce v Lcc)
1-2. Transcribe the following English sentences into sentences of predicate logic;
a. Cid is a cat.
b. Cid is taller than Adam.
c. Either Cid is a cat or he is taller than Adam.
d. If Cid is taller than Eve then he loves her.
e. Cid loves Eve if he is taller than she is.
f. Eve loves both Adam and Cid.
g. Eve loves either Adam or Cid.
h. Either Adam loves Eve or Eve loves Adam, but both love Cid.
i. Only if Cid is a cat does Eve love him.
j. Eve is taller than but does not love Cid.
3.1.1: We Need More Logical Form is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
1.2: Quantifiers and Variables
We still have not done enough to deal with arguments (1) and (2). The sentences in these arguments not only attribute properties
and relations to things, but they involve a certain kind of generality. We need to be able to express this generality, and we must be
careful to do it in a way which will make the relevant logical form quite clear. This involves a way of writing general sentences
which seems very awkward from the point of view of English. But you will see how smoothly everything works when we begin
proving the validity of arguments.
English has two ways of expressing general statements. We can say 'Everyone loves Adam.' (Throughout, 'everybody' would do as
well as 'everyone'.) This formulation puts the general word 'everyone' where ordinarily we might put a name, such as 'Eve'.
Predicate logic does not work this way. The second way of expressing general statements in English uses expressions such as
'Everyone is such that they love Adam.' or 'Everything is such that it loves Adam.' Predicate logic uses a formulation of this kind.
Read the symbol '(Vx)' as 'Every x is such that'. Then we transcribe 'Everyone loves Adam.' as '(Vx)Lxa'. In words, we read this as
"Every x is such that x loves Adam." '(Vx)' is called a Universal Quantifer. In other logic books you may see it written as (x).
We are going to need not only a way of saying that everyone loves Adam but also a way of saying that someone loves Adam.
Again, English does this most smoothly by putting the general word 'someone' where we might have placed a name like 'Eve'. And
again logic does not imitate this style. Instead, it imitates English expressions such as 'Someone is such that he or she loves Adam.',
or 'Some person is such that he or she loves Adam.', or 'Something is such that it loves Adam.' Read the symbol '(Ǝx)' as 'Some x is
such that'. Then we transcribe 'Someone loves Adam.' as '(Ǝx)Lxa'. '(Ǝx)' is called an Existent Quantifier.
In one respect, '(Ǝx)' corresponds imperfectly to English expressions which use words such as 'some', 'there is a', and 'there are'. For
example, we say 'Some cat has caught a mouse' and 'There is a cat which has caught a mouse' when we think that there is exactly
one such cat. We say 'Some cats have caught a mouse' or 'There are cats which have caught a mouse' when we think that there are
more than one. Predicate logic has only the one expression, '(Ǝx)', which does not distinguish between 'exactly one' and 'more than
one'. '(Ǝx)' means that there is one or more x such that. (In chapter 9 we will learn about an extension of our logic which will enable
us to make this distinction not made by '(Ǝx)'.)
In English, we also make a distinction by using words such as 'Everyone' and 'everybody' as opposed to words like 'everything'.
That is, English uses one word to talk about all people and another word to talk about all things which are not people. The universal
quantifier, '(Vx)', does not mark this distinction. If we make no qualification, '(Vx), means all people and things. The same
comments apply to the existential quantifier. English contrasts 'someone' and 'somebody' with 'something'. But in logic, if we make
no qualification, '(Ǝx)' means something, which can be a person or a thing. All this is very inconvenient when we want to transcribe
sentences such as 'Someone loves Adam.' and 'Everybody loves Eve.' into predicate logic.
Many logicians try to deal with this difficulty by putting restrictions on the things to which the 'x' in '(Vx)' and '(Ǝx)' can refer. For
example, in dealing with a problem which deals only with people, they say at the outset: For this problem 'x' will refer only to
people. This practice is called establishing a Universe of Discourse or Restricting the Domain of Discourse. I am not going to fill in
the details of this common logical practice because it really does not solve our present problem. If we resolved to talk only about
people, how would we say something such as 'Everybody likes something.'? In chapter 4 I will show you how to get the effect of
restricting the domain of discourse in a more general way which will also allow us to talk at the same time about people, things,
places, or whatever we like.
But until chapter 4 we will make do with the intuitive idea of restricting 'x' to refer only to people when we are transcribing
sentences using expressions such as 'anybody', 'no one', and 'someone'. In other words, we will, for the time being indulge in the not
quite correct practice of transcribing '(Vx)' as 'everyone', 'anybody', etc., and '(Ǝx)' as 'someone', 'somebody', or the like, when this
is the intuitively right way to proceed, instead of the strictly correct 'everything', 'something', and similar expressions.
The letter 'x' in '(Vx)' and '(Ǝx)' is called a Variable. Variables will do an amazing amount of work for us, work very similar to that
done by English pronouns, such as 'he', 'she', and 'it'. For example, watch the work 'it' does for me when I say the following: "I felt
something in the closed bag. It felt cold. I pulled it out." This little discourse involves existential quantification. The discourse
begins by talking about something without saying just which thing this something is. But then the discourse goes on to make
several comments about this thing. The important point is that all the comments are about the same thing. This is the work that 'it'
does for us. It enables us to cross-reference, making clear that we are always referring to the same thing, even though we have not
been told exactly what that thing is.
A variable in logic functions in exactly the same way. For example, once we introduce the variable 'x' with the existential
quantifier, '(Ǝx)' we can use 'x' repeatedly to refer to the same (unknown) thing. So I can say, 'Someone is blond and he or she loves
Eve' with the sentence '(Ǝx)(Bx &Lxe)'. Note the use of parentheses here. They make clear that the quantifier '(Ǝx)' applies to all of
the sentence 'Bx & Lxe'. Like negation, a quantifier applies to the shortest full sentence which follows it, where the shortest full
following sentence may be marked with parentheses. And the 'x' in the quantifier applies to, or is linked to, all the occurrences of 'x'
in this shortest full following sentence. We say that
A quantifier Governs the shortest full sentence which follows it and Binds the variables in the sentence it governs. The latter
means that the variable in the quantifier applies to all occurrences of the same variable in the shortest full following sentence.
Unlike English pronouns, variables in logic do not make cross-references between sentences.
These notions actually involve some complications in sentences which use two quantifiers, complications which we will study in
chapter 3. But this rough characterization will suffice until then.
Let us look at an example with the universal quantifier, '(Vx)'. Consider the English sentences 'Anyone blond loves Eve.', 'All
blonds love Eve.','Any blond loves Eve.', and 'All who are blond love Eve.' All these sentences say the same thing, at least so far as
logic is concerned. We can express what they say more painstakingly by saying, 'Any people are such that if they are blond then
they love Eve.' This formulation guides us in transcribing into logic. Let us first transcribe a part of this sentence, the conditional,
which talks about some unnamed people referred to with the pronoun 'they': 'If they are blond then they love Eve.' Using the
variable 'x' for the English pronoun 'they', this comes out as 'Bx ⊃ Lxe'. Now all we have to do is to say that this is true whoever
"they" may be. This gives us '(Vx)(Bx ⊃ Lxe)'. Note that I have enclosed 'Bx ⊃ Lxe' in parentheses before prefixing the quantifier.
This is to make clear that the quantifier applies to the whole sentence.
I have been using 'x' as a variable which appears in quantifiers and in sentences governed by quantifiers. Obviously, I would just as
well have used some other letter, such as 'y' or 'z'. In fact, later on, we will need to use more than one variable at the same time with
more than one quantifier. So we will take '(Vx)', '(Vy)', and '(Vz)' all to be universal quantifiers, as well as any other variable
prefixed with 'V' and surrounded by parentheses if we should need still more universal quantifiers. In the same way, '(Ǝx)', '(Ǝy)',
and '(Ǝz)' will all function as existential quantifiers, as will any similar symbol obtained by substituting some other variable for 'x',
'y', or 'z'.
To make all this work smoothly, we should clearly distinguish the letters which will serve as variables from other letters.
Henceforth, I will reserve lowercase 'w', 'x', 'y', and 'z' to use as variables. I will use lowercase 'a' through 'r' as names. If one ever
wanted more variables or names, one could add to these lists indefinitely by using subscripts. Thus 'a1,' and 'd17,' are both names,
and 'x1' and 'z34( are both variables. But in practice we will never need that many variables or names.
What happened to 's', 't', 'u', and 'v'? I am going to reserve these letters to talk generally about names and variables. The point is
this: As I have mentioned, when I want to talk generally in English about sentences in sentence logic, I use boldface capital 'X', 'Y',
and 'Z'. For example, when I stated the & rule I wrote, "For any sentences X and Y. . . ." The idea is that what I wrote is true no
matter what sentence you might write in for 'X' and 'Y'. I will need to do the same thing when I state the new rules for quantifiers. I
will need to say something which will be true no matter what names you might use and no matter what variables you might use. I
will do this by using boldface 's' and 't' when I talk about names and boldface 'u' and 'v' when I talk about variables.
To summarize our conventions for notation:
We will use lowercase letter 'a' through 'r' as names, and 'w', 'x', 'y' and '2' as variables. We will use boldface 's' and 't' to talk
generally about names and boldface 'u' and 'v' to talk generally about variables.
1.2: Quantifiers and Variables is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
We now have all the pieces for saying exactly which expressions are going to count as sentences of predicate logic. First, all the
sentences of sentence logic count as sentences of predicate logic. Second, we expand our stock of atomic sentences. I have already
said that we will include among the atomic sentences predicates followed by the right number of names (one name for one place
predicates, two names for two place predicates, and so on). We will do the same thing with variables and with variables mixed with
names. So 'Bx' will count as an atomic sentence, as will 'Lxx', 'Lxy', and 'Lxa'. In general, any predicate followed by the right
number of names and/or variables will count as an atomic sentence.
We get all the rest of the sentences of predicate logic by using connectives to build longer sentences from shorter sentences, starting
from atomic sentences. We use all the connectives of sentence logic. And we add to these '(Vx)', '(Vy)', '(Ǝx)', '(Ǝy)', and other
quantifiers, all of which count as new connectives. We use a quantifier to build a longer sentence from a shorter one in exactly the
same way that we use the negation sign to build up sentences. Just put the quantifier in front of any expression which is already
itself a sentence. We always understand the quantifier to apply to the shortest full sentence which follows the quantifier, as
indicated by parentheses. Thus, if we start with 'Lxa', '(Vx)Lxa' counts as a sentence. We could have correctly written '(Vx)(Lxa)',
though the parentheses around 'Lxa' are not needed in this case. To give another example, we can start with the atomic sentences
'Bx' and 'Lxe'. We build a compound by joining these with the conditional, '⊃', giving 'Bx ⊃ Lxe'. Finally, we apply '(Vx)' to this
compound sentence. We want to be clear that '(Vx)' applies to the whole of 'Bx ⊃ Lxe', so we have to put parentheses around it
before prefixing '(Vx)'. This gives '(Vx)(Bx ⊃ Lxe)'.
Here is a formal definition of sentences of predicate logic:
All sentence letters and predicates followed by the appropriate number of names andlor variables are sentences of predicate
logic. (These are the atomic sentences.) If X is any sentence of predicate logic and u is any variable, then (Vu)X (a
universally quantified sentence) and (Ǝu)X (an existentially quantified sentence) are both sentences of predicate logic. If X
and Y are both sentences of predicate logic, then any expression formed from X and Y using the connectives of sentence
logic are sentences of predicate logic. Finally, only these expressions are sentences of predicate logic.
Logicians often use the words Well Formed Formula (Abbreviated wff) for any expression which this definition classifies as a
predicate logic sentence.
You may have noticed something a little strange about the definition. It tells us that an expression such as '(Vx)Ba' is a predicate
logic sentence. If 'A' is a sentence letter, even '(Vx)A' is going to count as a sentence! But how should we understand '(Vx)Ba' and
'(Vx)A'? Since the variable 'x' of the quantifier does not occur in the rest of the sentence, it is not clear what these sentences are
supposed to mean.
To have a satisfying definition of predicate logic sentence, one might want to rule out expressions such as '(Vx)Ba' and '(Vx)A'. But
it will turn out that keeping these as official predicate logic sentences will do no harm, and ruling them out in the definition makes
the definition messier. It is just not worth the effort to rule them out. In the next chapter we will give a more exact characterization
of how to understand the quantifiers, and this characterization will tell us that "vacuous quantifiers," as in '(Vx)Ba' and '(Vx)A',
have no effect at all. These sentences can be understood as the sentences 'Ba' and 'A', exactly as if the quantifiers were not there.
The definition also counts sentences such as 'By', 'Lze', and 'Bx & Lxe' as sentences, where 'x' and 'z' are variables not governed by
a quantifier. Such sentences are called Open Sentences. Open sentences can be a problem in logic in the same way that English
sentences are a problem when they contain "open" pronouns. You fail to communicate if you say, 'He has a funny nose,' without
saying or otherwise indicating who "he" is. Many logicians prefer not to count open sentences as real sentences at all. Where I use
the expression 'open sentence', often logicians talk about 'open formulas' or 'propositional functions'. If you go on in your study of
logic, you will quickly get used to these alternative expressions, but in an introductory course I prefer to keep the terminology as
simple as possible.
Have you been wondering what the word 'syntax' means in the title of this chapter? The Syntax of a language is the set of rules
which tell you what counts as a sentence of the language. You now know what constitutes a sentence of predicate logic, and you
have a rough and ready idea of how to understand such a sentence. Our next job will be to make the interpretation of these
sentences precise. We call this giving the Semantics for predicate logic, which will be the subject of the next chapter. But, first, you
should practice what you have learned about the syntax of predicate logic to make sure that your understanding is secure.
 Exercise
1-3. Which of the following expressions are sentences of predicate logic?

a) Ca
b) Tab
C) aTb
d) Ca ⊃ Tab
e) (Ǝx)~Cx
f) (Vx)(Cx ⊃ Tax)
g) (Vx)Cx & Tax(Vx)
h) ~(Vx)(Txa v Tax)
i) [(Ǝx)Cx v(Ǝx)~Cx] ≡ (Vx)(Txa & Tax)
In the following exercises, use this transcription guide:
a: Adam
e: Eve
c: Cid
Bx: x is blond
Cx: x is a cat
Lxy: x loves y
Txy: x is taller than y
Before you begin, I should point out something about transcribing between logic and pronouns in English. I used the analogy
to English pronouns to help explain the idea of a variable. But that does not mean that you should always transcribe variables
as pronouns or that you should always transcribe pronouns as variables. For example, you should transcribe 'If Eve is a cat,
then she loves herself.' with the predicate logic sentence 'Ce ⊃ Lee'. Notice that 'she' and 'herself are both transcribed as 'e'.
That is because in this case we have been told who she and herself are. We know that they are Eve, and so we use the name for
Eve, namely, 'e' to transcribe these pronouns. How should we describe 'Ca ⊃ ~Ba'? We could transcribe this as 'If Adam is a
cat then Adam is not blond.' But a nicer transcription is simply 'If Adam is a cat then he is not blond.'
Now do your best with the following transcriptions.
1-4. Transcribe the following predicate logic sentences into English:
a) ~Laa
b) Laa ⊃ ~Taa
c) ~(Bc v Lce)
d) Ca ≡ (Ba v Lae)
e) (Ǝx)Txc
f) (Vx)Lax & (Vx)Lcx
g) (Vx)(Lax & Lcx)
h) (Ǝx)Txa v (Ǝx)Txc
i) (Ǝx)(Txa v Txc)
j) (Vx)(Cx ⊃ Lxe)
k) (Ǝx)(Cx & ~Lex)
l) ~(Vx)(Cx ⊃ Lex)
m) (Vx)[Cx ⊃ (Lcx v Lex)]
n) (Ǝx)[Cx & (Bx & Txc)]
1-5. Transcribe'the following English sentences into sentences of predicate logic:
a) Everyone loves Eve.
b) Everyone is loved by either Cid or Adam.
c) Either everyone is loved by Adam or everyone is loved by Cid.
d) Someone is taller than both Adam and Cid.
e) Someone is taller than Adam and someone is taller than Cid.
f) Eve loves all cats.
g) AU cats love Eve.
h) Eve loves some cats.
i) Eve loves no cats.
j) Anyone who loves Eve is not a cat.
k) No one who loves Eve is a cat.
l) Somebody who loves Adam loves Cid.
m) No one loves both Adam and Cid.
 Chapter summary Exercise
Provide short explanations for each of the following. Check against the text to make sure that your explanations are correct,
and keep your explanations in your notebook for reference and review.
a) Predicate Logic
b) Name
c) Predicate
d) One Place Predicate
e) Two Place Predicate
f) Relation
g) Variable
h) Universal Quantifier
i) Existential Quantifier
j) Universe, or Domain of Discourse
k) Govern
l) Bind
m) Open Sentence
n) Sentence of Predicate Logic
0) Well Formed Formula (wff)
p) syntax
q) Semantics
3.1.3: The Sentences of Predicate Logic is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
3.2: Predicate Logic - Semantics and Validity


3.2: Predicate Logic - Semantics and Validity is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Recall that we used truth tables to give very precise definitions of the meaning of '&', 'v' '~', '⊃', and '≡'. We would like to do the
same for the meaning of quantifiers. But, as you will see very soon, truth tables won't do the job. We need something more
complicated.
When we were doing sentence logic, our atomic sentences were just sentence letters. By specifying truth values for all the sentence
letters with which we started, we already fixed the truth values of any sentence which we could build up from these smallest pieces.
Now that we are doing predicate logic, things are not so easy. Suppose we are thinking about all the sentences which we could
build up using the one place predicate 'B', . the two place predicate 'L', the name 'a', and the name 'e'. We can form six atomic
sentences from these ingredients: 'Ba', 'Be', 'Laa', 'Lae', 'Lea', and 'Lee'. The truth table formed with these six atomic sentences
would have 64 lines. Neither you nor I are going to write out a 64-line truth table, so let's consider just one quite typical line from
the truth table:
Figure 2-1
Ba, Be, Laa, Lae, Lea, Lee
tfftftáé
Even such an elementary case in predicate logic begins to get quite complicated, so I have introduced a pictorial device to help in
thinking about such cases (see Figure 2-1). I have drawn a box with two dots inside, one labeled 'a' and the other labeled 'e'. This
box is very different from a Venn diagram. This box is supposed to picture just one way the whole world might be. In this very
simple picture of the world, there are just two things, Adam and Eve. The line of the truth table on the left gives you a completed
description of what is true and what is false about Adam and Eve in this very simple world: Adam is blond, Eve is not blond, Adam
does not love himself, Adam does love Eve, Eve does not love Adam, and Eve does love herself.
You can also think of the box and the description on the left as a very short novel. The box gives you the list of characters, and the
truth table line on the left tells you what happens in this novel. Of course, the novel is not true. But if the novel were true, if it
described the whole world, we would have a simple world with just Adam and Eve having the properties and relations described on
the left.
Now, in writing this novel, I only specified the truth value for atomic sentences formed from the one and two place predicates and
from the two names. What about the truth value of more complicated sentences? We can use our old rules for figuring out the truth
value of compounds formed from these atomic sentences using '&', 'v', '~', '⊃', and '≡'. For example, in this novel 'Ba & Lae' is true
because both the components are true.
What about the truth value of '(Ǝx)Bx'? Intuitively, '(Ǝx)Bx' should be true in the novel because in the novel there is someone,
namely Adam, who is blond. As another example, consider '(Ǝx)Lxa'. In this novel '(Ǝx)Lxa' is false because Eve does not love
Adam and Adam does not love Adam. And in this novel there isn't anyone (or anything) else. So no one loves Adam. In other
words, in this novel it is false that there is someone who loves Adam.
Let's move on and consider the sentence '(Vx)Lxe'. In our novel this sentence is true, because Adam loves Eve, and Eve loves
herself, and that's all the people there are in this novel. If this novel were true, it would be true that everyone loves Eve. Finally,
'(Vx)Bx' is false in the novel, for in this novel Eve is not blond. So in this novel it is false that everyone is blond.
Remember what we had set out to do: We wanted to give a precise account of the meaning of the quantifiers very like the precise
account which truth table definitions gave to '&' and the other sentence logic connectives. In sentence logic we did this by giving
precise rules which told us when a compound sentence is true, given the truth value of the compound's components.
We now have really done the same thing for '(Vx)' and '(Ǝx)' in one special case. For a line of a truth table (a "novel") that gives a
truth value for all atomic sentences using 'B', 'L', 'a', and 'e', we can say whether a universally quantified or an existentially
quantified sentence is true or false. For example, the universally quantified sentence '(Vx)Lxe' is true . just in case 'Lxe' is true for
all values of 'x' in the novel. At the moment we are considering a novel in which the only existing things are Adam and Eve. In
such a novel '(Vx)Lxe' is true if both 'Lxe' is true when we take 'x' to refer to Adam and 'Lxe' is also true when we take 'x' to refer
to Eve. Similarly, '(Ǝx)Bx' is true in such a novel just in case 'Bx' is true for some value of 'x' in the novel. As long as we continue
to restrict attention to a novel with only Adam and Eve as characters, '(Ǝx)Bx' is true in the novel if either 'Bx' is true when we take
'x' to refer to Adam or 'Bx' is true if we take 'x' to refer to Eve.
If the example seems a bit complicated, try to focus on this thought: All we are really doing is following the intuitive meaning of
"all x" and "some x" in application to our little example. If you got lost in the previous paragraph, go back over it with this thought
in mind.
Now comes a new twist, which might not seem very significant, but which will make predicate logic more interesting (and much
more complicated) than sentence logic. In sentence logic we always had truth tables with a finite number of lines. Starting with a
fixed stock of atomic sentence letters, we could always, at least in principle, write out all possible cases to consider, all possible
assignments of truth values to sentence letters. The list might be too long to write out in practice, but we could at least understand
everything in terms of such a finite list of cases.
Can we do the same thing when we build up sentences with predicates and names? If, for example, we start with just 'B', 'L', 'a', and
'e', we can form six atomic sentences. We can write out a 64-line truth table which will give us the truth value for any compound
built up from these six atomic sentences, for any assignment of truth values to the atomic sentences. But the fact that we are using
quantifiers means that we must also consider further possibilities.
Consider the sentence '(Vx)Bx'. We know this is false in the one case we used as an example (in which 'Ba' is true and 'Be' is false).
You will immediately think of three alternative cases (three alternative "novels") which must be added to our list of relevant
possible cases: the case in which Eve is blond and Adam is not, the case in which Adam and Eve are both blond, and the case in
which both are not blond. But there are still more cases which we must include in our list of all possible cases! I can generate more
cases by writing new novels with more characters. Suppose 1 write a new novel with Adam, Eve, and Cid. I now have eight
possible ways of distributing hair color (blond or not blond) among my characters, which can be combined with 512 different
possible combinations of who does or does not love whom! And, of course, this is just the beginning of an unending list of novels
describing possible cases in which '(Vx)Bx' will have a truth value. 1 can always expand my list of novels by adding new
characters. I can even describe novels with infinitely many characters, although I would not be able to write such a novel down.
How are we going to manage all this? In sentence logic we always had, for a given list of atomic sentence, a finite list of possible
cases, the finite number of lines of the corresponding truth table. Now we have infinitely many possible cases. We can't list them
all, but we can still say what any one of these possible cases looks like. Logicians call a possible case for a sentence of predicate
logic an Interpretation of the sentence. The example with which we started this chapter is an example of an interpretation, so
actually you have already seen and understood an example of an interpretation. We need only say more generally what
interpretations are.
We give an interpretation, first, by specifying a collection of objects which the interpretation will be about, called the Domain of
the interpretation. A domain always has at least one object. Then we give names to the objects in the domain, to help us in talking
about them. Next, we must say which predicates will be involved. Finally, we must go through the predicates and objects and say
which predicates are true of which objects. If we are concerned with a one place predicate, the interpretation specifies a list of
objects of which the object is true. If the predicate is a two place predicate, then the interpretation specifies a list of pairs of objects
between which the two place relation is supposed to hold, that is, pairs of objects of which the two place relation is true. Of course,
order is important. The pair a-followed-by-b counts as a different pair from the pair b-followed-by-a. Also, we must consider
objects paired with themselves. For example, we must specify whether Adam loves himself or does not love himself. The
interpretation deals similarly with three and more place predicates.
In practice, we often specify the domain of an interpretation simply by giving the interpretation's names for those objects. I should
mention that in a fully developed predicate logic, logicians consider interpretations which have unnamed objects. In more advanced
work, interpretations of this kind become very important. But domains with unnamed objects would make it more difficult to
introduce basic ideas and would gain us nothing for the work we will do in part I of this volume. So we won't consider
interpretations with unnamed objects until part 11.
The following gives a summary and formal definition of an interpretation:
Definition: Interpretation
An Interpretation consists of
a. A collection of objects, called the interpretation's Domain. The domain always has at least one object.
b. A name for each object in the domain. An object may have just one name or more than one name. (In part I1 we will expand the
definition to allow domains with unnamed objects.)
c. A list of predicates.
d. A specification of the objects of which each predicate is true and the objects of which each predicate is false-that is, which one
place predicates apply to which individual objects, which two place predicates apply to which pairs of objects, and so on. In this
way every atomic sentence formed from predicates and names gets a truth value.
e. An interpretation may also include atomic sentence letters. The interpretation specifies a truth value for any included atomic
sentence letter.
By an Interpretation of a Sentence, we mean an interpretation which is sure to have enough information to determine whether or
not the sentence is true or false in the interpretation.
Definition: Interpretation of a Sentence
An Interpretation of a Sentence is an interpretation which includes all the names and predicates which occur in the sentence and
includes truth values for any atomic sentence letters which occur in the sentence.
For example, the interpretation of Figure 2-1 is an interpretation of 'Ba' and of '(Vx)Lxx'. In this interpretation 'Ba' is true and
'(Vx)Lxx' is false. Note that for each of these sentences, the interpretation contains more information than is needed to determine
whether the sentence is true or false. This same interpretation is not an interpretation of 'Bc' or of '(Ǝx)Txe'. This is because the
interpretation does not include the name 'c' or the two place predicate 'T', and so can't tell us whether sentences which use these
terms are true or false.
 Exercise
2-1. I am going to ask you to give an interpretation for some sentences. You should use the following format. Suppose you are
describing an interpretation with a domain of three objects named 'a', 'b', and 'c'. Specify the domain in this way: D ≡ {a,b,c}.
That is. specify the domain by giving a list of the names of the objects in the domain. Then specify what is true about the
objects in the domain by using a sentence of predicate logic. Simply conjoin all the atomic and negated atomic sentences which
say which predicates are true of which objects and which are false. Here is an example. The following is an interpretation of
the sentence 'Tb & Kbd':
D ≡ {b,d); Tb & Td & Kbb & Kbd & Kdb & Kdd.
In this interpretation all objects have property T and everything stands in the relation K to itself and to everything else. Here is
another interpretation of the same sentence:
D ≡ (b,d); -Tb & Td & Kbb & -Kbd & -Kdb & Kdd.
2-2. Sometimes students have trouble understanding what I want in this exercise. They ask, How am I supposed to decide
which interpretation to write down? You can write down any interpretation you want as long as it is an interpretation of the
sentence I give you. In every case you have infinitely many interpretations to choose from because you can always get more
interpretations by throwing in more objects and then saying what is true for the new objects. Choose any you like. Just make
sure you are writing down an interpretation of the sentence I give you.
a. Lab
b. Lab ⊃ Ta
c. Lab v ~Lba
d. (Vx)(Fx≡Rxb)
e. Ga & (Ǝx)(Lxb v Rax)
f. (Kx & (Vx)Rax) ⊃ (Ǝx)(Mx v Rcx)
3.2.1: Interpretations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Just like a line of a truth table, an interpretation tells us whether each atomic sentence formed from predicates and names is true or
false. What about compound sentences? If the main connective of a compound sentence does not involve a quantifier, we simply
use the old rules for the connectives of sentence logic. We have only one more piece of work to complete: We must make more
exact our informal description of the conditions under which a quantified sentence is true or is false in an interpretation.
Intuitively, a universally quantified sentence is going to be true in an interpretation if it is true in the interpretation for everything to
which the variable could refer in the interpretation. (Logicians say, "For every value of the universally quantified variable.") An
existentially quantified sentence will be true in an interpretation if it is true for something to which the variable could refer in the
interpretation (that is, "for some value of the existentially quantified variable.") What we still need to do is to make precise what it
is for a quantified sentence to be true for a value of a variable. Let's illustrate with the same example we have been using, the
interpretation given in figure 2-1.
Consider the sentence '(Vx)Bx'. In the interpretation we are considering, there are exactly two objects, a, and e. '(Vx)Bx' will be
true in the interpretation just in case, roughly speaking, it is true both for the case of 'x' referring to a and the case of 'x' referring to
e. But when 'x' refers to a, we have the sentence 'Ba'. And when 'x' refers to 'e', we have the sentence 'Be'. Thus '(Vx)Bx' is true in
this interpretation just in case both 'Ba' and 'Be' are true. We call 'Ba' the Substitution Instance of '(Vx)Bx' formed by substituting 'a'
for 'x'. Likewise, we call 'Be' the substitution instance of '(Vx)Bxl formed by substituting 'e' for 'x'. Our strategy is to explain the
meaning of universal quantification by defining this notion of substitution - instance and then specifying that a universally
quantified sentence is true in an interpretation just in case it is true for all substitution instances in the interpretation:
Definition (Incomplete): the Substitution Instance
For any universally quantified sentence (Vu)(. . . u . . .), the Substitution Instance of the sentence with the name s substituted for
the variable u is (. . . s . . .), the sentence formed by dropping the initial universal quantifier and writing s wherever u had occurred.
A word of warning: This definition is not yet quite right. It works only as long as we don't have multiple quantification, that is, as
long as we don't have sentences which stack one quantifier on top of other quantifiers. But until chapter 3 we are going to keep
things simple and consider only simple sentences which do not have one quantifier applying to a sentence with another quantifier
inside. When we have the basic concepts we will come back and give a definition which is completely general. Now we can easily
use this definition of substitution instance to characterize truth of a universally quantified sentence in an interpretation:
Definition: (Incomplete Definition)
A universally quantified sentence is tw in an interpretation just in case all of the sentence's substitution instances, formed with
names in the interpretation, are true in the interpretation.
Another word of warning: As with the definition of substitution instance, this definition is not quite right. Again, Chapter 3 will
straighten out the details.
To practice, let's see whether '(Vx)(Bx ⊃ Lxe)' is true in the interpretation of figure 2-1. First we form the substitution instances
with the names of the interpretation, 'a', and 'e'. We get the first substitution instance by dropping the quantifier and writing in 'a'
everywhere we see 'x'. This gives
Ba ⊃ Lae. (3.2.2.1)
Note that because 'Ba' and 'Lae' are both true in the interpretation, this first substitution instance is true in the interpretation. Next
we form the second substitution instance by dropping the quantifier and writing in 'e' wherever we see 'x':
Be ⊃ Lee. (3.2.2.2)
Because 'Be' is false and 'Lee' is true in the interpretation, the conditional 'Be ⊃ Lee' is true in the interpretation. We see that all the
substitution instances of '(Vx)(Bx ⊃ Lxe)' are true in the interpretation. So this universally quantified sentence is true in the
interpretation.
To illustrate further our condition for truth of a universally quantified sentence, consider the sentence '(Vx)(Bx ⊃ Lxa)'. This has
the substitution instance 'Ba ⊃ Laa'. In this interpretation 'Ba' is true and 'Laa' is false, so 'Ba 3 Laa' is false in the interpretation.
Because '(Vx)(Bx ⊃ Lxa)' has a false substitution instance in the interpretation, it is false in the interpretation.
You may have noticed the following fact about the truth of a universally quantified sentence and the truth of its substitution
instances. By definition '(Vx)(Bx ⊃ Lxe)' is true in the interpretation just in case all of its instances are true in the interpretation.
But its instances are all true just in case the conjunction of the instances is true. That is, '(Vx)(Bx ⊃ Lxe)' is true in the
interpretation just in case the conjunction
(Ba ⊃ Lae) & (Be ⊃ Lee) (3.2.2.3)
is true in the interpretation. If you think about it, you will see that this will hold in general. In the interpretation we have been
discussing (or any interpretation with two objects named 'a' and 'e'), any universally quantified sentence, '(Vx)(. . . x . . .)', will be
true just in case the conjunction of its substitution instance, '(. . . a . . .)&(. . . e . . .)', is true in the interpretation.
It's looking like we can make conjunctions do the same work that the universal quantifier does. A universally quantified sentence is
true in an interpretation just in case the conjunction of all its substitution instances is true in the interpretation. Why, then, do we
need the universal quantifier at all?
To answer this question, ask yourself what happens when we shift to a new interpretation with fewer or more things in its domain.
In the new interpretation, what conjunction will have the same truth value as a given universally quantified sentence? If the new
interpretation has a larger domain, our conjunction will have more conjuncts. If the new interpretation has a smaller domain, our
conjunction will have fewer conjuncts. In other words, when we are looking for a conjunction of instances to give us the truth value
of a universally quantified sentence, the conjunction will change from interpretation to interpretation. You can see in this way that
the universal quantifier really does add something new. It acts rather like a variable conjunction sign. It has the effect of forming a
long conjunction, with one conjunct for each of the objects in an interpretation's domain. If an interpretation's domain has infinitely
many objects, a universally quantified sentence has the effect of an infinitely long conjunction!
What about existentially quantified sentences? All the work is really done. We repeat everything we said for universal
quantification, replacing the word 'all' with 'some':
(Incomplete) Definition: Substitution Instance
For any existentially quantified sentence (Ǝ)(. . . u. . .), the Substitution Instance of the sentence, with the name s substituted for the
variable u is (. . . s . . .), the sentence formed by dropping the initial - existential quantifier and writing s wherever u had occurred.
(incomplete) Definition: existentially quantified sentence
An existentially quantified sentence is true in an interpretation just in case some (i.e., one or more) of the sentence's substitution
instances, formed with names in the interpretation, are true in the interpretation.
As with the parallel definitions for universally quantified sentences, these definitions will have to be refined when we get to chapter
3.
To illustrate, let's see whether the sentence '(Ǝx)(Bx & Lxe)' is true in the interpretation of figure 2-1. We will need the sentence's
substitution instances. We drop the quantifier and write in 'a' wherever we see 'x', giving 'Ba & Lae', the instance with 'a' substituted
for 'x'. In the same way, we form the instance with 'e' substituted for 'x', namely, 'Be & Lee'. '(Ǝx)(Bx & Lxe)' is true in the
interpretation just in case one or more of its substitution instances are true in the interpretation. Because 'Ba' and 'Lae' are true in
the interpretation, the first instance, 'Ba & Lae', is true, and so '(Ǝx)(Bx & Lxe)' is true.
Have you noticed that, just as we have a connection between universal quantification and conjunction, we have the same
connection between existential quantification and disjunction: '(Ǝx)(Bx & Lxe)' is true in our interpretation just in case one or
more of its instances are true. But one or more of its instances are true just in case their disjunction
(Ba & Lae) v (Be & Lee)
is true. In a longer or shorter interpretation we will have the same thing with a longer or shorter disjunction. Ask yourself, when is
an existentially quantified sentence true in an interpretation? It is true just in case the disjunction of all its substitution instances in
that interpretation is true in the interpretation. Just as the universal quantifier acted like a variable conjunction sign, the existential
quantifier acts like a variable disjunction sign. In an interpretation with an infinite domain, an existentially quantified sentence even
has the effect of an infinite disjunction.
I hope that by now you have a pretty good idea of how to determine whether a quantified sentence is true or false in an
interpretation. In understanding this you also come to understand everything there is to know about the meaning of the quantifiers.
Remember that we explained the meaning of the sentence logic connectives '~', '&', 'v', '⊃', and '≡' by giving their truth table
definitions. For example, explaining how to determine whether or not a conjunction is true in a line of a truth table tells you
everything there is to know about the meaning of '&'. In the same way, our characterization of truth of a quantified sentence in an
interpretation does the same kind of work in explaining the meaning of the quantifiers.
This point about the meaning of the quantifiers illustrates a more general fact. By a "case" in sentence logic we mean a line of a
truth table, that is, an assignment of truth values to sentence letters. The interpretations of predicate logic generalize this idea of a
case. Keep in mind that interpretations do the same kind of work in predicate logic that assignments of truth values to sentence
letters do in sentence logic, and you will easily extend what you already know to understand validity, logical truth, contradictions,
and other concepts in predicate logic.
By now you have also seen how to determine the truth value which an interpretation gives to any sentence, not just to quantified
sentences. An interpretation itself tells you which atomic sentences are true and which are false. You can then use the rules of
valuation for sentence logic connectives together with our two new rules for the truth of universally and existentially quantified
sentences to determine the truth of any compound sentence in terms of the truth of shorter sentences. Multiple quantification still
calls for some refinements, but in outline you have the basic ideas.
 Exercise
2-2. Consider the interpretation.

D ≡ {a,b}; ~Ba & Bb & Laa & ~Lab & Lba & ~Lbb.
For each of the following sentences, give all of the sentence's substitution instances in this interpretation, and for each
substitution instance say whether the instance is true or false in the interpretation. For example, for the sentence '(Vx)Bx', your
answer should look like this:
GIVEN SENTENCE SUBSTITUTION INSTANCES
(Vx)Bx Ba, false in the interpretation
Bb, true
a) (Ǝx)Bx b) (Ǝx)~Lxa c) (Vx)Lxa
d) (Ǝx)Lbx e) (V)(Bx v Lax) f) (Ǝx)(Lxa & Lbx)
g) (Vx)(Bx ⊃ Lbx) h) (Ǝx)[(Lbx &Bb) vBx] i) (Ǝx)[Bx ⊃ (Lxx ⊃ Lxa)]
j) (Vx)[(Bx v Lax) ⊃ (Lxb v~Bx)] k) (Ǝx)[(Lax & Lxa) ≡ (Bx v Lxb)]
2-3. For each of the sentences in exercise 2-2, say whether the sentence is true or false in the interpretation of exercise 2-2.
2-4. For each of the following sentences, determine whether the sentence is true or false in the interpretation of exercise 2-2. In
this exercise, you must carefully determine the main connective of a sentence before applying the rules to determine its truth in
an interpretation. Remember that a quantifier is a connective which applies to the shortest full sentence which follows it.
Remember that the main connective of a sentence is the last connective that gets used in building the sentence up from its
parts. To determine whether a sentence is true in an interpretation, first determine the sentence's main connective. If the
connective is '&', 'v', '~', '⊃', or '≡', you must first determine the truth value of the components, and then apply the rules for the
main connective (a conjunction is true just in case both conjuncts are true, and so on). If the main connective is a quantifier,
you have to determine the truth value of the substitution instances and then apply the rule for the quantifier, just as you did in
the last exercise.
a) (Ǝx)Lxx ⊃ (Vx)(Bx v Lbx)
b) ~(Ǝx)(Lxx ⊃ Bx) & (Vx)(Bx ⊃ Lxx)
c) (Ǝx)[Bx ≡ (Lax v Lxb)]
d) (Ǝx)(Lxb v Bx) ⊃ (Lab v ~Ba)
e) ~(Vx)(~Lxx v Lxb) ⊃ (Lab v ~Lba)
f) (Ǝx)[(Lbx v Bx) ⊃ (Lxb & -Bx)]
g) (Vx)~[(~Lxx ≡ Bx) ⊃ (Lax ≡ Lxa)]
h) (Vx)(Lax v Lxb) v (3x)(Lax v Lxb)
i) (Ǝx)[Lxx & (Bx Ǝ Laa)] & (Ǝx)~(Lab ≡ Lxx)
j) (Vx){[Bx v (Lax & ~Lxb)] ⊃ (Bx ⊃ Lxx)}
2-5. In the past exercises we have given interpretations by explicitly listing objects in the domain and explicitly saying which
predicates apply to which things. We can also describe an interpretation in more general terms. For example, consider the
interpretation given by
i) Domain: All U.S. citizens over the age of 21.
ii) Names: Each person in the domain is named by 'a' subscripted by his or her social security number.
iii) Predicates: Mx: x is a millionaire.
Hx: x is happy.
(That is, a one place predicate 'Mx' which holds of someone just in case that person is a millionaire and a one place
predicate 'Hx' which holds of someone just in case that person is happy.)
a) Determine the truth value of the following sentences in this interpretation. In each case explain your answer. Since you can't
write out all the substitution instances, you will have to give an informal general statement to explain your answer, using
general facts you know about being a millionaire, being happy, and the connection (or lack of connection) between these.
a1) (Ǝx)Mx a2) (Vx)Hx a3) (Vx)(Hx Mx) a4) (Ǝx)(Mx & ~Hx)
a5) (Vx)[(Mx ⊃ Hx) & (~Mx ⊃ ~Hx)]
a6) (Ǝx)[(Hx & Mx) v (~Hx & ~Mx)]
a7) (Ǝx)(Mx & Hx) & (Ǝx)(Mx & ~Hx)
a8) (Vx)(Hx ⊃ Mx) ⊃ ~(Ǝx)Mx
i) Domain: All integers, 1, 2, 3, 4, . . .
ii) Names: Each integer is named by 'a' subscripted by that integer's numeral. For example, 17 is named by 'a17'.
iii) Predicates: Ox: x is odd.
Kxy: x is equal to or greater than y.
b) Again, figure out the truth value of the following sentences, and explain informally how you arrived at your answer.
b1) (Ǝx)Ox b2) (Vx)~Ox b3) (Ǝx)(Ox & Kxx)
b4) (Vx)Kxa17 b5) (Vx)(Ox v ~Ox)
b6) (Ǝx)(Ox & Kxa17)
b7) (Vx)[Ox ≡ (~Kxa18 & Kxa17)]
b8) (Ǝx)(Kxa17 ⊃ Kxa18) & (Vx)(~Kxa17 v Kxa18)
b9) (Vx)(Ox ⊃ Kxa17) & (Vx)(~Ox ⊃ ~ Kxa17)
3.2.2: Truth in an Interpretation is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
In sentence logic, we said that an argument is valid if and only if, for all possible cases in which all the premises are true, the
conclusion is true also. In predicate logic, the intuitive notion of validity remains the same. We change things only by generalizing
the notion of possible case. Where before we meant that all lines in the truth table which made all premises true also make the
conclusion true, now we mean that all interpretations which make all the premises true also make the conclusion true:
An argument expressed with sentences in predicate logic is valid if and only if the conclusion is true in every interpretation
in which all the premises are true.
You may remember that we got started on predicate logic at the beginning of chapter 1 because we had two arguments which
seemed valid but which sentence logic characterized as invalid. To test whether predicate logic is doing the job it is supposed to do,
let us see whether predicate logic gives us the right answer for these arguments;
Everyone loves Eve. (Vx)Lxe
Adam loves Eve. Lae
Suppose we have an interpretation in which '(Vx)Lxe' is true. Will 'Lae' have to be true in this interpretation also? Notice that 'Lae'
is a substitution instance of '(Vx)Lxe'. A universally quantified sentence is true in an interpretation just in case all its substitution
instances are true in the interpretation. So in any interpretation in which '(Vx)Lxe' is true, the instance 'Lae' will be true also. And
this is just what we mean by the argument being valid.
Let's examine the other argument:
Adam loves Eve. Lae
Someone loves Eve. (Ǝx)Lxe
Suppose we have an interpretation in which 'Lae' is true. Does '(Ǝx)Lxe' have to be true in this interpretation? Notice that 'Lae' is an
instance of '(Ǝx)Lxe'. We know that '(Ǝx)Lxe' is true in an interpretation if even one of its instances is true in the interpretation.
Thus, if 'Lae' is true in an interpretation, '(Ǝx)Lxe' will also be true in that interpretation. Once again, the argument is valid.
Along with validity, all our ideas about counterexamples carry over from sentence logic. When we talked about the validity of a
sentence logic argument, we first defined it in this way: An argument is valid just in case any line of the truth table which makes all
the premises true makes the conclusion true also. Then we reexpressed this by saying: An argument is valid just in case it has no
counterexamples; that is, no lines of the truth table make all the premises true and the conclusion false. For predicate logic, all the
ideas are the same. The only thing that has changed is that we now talk about interpretations where before we talked about lines of
the truth table:
A Counterexample to a predicate logic argument is an interpretation in which the premises are all true and the conclusion is
false.
A predicate logic argument is Valid if and only if it has no counterexamples.
Let's illustrate the idea of counterexamples in examining the validity of
Lae
(Ǝx)Lxe
Is there a counterexample to this argument? A counterexample would be an interpretation with 'Lae' true and '(Ǝx)Lxe' false. But
there can be no such interpretation. 'he' is an instance of '(Ǝx)Lxe', and '(Ǝx)Lxe' is true in an interpretation if even one of its
instances is true in the interpretation. Thus, if 'he' is true in an interpretation, '(Ǝx)Lxe' will also be true in that interpretation. In
other words, there can be no interpretation in ' which 'Lae' is true and '(3x)Lxe' is false, which is to say that the argument has no
counterexamples. And that is just another way of saying that the argument is valid.
For comparison, contrast the last case with the argument
(Ǝx)Bx
Ba
It's easy to construct a counterexample to this argument. Any case in which someone other than Adam is blond and Adam is not
blond will do the trick. So an interpretation with Adam and Eve in the domain and in which Eve is blond and Adam is not blond
gives us a counterexample, showing the argument to be invalid.
This chapter has been hard work.. But your sweat will be repaid. The concepts of interpretation, substitution instance, and truth in
an interpretation provide the essential concepts you need to understand quantification. In particular, once you understand these
concepts, you will find proof techniques for predicate logic to be relatively easy.
 Exercise
2-6. For each of the following arguments, determine whether the argument is valid or invalid. If it is invalid, show this by
giving a counterexample. If it is valid, explain your reasoning which shows it to be valid. Use the kind of informal reasoning
which I used in discussing the arguments in this section.
You may find it hard to do these problems because I haven't given you any very specific strategies for figuring out whether an
argument is valid. But don't give up! If you can't do one argument, try another first. Try to think of some specific, simple
interpretation of the sentences in an argument, and ask yourseIf-"Are the premise and conclusion both true in that
interpretation?' Can I change the interpretation so as to make the premise true and the conclusion false? If you succeed in doing
that, you will have worked the problem because you will have constructed a counterexample and shown the argument to be
invalid. If you can't seem to be able to construct a counterexample, try to understand why you can't. If you can see why you
can't and put this into words, you will have succeeded in showing that the argument is valid. Even if you might not succeed in
working many of these problems, playing around in this way with interpretations, truth in interpretations, and counterexamples
will strengthen your grasp of these concepts and make the next chapter easier.
a) (Vx)Lxe b) Lae c) (Ǝx)Lxe
(Ǝx)Lxe (Vx)Lxe Lae
d) (Vx)(Bx & Lxe) e) (Vx)(Bx ⊃ Lxe)
(VX)BX (Ǝx)Bx
f) (Ǝx)Bx & (Ǝx)Lxa g) (Vx)(Bx ⊃ LxeL) & (Vx)(~Bx ⊃ Lxa)
(Ǝx)(Bx & Lxa) (Vx)[(Bx ⊃ Lxe) & (~Bx ⊃ Lxa)]
 Chapter SUmmary Exercise
Provide short explanations for each of the following, checking against the text to make sure you understand each term clearly
and saving your answers in your notebook for reference and review.
a) Interpretation
b) Interpretation of a Sentence
c) Substitution Instance
d) Truth in an Interpretation
e) Validity of a Predicate Logic Argument
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SECTION OVERVIEW
3.3: More about Quantifiers


3.3: More about Quantifiers is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
3.1: Some Examples of Multiple Quantification
All of the following are sentences of predicate logic:
(1) (Vx)(Vy)Lxy
(2) (Ǝx)(Ǝy)Lxy
(3) (Ǝx)(Vy)Lxy
(4) (Ǝx)(Vy)Lyx
(5) (Vx)(Ǝy)Lxy
(6) (Vx)(Ǝy)Lyx
Let's suppose that 'L' stands for the relation of loving. What do these sentences mean?
Sentence (1) says that everybody loves everybody (including themselves). (2) says that somebody loves somebody. (The somebody
can be oneself or someone else.) Sentences (3) to (6) are a little more tricky. (3) says that there is one person who is such that he or
she loves everyone. (There is one person who is such that, for all persons, the first loves the second-think of God as an example.)
We get (4) from (3) by reversing the order of the 'x' and 'y' as arguments of 'L'. As a result, (4) says that there is one person who is
loved by everyone. Notice what a big difference the order of the 'x' and 'y' makes.
Next, (5) says that everyone loves someone: Every person is such that there is one person such that the first loves the second. In a
world in . which (5) is true, each person has an object of their affection. Finally we get (6) out of (5) by again reversing the order of
'x' and 'y'. As a result, (6) says that everyone is loved by someone or other. In a world in which (6) is true no one goes unloved. But
(6) says something significantly weaker than (3). (3) say that there is one person who loves everyone. (6) says that each person gets
loved, but Adam might be loved by one person, Eve by another, and so on.
Can we say still other things by further switching around the order of the quantifiers and arguments in sentences (3) to (6)? For
example, switching the order of the quantifiers in (6) gives
(7) (Ǝy)(Vx)Lyx
Strictly speaking, (7) is a new sentence, but it does not say anything new because it is logically equivalent to (3). It is important to
see why this is so:
These diagrams will help you to see that (7) and (3) say exactly the same thing. The point is that there is nothing special about the
variable 'x' or the variable 'y'. Either one can do the job of the other. What matters is the pattern of quantifiers and variables. These
diagrams show that the pattern is the same. All that counts is that the variable marked at position 1 in the existential quantifier is
tied to, or, in logicians' terminology, Binds the variable at position 3; and the variable at position 2 in the universal quantifier binds
the variable at position 4. Indeed, we could do without the variables altogether and indicate what we want with the third diagram.
This diagram gives the pattern of variable binding which (7) and (3) share.
3.1: Some Examples of Multiple Quantification is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
In the last example we saw that the variable at 3 is bound by the quantifier at 1 and the variable at 4 is bound by the quantifier at 2.
This case contrasts with that of a variable which is not bound by any quantifier, for example
In (8), the occurrence of 'x' at 3 is bound by the quantifier at 2. However, the occurrence of 'x' at 1 is not bound by any quantifier.
Logicians say that the occurrence of 'x' at 1 is Free. In (9), the occurrence of 'x' at 3 is free because the quantifier at 1 binds only
variables in the shortest full sentence which follows it. Logicians call the shortest full sentence following a quantifier the
quantifier's Scope. In (9), the 'x' at 3 is not in the scope of the quantifier at 1. Consequently, the quantifier does not bind 'x' at 3.
All the important ideas of this section have now been presented. We need these ideas to understand clearly how to apply the
methods of derivations and truth trees when quantifiers get stacked on top of each other. All we need do to complete the job is to
give the ideas an exact statement and make sure you know how to apply them in more complicated situations.
Everything can be stated in terms of the simple idea of scope. A quantifier is a connective. We use a quantifier to build longer
sentences out of shorter ones. In building up sentences, a quantifier works just like the negation sign: It apples to the shortest full
sentence which follows it. This shortest full following sentence is the quantifier's scope:
Definition: scope
The Scope of a quantifier is the shortest full sentence which follows it. Everything inside this shortest full following sentence is
said to be in the scope of the quantifier.
We can now define 'bound' and 'free' in terms of scope:
A variable, u, is Bound just in case it occurs in the scope of a quantifier, (Vu) or (Ǝu).
A variable, u, is Free just in case it is not bound; that is, just in case it does not occur in the scope of any quantifier, (Vu) or
(Ǝu).
Clearly, a variable gets bound only by using a quantifier expressed with the same variable. 'x' can never be bound by quantifiers
such as '(Vy)' or '(Ǝz)'.
Occasionally, students ask about the variables that occur within the quantifiers-the 'x' in '(Ǝx)' and in '(Vx)'. Are they bound? Are
they free? The answer to this question is merely a matter of convention on which nothing important turns. I think the most sensible
thing to say is that the variable within a quantifier is part of the quantifier symbol and so does not count as either bound or free.
Only variables outside a quantifier can be either bound or free. Some logicians prefer to deal with this question by defining the
scope of a quantifier to include the quantifier itself as well as the shortest full sentence which follows it. On this convention one
would say that a variable within a quantifier always binds itself.
These definitions leave one fine point unclear. What happens if the variable u is in the scope of two quantifiers that use u? For
example, consider
(10) (Ǝx)[(Vx)Lxa ⊃ Lxb]
1 234
The occurrence of 'x' at 3 is in the scope of both the 'x' quantifiers. Which quantifier binds 'x' at 3?
To get straight about this, think through how we build (10) up from atomic constituents. We start with the atomic sentences 'Lxa'
and 'Lxb'. Because atomic sentences have no quantifiers, 'x' is free in both of these atomic sentences. Next we apply '(Vx)' to 'Lxa',
forming '(Vx)Lxa', which we use as the antecedent in the conditional
(11) (Vx)Lxa ⊃ Lxb
23 4
In (11), the occurrence of 'x' at 3 is bound by the quantifier at 2. The occurrence of 'x' at 4 is free in (11).
Finally, we can clearly describe the effect of '(Ǝx)' when we apply it to (11). '(Ǝx)' binds just the free occurrences of 'x' in (11). The
occurrence at 4 is free and so gets bound by the new quantifier. The occurrence at 3 is already bound, so the new quantifier can't
touch it. The following diagram describes n the overall effect:
First, the occurrence at 3 is bound by the quantifier at 2. Then the occurrence at 4 is bound by the quantifier at 1. The job being
done by the 2-3 link is completely independent of the job being done by the 1-4 link.
Let's give a general statement to the facts we have uncovered:
A quantifier (Vu) or (Ǝu) binds all and only all free occurrences of u in its scope. Such a quantifier does not bind an
occurrence of u in its scope which is already bound by some other quantifier in its scope.
We can make any given case even clearer by using different variables where we have occurrences of a variable bound by different
quantifiers. So, for example, (10) is equivalent to
In (12), there can be no confusion about which quantifier binds which variable-we keep track of everything by using different
variables. Why, then, didn't we just resolve to use different variables from the beginning and save ourselves a lot of trouble? We
could have done that, but then the definition of the sentences of predicate logic would have been much more complicated. Either
way, we have work to do. Besides, the formulation I have presented here is the one traditionally used by logicians and so the one
you will need to know if you study more logic.
Let's look at another, slightly more complicated, example to make sure you have put this all together. Draw in the lines which show
which quantifier binds which variable in the following:
(13) (Ǝx)[(Ǝx)(Bx v Lxa) ⊃ (Bx & Lxb)]
If you are having trouble, think through how (13) gets built up from its parts. In
the quantifier at 2 applies only to the shortest full sentence which follows it, which ends before the '⊃'. So the occurrences of 'x' at
3 and 4 are both bound by the quantifier at 2. The two occurrences of 'x' at 5 and 6 are not in the scope of a quantifier and are both
free. So when we apply the second '(Ǝx)' to all of (14), the new '(Ǝx)' binds only the 'x's which are still free in (14), namely, the 'x's
which occur at 5 and 6. In sum, the pattern of binding is
We can make this pattern even clearer by writing the sentence equivalent to (13):
(15) (Ǝx)[(Ǝz)(Bz v Lza) ⊃ (Bx & Lxb)]
In practice, of course, it is much better to use sentences such as (15) and (12) instead of the equivalent (13.) and (lo), which are
more difficult to interpret.
 Exercise
3-1. In the following sentences draw link lines to show which quantifiers bind which variables and say which occurrences of
the variables are bound and which are free:
a) L z z b) (Vy)(Vz)L z y c) (Vz)(Bz ⊃ L x z)
1 21 2 12 3
d) (Ǝx)[Lxz & (Vy)(Lxy v Lzx)]
12 34 56
e) (Vx)(Lyx ⊃ (Bx ⊃ (Ǝx)Lyx)]
12 3 45
3.3.2: Quantifier Scope, Bound Variables, and Free Variables is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
In chapter 2 I gave an incorrect definition of 'substitution instance.' I said that we get the substitution instance of (Vu) ( . . . u . . .)
with s substituted for u by simply dropping the initial (u) and writing in s wherever we find u in (. . . u . . .). This is correct as long
as neither a second (Vu) nor a (Ǝu) occurs within the scope of the initial (Vu), that is, within the sentence ( . . . u . . .). Since I used
only this kind of simple sentence in chapter 2, there we could get away with the simple but incorrect defintion. But now we must
correct our definition so that it will apply to any sentence. Before reading on, can you see how multiple quantification can make
trouble for the simple definition of substitution instance, and can you see how to state the definition correctly?
To correct the definition of substitution instance, all we have to do is to add the qualification that the substituted occurrences of the
variable be free:
For any universally quantified sentence (Vu) ( . . . u . . .), the Substitution Instance of the sentence, with the name s
substituted for the variable u, is (. . . s . . .), the sentence formed by dropping the initial universal quantifier and writing s for
all free occurrences of u in (. . . u . . .).
For any existentially quantified sentence (Ǝu) (. . . u . . .), the Substitution Instance of the sentence, with the name s
substituted for the variable u, is (. . . s . . .), the sentence formed by dropping the initial existential quantifier and writing s for
all free occurrences of u in (. . . u . . .).
For example, look back at (13). Its substitution instance with 'c' substituted for 'x' is
(16) (Ǝx)(Bx v Lxa) ⊃ (Bc & Lcb)
23456
The occurrences of 'x' at 3 and 4 are not free in the sentence which results from (13) by dropping the initial quantifier. So we don't
substitute 'c' for 'x' at 3 and 4. We substitute 'c' only at the free occurrences, which were at 5 and 6.
Can you see why, when we form substitution instances, we pay attention only to the occurrences which are free after dropping the
outermost quantifier? The occurrences at 3 and 4, bound by the '(Ǝx)' quantifier at 2, have nothing to do with the outermost
quantification. When forming substitution instances of a quantified sentence, we are concerned only with the outermost quantifier
and the occurrences which it binds.
To help make this clear, once again consider (15), which is equivalent to (13). In (15), we have no temptation to substitute 'c' for '2'
when forming the 'c'-substitution instance for the sentence at a whole. (15) says that there is some x such that so on and so forth
about x. In making this true . for some specific x, say c, we do not touch the occurrences of '2'. The internal 'z'-quantified sentence
is just part of the so on and so forth which is asserted about x in the quantified form of the sentence, that is, in (15). So the internal
'z'-quantified sentence is just part of the so on and so forth which is asserted about c in the substitution instance of the sentence.
Finally, (13) says exactly what (15) says. So we treat (13) in the same way.
Now let's straighten out the definition of truth of a sentence in an interpretation. Can you guess what the problem is with our old
definition? I'll give you a clue. Try to determine the truth value of 'Lxe' in the interpretation of figure 2-1. You can't do it! Nothing
in our definition of an interpretation gives us a truth value for an atomic sentence with a free variable. An interpretation only gives
truth values for atomic sentences which use no variables. You will have just as much trouble trying to determine the truth value of
'(Vx)Lxy' in any interpretation. A substitution instance of '(Vx)Lxy' will still have the free variable 'y', and no interpretation will
assign such a substitution instance a truth value.
Two technical terms (mentioned in passing in chapter 1) will help us in talking about our new problem:
A sentence with one or more free variables is called an Open Sentence.
A sentence which is not open (i.e., a sentence with no free variables) is called a Closed Sentence
In a nutshell, our problem is that our definitions of truth in an interpretation do not specify truth values for open sentences. Some
logicians deal with this problem by treating all free variables in an open sentence as if they were universally quantified. Others do
what I will do here: We simply say that open sentences have no truth value.
If you think about it, this is really very natural What, anyway, is the truth value of the English "sentence" 'He is blond.', when
nothing has been said or done to give you even a clue as to who 'he' refers to? In such a situation you can't assign any truth value to
'He is blond.' 'He is blond.' functions syntactically as a sentence-it has the form of a sentence. But there is still something very
problematic about it. In predicate logic we allow such open sentences to function syntactically as sentences. Doing this is very
useful in making clear how longer sentences get built up from shorter ones. But open sentences never get assigned a truth value,
and in this way they fail to be full-fledged sentences of predicate logic.
It may seem that I am dealing with the problem of no truth value for open sentences by simply ignoring the problem. In fact, as
long as we acknowledge up-front that this is what we are doing, saying that open sentences have no truth value is a completely
adequate way to proceed.
We have only one small detail to take care of. As I stated the definitions of truth of quantified sentences in an interpretation, the
definitions were ,said to apply to any quantified sentences. But they apply only to closed sentences. So we must write in this
restriction:
A universally quantified closed sentence is true in an interpretation just in case all of the sentence's substitution instances,
formed with names in the interpretation, are true in the interpretation.
An existentially quantified closed sentence is true in an interpretation just in case some (i.e., one or more) of the sentence's
substitution instances, formed with names in the interpretation, are true in the interpretation.
These two definitions, together with the rules of valuation given in chapters 1 and 4 of volume I for the sentence logic connectives,
specify a truth value for any closed sentence in any of our interpretations.
You may remember that in chapter 1 in volume I we agreed that sentences of logic would always be true or false. Sticking by that
agreement now means stipulating that only the closed sentences of predicate logic are real sentences. As I mentioned in chapter 1 in
this volume, some logicians use the phrase Formulas, or Propositional Functions for predicate logic open sentences, to make the
distinction clear. I prefer to stick with the word 'sentence' for both open and closed sentences, both to keep terminology to a
minimum and to help us keep in mind how longer (open and closed) sentences get built up from shorter (open and closed)
sentences. But you must keep in mind that only the closed sentences are fullfledged sentences with truth values.
 Exercise
3-2. Write a substitution instance using 'a' for each of the following sentences:
a) (Vy)(Ǝx)Lxy b) (Ǝz)[(Vx)Bx v Bz]
c) (Ǝx)[Bx ≡ (Vx)(Lax v Bx)]
d) (Vy){(Ǝx)(Bx v [(Ǝu)By ⊃ Bx)} & (Vx)(By ⊃ Bx)]
e) (Vy){(Ǝx)Bx v [(Ǝy)By ⊃ Lyy]}
f) (Vy)(Ǝx)[(Rxy ⊃ Dy) ⊃ Ryx]
g) Wx)(Vy)Wz){[Sxy v (Hz ⊃ Lxz)] ≡ (Scx & Hy))
h) (Ǝx)(Vz){(Pxa ⊃ Kz) & (Ǝy)[(Pxy v Kc) & Pxx])
i) (Ǝz)(Vy){[(Ǝx)Mzx v (Ǝx)(Mxy ⊃ Myz)] & (Ǝx)Mzx)
j) (Vx){[(Vx)Rxa ⊃ Rxb] v [(Ǝx)(Rcx v Rxa) ⊃ Rxx])
3-3. If u does not occur free in X, the quantifiers (Vu) and (Ǝu) are said to occur Vacuously in (Vu)X and (Ǝu)X. Vacuous
quantifiers have no effect. Let's restrict our attention to the special case in which X is closed, so that it has a truth value in any
of its interpretations. The problem I want to raise is how to apply the definitions for interpreting quantifiers to vacuously
occurring quantifiers. Because truth of a quantified sentence is defined in terms of substitution instances of (Vu)X and (Ǝu)X,
when u does not occur free in X, we most naturally treat this vacuous case by saying that X counts as a substitution instance of
(Vu)(X) and (Ǝu)(X). (If you look back at my definitions of 'substitution instance', you will see that they really say this if by
'for all free occurrences of u' you understand 'for no occurrences of u' when u does not occur free in X at all. In any case, this is
the way you should understand these definitions when u does not occur free in X.) With this understanding, show that (Vu)X,
(Ǝu)X, and X all have the same truth value in any interpretation of X.
3-4. a) As I have defined interpretation, every object in an interpretation has a name. Explain why this chapter's definitions of
truth of existentially and universally quantified sentences would not work as intended if interpretations were allowed to have
unnamed objects.
b) Explain why one might want to consider interpretations with unnamed objects. In part I1 we will consider interpretations
with unnamed objects and revise the definitions of truth of quantified sentences accordingly.
3.3.3: Correct Definitions of Substitution Instance and Truth in an Interpretation is shared under a not declared license and was authored, remixed,
and/or curated by LibreTexts.
The idea of logical equivalence transfers from sentence logic to predicate logic in the obvious way. In sentence logic two sentences
are logically equivalent if and only if in all possible cases the sentences have the same truth value, where a possible case is just a
line of the truth table for the sentence, that is, an assignment of truth values to sentence letters. All we have to do is to redescribe
possible cases as interpretations:
Two closed predicate logic sentences are Logically Equivalent if and only if in each of their interpretations the two sentences
are either both true or both false.
Notice that I have stated the definition only for closed sentences. Indeed, the definition would not make any sense for open
sentences because open sentences don't have truth values in interpretations. Nonetheless, one can extend the idea of logical
equivalence to apply to open sentences. That's a good thing, because otherwise the law of substitution of logical equivalents would
break down in predicate logic. We won't be making much use of these further ideas in this book, so I won't hold things up with the
details. But you might amuse yourself by trying to extend the definition of logical equivalence to open sentences in a way which
will make the law of substitution of logical equivalents work in just the way you would expect.
Let us immediately take note of two equivalences which will prove very useful later on. By way of example, consider the sentence,
'No one loves Eve', which we transcribe as '~(Ǝx)Lxe', that is, as 'It is not the case that someone loves Eve'. How could this
unromantic situation arise? Only if everyone didn't love Eve. In fact, saying '~(Ǝx)Lxe' comes to the same thing as saying
'(Vx)~Lxe'. If there is not a single person who does love Eve, then it has to be that everyone does not love Eve. And conversely, if
positively everyone does not love Eve, then not even one person does love Eve.
There is nothing special about the example I have chosen. If our sentence is of the form ~(Ǝu)(. . . u . . .), this says that there is not
a single u such that so on and so forth about u. But this comes to the same as saying about each and every u that so on and so forth
is not true about u, that is, that (Vu)~(. . . u . . .).
We can easily prove the equivalence of ~(Ǝu)(. . . u . . .) and (Vu)~(. . . u . . .) by appealing to De Morgan's laws. We have to prove
that these two sentences have the same truth value in each and every interpretation. In any one interpretation, ~(Ǝu)(. . . u . . .) is
true just in case the negation of the disjunction of the instances
~[(. . . a . . .) v (. . . b . . .) v (. . . c . . .) v . . .]
is true in the interpretation, where we have included in the disjunction all the instances formed using names which name things in
the interpretation. By De Morgan's laws, this is equivalent to the conjunction of the negation of the instances
~(. . . a . . .) & ~(. . . b . . .) & ~(. . . c . . .) & . . .
which is true in the interpretation just in case (Vu)~(. . . u . . .) is true in the interpretation. Because this is true in all interpretations,
we see that
Rule ~Ǝ: ~(Ǝu)(. . . u . . .) is logically equivalent to (Vu)~(. . . u . . .).
Now consider the sentence 'Not everyone loves Eve,' which we transcribe as '~(Vx)Lxe'. If not everyone loves Eve, then there must
be someone who does not love Eve. And if there is someone who does not love Eve, then not everyone loves Eve. So '~(Vx)Lxe' is
logically equivalent to '(Ǝx)~Lxe'.
Pretty clearly, again there is nothing special about the example. ~(Vu)(. . . u . . .) is logically equivalent to (Ǝu)~(. . . u . . .). If it is
not the case that, for all u, so on and so forth about u, then there must be some u such that not so on and so forth about u. And,
conversely, if there is some u such that not so on and so forth about u, then it is not the case that for all u, so on and so forth about
u. In summary
Rule ~V: ~(Vu)(. . . u . . .) is logically equivalent to (Ǝu)~(. . . u . . .).
You can easily give a proof of this rule by imitating the proof of the rule ~Ǝ. But I will let you write out the new proof as an
exercise.
 Exercise
3-5. a) Give a proof of the rule of logical equivalence, ~V. Your proof will be very similar to the proof given in the text for the
rule ~Ǝ.
b) The proof for the rule ~Ǝ is flawed! It assumes that all interpretations have finitely many things in their domain. But not all
interpretations are finite in this way. (Exercise 2-5 gives an example of an infinite interpretation.) The problem is that the proof
tries to talk about the disjunction of all the substitution instances of a quantified sentence. But if an interpretation is infinite,
there are infinitely many substitution instances, and no sentence can be infinitely long. Since I instructed you, in part (a) of this
problem, to imitate the proof in the text, probably your proof has the same problem as mine.
Your task is to correct this mistake in the proofs. Give informal arguments for the rules ~Ǝ and ~V which take account of the
fact that some interpretations have infinitely many things in their domain.
3-6. In the text I defined logical equivalence for closed sentences of predicate logic. However, this definition is not broad
enough to enable us to state a sensible law of substitution of logical equivalents for predicate logic. Let me explain the problem
with an example. The following two sentences are logically equivalent:
(1) ~(Vx)(Vy)Lxy
(2) (Ǝx)(Ǝy)~Lxy
But we cannot prove that (1) and (2) are logically equivalent with the rule ~V as I have stated it. Here is the difficulty. The rule
~V tells us that (1) is logically equivalent to
(3) (Ǝx)~(Vy)Lxy
What we would like to say is that ~(Vy)Lxy is logically equivalent to (Ǝy)~Lxy, again by the rule ~V. But the rule ~V does not
license this because I have defined logical equivalence only for closed sentences and '~(Vy)Lxy' and '(Ǝy)~Lxy' are open
sentences. (Strictly speaking, I should have restricted the ~V and ~Ǝ rules to closed sentences. I didn't because I anticipated the
results of this exercise.) Since open sentences are never true or false, the idea of logical equivalence for open sentences does
not make any sense, at least not on the basis of the definitions I have so far introduced.
Here is your task:
a) Extend the definition of logical equivalence for predicate logic sentences so that it applies to open as well as closed
sentences. Do this in such a way that the law of substitution of logical equivalents will be correct when one open sentence is
substituted for another when the two open sentences are logically equivalent according to your extended definition.
b) Show that the law of substitution of logical equivalents works when used with open sentences which are logically equivalent
according to your extended definition.
Here are this chapter's important terms. Check your understanding by writing short explanations for each, saving your results
in your notebook for reference and review.
a) Bound Variables
b) Free Variables
c) Scope
d) Closed Sentence
e) Open Sentence
f) Truth of a Sentence in an Interpretation
g) Rule ~Ǝ
h) Rule ~V
3.3.4: Some Logical Equivalences is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
3.4: Transcription

3.4: Transcription is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
For three chapters now I have been merrily transcribing '(Ǝx)' both as 'something' and 'someone', and I have been transcribing '(Vx)'
both as 'everything' and 'everyone.' I justified this by saying that when we talked only about people we would restrict the variables
'x', 'y', etc. to refer only to people, and when we talked about everything, we would let the variables be unrestricted. It is actually
very easy to make precise this idea of restricting the universe of discourse. If we want the universe of discourse to be restricted to
people, we simply declare that all the objects in our interpretations must be people. If we want a universe of discourse consisting
only of cats, we declare that all the objects in our interpretations must be cats. And so on.
As I mentioned, this common practice is not fully satisfactory. What if we want to talk about people and things, as when we assert,
'Everyone likes sweet things.'? Restricted quantifiers will help us out here. They also have the advantage of getting what we need
explicitly stated in the predicate logic sentences themselves.
We could proceed by using '(Ǝx)' and '(Vx)' to mean 'something' and 'everything' and introduce new quantifiers for 'someone' and
'everyone'. To see how to do this, let's use the predicate 'P' to stand for 'is a person.' Then we can introduce the new quantifier
'(Ǝx)p' to stand for some x chosen from among the things that are P, that is, chosen from among people. We call this a restricted
quantifier. You should think of a restricted quantifier as saying exactly what an unrestricted quantifier says except that the variable
is restricted to the things of which the subscripted predicate is true. With 'P' standing for 'is a person', '(Ǝx)p' has the effect of
'someone' or 'somebody'. We can play the same game with the universal quantifier. '(Vx)p' will mean all x chosen from among the
things that are P. With 'P' standing for 'is a person', ('Vx)p' means, not absolutely everything, but all people, that is, everyone or
everybody or anyone or anybody.
This notion of a restricted quantifier can be useful for other things. Suppose we want to transcribe 'somewhere' and 'everywhere' or
'sometimes' and 'always'. Let's use 'N' stand for 'is a place' or 'is a location'. 'Somewhere' means 'at some place' or 'at some location'.
So we can transcribe 'somewhere' as '(Ǝx)N' and'everywhere' as '(Vx)N'. For example, to transcribe 'There is water everywhere', I
would introduce the predicate 'Wx' to stand for 'there is water at x'. Then '(Vx)NWx' says that there is water everywhere.
Continuing the same strategy, let's use 'Q' to stand for 'is a time'. Then '(Ǝx)Q' stands for 'sometime(s)' and '(Vx)Q' stands for
'always' ('at all times').
In fact, we can also use the same trick when English has no special word corresponding to the restricted quantifier. Suppose I want
to say something about all cats, for example, that all cats are furry. Let 'Cx' stand for 'x is a cat' and 'Fx' stand for 'x is furry'. Then
'(Vx)CFx' says that all things which are cats are furry; that is, all cats are furry. Suppose I want to say that some animals have tails.
Using 'Ax' for 'x is an animal' and 'Txy' for 'x is a tail of y', I write '(Ǝx)A(Ǝy)Tyx': There is an animal, x, and there is a thing, y,
such that y is a tail of x.
As you will see, restricted quantifiers are very useful in figuring out transcriptions, but there is a disadvantage in introducing them
as a new kind of quantifier in our system of logic. If we have many different kinds of quantifiers, we will have to specify a new rule
for each of them to tell us the conditions under which a sentence formed with the quantifier is true. And when we get to checking
the validity of arguments, we will have . to have a new rule of inference to deal with each new quantifier. We could state the
resulting mass of new rules in a systematic way. But the whole business would still require a lot more work. Fortunately, we don't
have to do any of that, for we can get the full effect of restricted quantifiers with the tools we already have.
Let's see how to rewrite subscripted quantifiers. Consider the restricted quantifier '(Ǝx)C' which says that there is cat such that, or
there are cats such that, or some cats are such that. We say 'some cats are furry' (or 'there is a furry cat' or the like) with '(Ǝx)CFx'.
Now what has to be true for it to be true that some cats are furry, or that there is a furry cat? There has to be one or more things that
is both a cat and is furry. If there is not something which is both a cat and is furry, it is false that there is a furry cat. So we can say
that some cats are furry by writing '(Ǝx)(Cx & Fx)'. In short, we can faithfully rewrite '(Ǝx)CFx' as '(Ǝx)(Cx & Fx)'. This strategy
will work generally:
Rule for rewriting Subscripted Existential Quantifiers: For any predicate S, any sentence of the form (Ǝu)s(. . . u. . .) is
shorthand for (Ǝu)[Su & (. . . u . . .)I.
Here are some examples:
Some cats are blond. (Ǝx)CBx (Ǝx)(Cx & Bx)
Eve loves a cat. (Ǝx)CLex (Ǝx)(Cx & Lex)
Eve loves a furry cat. (Ǝx)C(Fx & Lex) (Ǝx)[Cx & (Fx & Lex)]
Clearly, we can proceed in the same way with 'someone' and 'somebody':
Someone loves Eve. (Ǝx)pLxe (Ǝx)(Px & Lxe)
Somebody loves Eve or Adam. (Ǝx)p(Lxe v Lxa) (Ǝx)[Px &(Lxe v Lxa)]
If somebody loves Eve, then Eve loves somebody.
(Ǝx)pLxe ⊃ (Ǝx)(Px & Lxe) ⊃ (Ǝx)(Px & Lex)
(Ǝx)pLex
Notice that in the last example I used the rule for rewriting the subscript on each of two sentences X and Y, inside a compound
sentence, X ⊃ Y.
How should we proceed with restricted universal quantifiers? This is a little tricky. Let's work on '(Vx)CFx'-that is, 'All cats are
furry'. Under what conditions is this sentence true? To answer the question, imagine that everything in the world is lined up in front
of you: All the cats, dogs, people, stones, basketballs, everything. You go down the line and examine the items, one by one, to
determine whether all cats are furry. If the first thing in line is a dog, you don't have to determine whether or not it is furry. If the
second thing is a basketball, you don't have to worry about it either. But as soon as you come to a cat you must examine it further to
find out if it is furry. When you finally come to the end of the line, you will have established that all cats are furry if you have
found of each thing that, if it is a cat, then it is furry. In short, to say that all cats are furry is to say '(Vx)(Cx ⊃ Fx)'.
At this point, many students balk. Why, they want to know, should we rewrite a restricted universal quantifier with the '⊃' when we
rewrite a restricted existential quantifier with the '&'? Shouldn't '&' work also for restricted universal quantifiers? Well, I'm sorry. It
doesn't. That is just not what restricted universal quantifiers mean.
You can prove this for yourself by trying to use '&' in rewriting the subscripted 'C' in our transcription of 'All cats are furry.' You get
(1) (Vx)(Cx & Fx)
What does (1) say? It says that everything is a furry cat, and in particular that everything is a cat! That's much too strong. All cats
could be furry even though there are lots of things which are not cats. Thus 'All cats are furry' could be true even when (1) is false,
so that (1) cannot be the right way to rewrite '(Vx)CFx'.
What has gone wrong? The unrestricted universal quantifier applies to everything. So we can use conjunction in expressing the
restriction of cats only if we somehow disallow or except the cases of noncats. We can do this by saying that everything is either
not a cat or is a cat and is furry:
(2) (Vx)[~Cx v (Cx & Fx)]
(2) does indeed say what 'All cats are furry' says. So (2) should satisfy your feeling that an '&' also comes into the restricted
universal quantifier in some way. But you can easily show that (2) is logically equivalent to '(Vx)(Cx ⊃ Fx)'! As the formulation
with the '⊃' is more compact, and is also traditional, it is the one we will use.
In general, we rewrite restricted universal quantifiers according to the rule
Rule for rewriting Subscripted Universal Quantifers: For any predicate S, any sentence of the form (Vu)s(. . . u . . .) is
shorthand for (Vu)[Su ⊃ (. . . u . . .)].
Here are some examples to make sure you see how this rule applies:
Eve loves all cats. (Vx)C(Lex) (Vx)(Cx ⊃ Lex)
Everybody loves Eve. (Vx)PLxe (Vx)(Px ⊃ Lxe)
Everyone loves either Adam or Eve.
(VX)P(Lxa v Lxe) (Vx)[Px ⊃ (Lxa v Lxe)]
Not everyone loves both Adam and Eve.
~(Vx)P(Lxa & Lxe) ~(Vx)[Px ⊃ (Lxa & Lxe)]
In the last example, I used the rewriting rule on a sentence, X, inside a negated sentence of the form ~X.
If you are still feeling doubtful about using the '⊃' to rewrite restricted universal quantifiers, I have yet another way to show you
that this way of rewriting must be right. I am assuming that you agree that our way of rewriting restricted existential quantifiers is
right. And I will use a new rule of logical equivalence. This rule tells us that the same equivalences that hold for negated
unrestricted universal quantifiers hold for negated restricted universal quantifiers. In particular, saying that not all cats are furry is
clearly the same as saying that some cat is not furry. In general
Rule ~VS: A sentence of the form ~(Vu)S(. . . u . . .) is logically equivalent to (Ǝu)S(. . . u . . .).
You can prove this new rule along the same lines we used in proving the rule ~V.
Now, watch the following chain logical equivalents:
1 (VU)S(. . . u . . .)
2 ~~(VU)S(. . . u . . .) DN
3 ~(Ǝ~)S~(. . . u . . .) ~VS
4 ~(Ǝu)[Su & ~(. . . u . . .)] Rule for rewriting (Ǝu)S
5 ~(Ǝu)~~[Su & ~(. . . u . . .)] DN
6 ~(Ǝu)~[~Su v (. . . u . . .)] DM, DN
7 ~(Ǝu)~[Su ⊃ (. . . u . . .)] C
8 ~~(Vu)[Su ⊃ (. . . u . . .)] ~Ǝ
9 (Vu)[Su ⊃ (. . . u . . .)] DN
Since the last line is logically equivalent to the first, it must be a correct way of rewriting the first.
If you are having a hard time following this chain of equivalents, let me explain the strategy. Starting with a restricted universal
quantifier, I turn it into a restricted existential quantifier in lines 2 and 3 by using double denial and pushing one of the two
negation signs through the restricted quantifier. I then get line 4 by using the rule we have agreed on for rewriting restricted
existential quantifiers. Notice that 1. am applying this rule inside a negated sentence, so that here (and below) I am really using
substitution of logical equivalents. In lines 5, 6, and 7 I use rules of logical equivalence to transform a conjunction into a
conditional. These steps are pure sentence logic. They involve no quantifiers. Line 8 comes from line 7 by pushing the negation
sign back out through what is now an unrestricted existential quantifier, changing it into an unrestricted universal quantifier.
Finally, in line 9, I drop the double negation. It's almost like magic!
 Exercise
4-1. Give an argument which shows that the rule ~VS is correct. Similarly, show that
Rule ~ƎX: a sentence of the form ~(Ǝu)S(. . . u . . .) is logically equivalent to (VU)S~(. . . u . . .).
is also correct.
4-2. Use the rule ~ƎS to show that, starting from the rule for rewriting subscripted universal quantifiers, you can derive the rule
for rewriting subscripted existential quantifiers. Your argument will closely follow the one given in the text for arguing the rule
for rewriting subscripted universal quantifiers from the rule for rewriting subscripted existential quantifiers.
4-3. Transcribe the following English sentences into the language of predicate logic. Use this procedure: In a first step,
transcribe into a sentence using one or more subscripted quantifiers. Then rewrite the resulting sentence using the rules for
rewriting subscripted quantifiers. Show both your first and second steps. Here are two examples of sentences to transcribe and
the two sentences to present in presenting the problem:
Someone loves Eve. All cats love Eve.
(Ǝ~)PLxe (Vx)CLxe
(Ǝx)(Px & Lxe) (Vx)(Cx ⊃ Lxe)
Transcription Guide
e: Eve Dx: x is a dog
Px: x is a person Bx: x is blond
Cx: x is a cat Lxy: x loves y
a) Everyone loves Eve.
b) Eve loves somebody.
c) Eve loves everyone.
d) Some cat loves some dog.
e) Somebody is neither a cat nor a dog.
f) Someone blond loves Eve.
g) Some cat is blond.
h) Somebody loves all cats.
i) No cat is a dog.
j) Someone loves someone.
k) Everybody loves everyone.
l) Someone loves everyone.
m) Someone is loved by everyone.
n) Everyone loves someone.
o) Everyone is loved by somebody.
3.4.1: Restricted Quantifiers is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Transcribing into the language of predicate logic can be extremely difficult. Actually, one can do logic perfectly well without
getting very good at transcription. But transcriptions into logic provide one of predicate logic's important uses. This is because,
when it comes to quantification, English is often extremely confusing, ambiguous, and even downright obscure. Often we can
become clearer about what is being said if we put a statement into logic. Sometimes transcribing into logic is a must for clarity and
precision. For example, how do you understand the highly ambiguous sentence, 'All of the boys didn't kiss all of the girls.'? I, for
one, am lost unless I transcribe into logic.
Before we get started, I should mention a general point. Just as in the case of sentence logic, if two predicate logic sentences are
logically equivalent they are both equally good (or equally bad!) transcriptions of an English sentence. Two logically equivalent
sentences share the same truth value in all possible cases (understood as all interpretations), and in this sense two logically
equivalent sentences "say the same thing." But if two predicate logic sentences say the same thing, then to the extent that one of
them says what an English sentence says, then so does the other.
We are going to be looking at quite a few examples, so let's agree on a transcription guide:
Transcription Guide
a: Adam Px: x is a person
J: The lights will be on Rx: x is a registered voter
Ax: x is an adult Vx: x has the right to vote
Bx:x is a boy Kxy: x kissed y
Cx: x is a cat Lxy: x love y
Dx: x is a dog Mxy: x is married to y
Fx: x can run a 3.45 mile Oxy: x owns y
Gx: x is a girl Txy: x is a tail of y
Hx: x is at home
Take note of the fact that in giving you a transcription guide, I have been using open sentences to indicate predicates. For example,
I am using the open sentence 'Px' to indicate the predicate 'is a person.' The idea of using an open sentence to indicate a predicate
will soon become very useful.
To keep us focused on the new ideas, I will often use subscripts on restricted quantifiers. However, you should keep in mind that
complete transcriptions require you to rewrite the subscripts, as explained in the last section.
Now let's go back and start with the basics. '(Vx)(Cx ⊃ Fx)' transcribes 'all cats are furry,' 'Every cat is furry,' 'Any cat is furry,' and
'Each cat is furry.' This indicates that
Usually, the words 'all', 'every', 'any', and 'each' signal a universal quantifier.
Let's make a similar list for the existential quantifier. '(Ǝx)(Cx & Fx)' transcribes 'Some cat is furry', 'Some cats are furry,' 'At least
one cat is furry', 'There is a furry cat,' and 'There are furry cats':
Usually, the expressions 'some', 'at least one', 'there is', and 'there are' signal an existential quantifier.
These lists make a good beginning, but you must use care. There are no hard and fast rules for transcribing English quantifier
words into predicate logic. For starters, 'a' can easily function as a universal or an existential quantifier. For example, 'A car can go
very fast.' is ambiguous. It can be used to say either that any car can go very fast or that some car can go very fast.
To make it clearer that 'a' can function both ways, consider the following examples. You probably understand 'A man is wise.' to
mean that some man is wise. But most likely you understand 'A dog has four legs.' to mean that all dogs have four legs. Actually,
both of these sentences are ambiguous. In both sentences, 'a' can correspond to 'all' or 'some'. You probably didn't notice that fact
because when we hear an ambiguous sentence we tend to notice only one of the possible meanings. If a sentence is obviously true
when understood with one of its meanings and obviously false when understood with the other, we usually hear the sentence only
as making the true statement. So if all the men in the world were wise, we would take 'A man is wise.' to mean that all men are
wise, and if only one dog in the world had four legs we would take 'A dog has four legs.' to mean that some dog has four legs.
It is a little easier to hear 'A car can go very fast.' either way. This is because we interpret this sentence one way or the other,
depending on how fast we take 'fast' to be. If 'fast' means 30 miles an hour (which is very fast by horse and buggy standards), it is
easy to hear 'A car can go very fast.' as meaning that all cars can go very fast. If "fast' means 180 miles an hour it is easy to hear 'a
car can go very fast.' as meaning that some car can go very fast.
'A' is not the only treacherous English quantifier word. 'Anyone' usually gets transcribed with a universal quantifier. But not
always. Consider
(3) If anyone is at home, the lights will be on.
(4) If anyone can run a 3:45 mile, Adam can.
We naturally hear (3), not as saying that if everyone is at home the lights will be on, but as saying that if someone is at home the
lights will be on. So a correct transcription is
(3a) (Ǝx)PHx ⊃ J
Likewise, by (4), we do not ordinarily mean that if everyone can run a 3:43 mile, Adam can. We mean that if someone can run that
fast, Adam can:
(4a) (Ǝx)PFx ⊃ Fa
At least that's what one would ordinarily mean by (4). However, I think that (4) actually is ambiguous. I think 'anyone' in (4) could
be understood as 'everyone'. This becomes more plausible if you change the '3:45 mile' to 'lo-minute mile'. And it becomes still
more plausible after you consider the following example: 'Anyone can tie their own shoe laces. And if anyone can, Adam can.'
Going back to (3), one would think that if (4) is ambiguous, (3) should be ambiguous in the same way. I just can't hear an
ambiguity in (3). Can you?
'Someone' can play the reverse trick on us. Usually, we transcribe it with an existential quantifier. But consider
(5) Someone who is a registered voter has the right to vote.
We naturally hear this as the generalization stating that anyone who is a registered voter has the right to vote. Thus we transcribe it
as
(5a) (Vx)P(Rx ⊃ Vx)
As in the case of (4), which uses 'anyone', we can have ambiguity in sentences such as (5), which uses 'someone'. If you don't
believe me, imagine that you live in a totalitarian state, called Totalitarania. In Totalitarania, everyone is a registered voter. But
voter registration is a sham. In fact, only one person, the boss, has the right to vote. As a citizen of Totalitarania, you can still
truthfully say that someone who is a registered voter (namely, the boss) has the right to vote. (You can make this even clearer by
emphasizing the word 'someone: 'someone who is a registered voter has the right to vote.') In this context we hear the sentence as
saying
(5b) (Ǝx)P(Rx ⊃ Vx)
Ambiguity can plague transcription in all sorts of ways. Consider an example traditional among linguists:
(6) All the boys kissed all the girls
This can easily mean that each and every one of the boys kissed each and every one of the girls:
(6a) (Vx)B(Vy)CKxy
But it can also mean that each of the boys kissed some girls so that, finally, each and every girl got kissed by some boy:
(6b) (Vx)B(Ǝy)GKxY & (Vy)G(Ǝx)BKxY
If you think that was bad, things get much worse when multiple quantifiers get tangled up with negations. Consider
(7) All the boys didn't kiss all the girls.
Everytime I try to think this one through, I blow a circuit. Perhaps the most natural transcription is to take the logical form of the
English at face value and take the sentence to assert that of each and every boy it is true that he did not kiss all the girls; that is, for
each and every boy there is at least one girl not kissed by that boy:
(7a) (Vx)B~(Vy)CKxy, or (Vx)B(Ǝy)C~Kxy
But one can also take the sentence to mean that each and every boy refrained from kissing each and every girl, that is, didn't kiss
the first girl and didn't kiss the second girl and not the third, and so on. In yet other words, this says that for each and every boy
there was no girl whom he kissed, so that nobody kissed anybody:
(7b) (Vx)B~(Vy)C~Kxy, or (Vx)B~(Ǝy)CKxy, or ~(Ǝx)B(Ǝy)CKxy
We are still not done with this example, for one can also use (7) to mean that not all the boys kissed every single girl-that is, that
some boy did not kiss all the girls, in other words that at least one of the boys didn't kiss at least one of the girls:
(7c) ~(Vx)B(Vy)CKxy, or (Ǝx)B~(Vy)CKxy , or (Ǝx)B(Ǝy)C~Kxy
It's worth an aside to indicate how it can happen that an innocent-looking sentence such as (7) can turn out to be so horribly
ambiguous. Modern linguistics postulates that our minds carry around more than one representation of a given sentence. There is
one kind of structure that . represents the logical form of a sentence. Another kind of structure represents sentences as we speak and
write them. Our minds connect these (and other) representations of a given sentence by making all sorts of complicated
transformations. These transformations can turn representations of different logical forms into the same representation of a spoken
or written sentence. Thus one sentence which you speak or write can correspond to two, three, or sometimes quite a few different
structures that carry very different meanings. In particular, the written sentence (7) corresponds to (at least!) three different logical
forms. (7a), (7b), and (7c) don't give all the details of the different, hidden structures that can be transformed into (7). But they do
describe the differences which show up in the language of predicate logic.
You can see hints of all this if you look closely at (7), (7a), (7b), and (7c). In (7) we have two universal quantifier words and a
negation. But since the quantifier words appear on either side of 'kissed', it's really not all that clear where the negation is meant to
go in relation to the universal quantifiers. We must consider three possibilities. We could have the negation between the two
universal quantifiers. Indeed, that is what you see in (7a), in the first of its logically equivalent forms. Or we could have the
negation coming after the two universal quantifiers, which is what you find in the first of the logically equivalent sentences in (7b).
Finally, we could have the negation preceding both universal quantifiers. You see this option in (7c). In sum, we have three similar,
but importantly different, structures. Their logical forms all have two universal quantifiers and a negation, but the three differ, with
the negation coming before, between, or after the two quantifiers. The linguistic transformations in our hinds connect all three of
these structures with the same, highly ambiguous English sentence, (7).
Let's get back to logic and consider some other words which you may find especially difficult to transcribe. I am always getting
mixed up by sentences which use 'only', such as 'Only cats are furry.' So I use the strategy of first transcribing a clear case (it helps
to use a sentence I know is true) and then using the clear case to figure out a formula. I proceed in this way: Transcribe
(8) Only adults can vote.
This means that anyone who is not an adult can't vote, or equivalently (using the law of contraposition), anyone who can vote is an
adult. So either of the following equivalent sentences provides a correct transcription:
(8a) (Vx)P(~Ax ⊃ ~Vx)
(8b) (Vx)P(Vx ⊃ Ax)
This works in general. (In the following I used boldface capital P and Q to stand for arbitrary predicates.) Transcribe
(9) Only Ps are Qs
either as
(9a) (Vx)(~Px ~Qx)
(9b) (Vx)(Qx ⊃ Px)
Thus 'Only cats are furry' becomes (Vx)(Fx ⊃ Cx).'
Nothing' and 'not everything' often confuse me also. We must carefully distinguish
(10) Nothing is furry: (Vx)~Fx, or ~(Ǝx)Fx
and
(11) Not everything is furry: ~(Vx)Fx, or (Ǝx)~Fx
(The alternative transcriptions given in (10) and (11) are logically equivalent, by the rules ~(Vx) and ~(Ǝx) for logical equivalence
introduced in section 3-4.) 'Not everything' can be transcribed literally as 'not all x . . .'. 'Nothing' means something different and
much stronger. 'Nothing' means 'everything is not . . . .' Be careful not to confuse 'nothing' with 'not everything.' If the distinction is
not yet clear, make up some more examples and carefully think them through. 'None' and 'none but' can also cause confusion:
(12) None but adults can vote: (Vx)(~Ax ⊃ ~Vx)
(13) None love Adam: (Vx)~Lxe '
None but' simply transcribes as 'only.' When 'none' without the 'but' fits in grammatically in English you will usually be able to treat
it as you do 'nothing'. 'Nothing' and 'none' differ in that we tend to use 'none' when there has been a stated or implied restriction of
domain: "How many cats does Adam love? He loves none." In this context a really faithful transcription of the sentence 'Adam
loves none.' would be '(Vx)C~Lax', or, rewriting the subscript, '(Vx)(Cx ⊃ ~Lax).
Perhaps the most important negative quantifier expression in English is 'no', as in
(14) No cats are furry.
To say that no cats are furry is to say that absolutely all cats are not furry, so that we transcribe (18) as
(15) (Vx)C~Fx, that is, (Vx)(Cx ⊃ ~Fx)
In general, transcribe
(16) No Ps are Qx.
(17) (Vx)P~Q, that is, (Vx)(P ⊃ ~Q)
 Exercise
4-4. Transcribe the following English sentences into the language of predicate logic. Use subscripts if you find them helpful in
figuring out your answers, but no subscripts should appear in your final answers.
Transcription Guide
a: Adam Fx: x is furry
e: Eve Px: x is a person
Ax: x is an animal Qx: x purrs
Bx: x is a blonde Lxy: x loves y
Cx: x is a cat Sxy: x is son of y
Dx: x is a dog Txy: x is tickling y
a) Anything furry loves Eve.
b) No cat is furry.
c) If anyone loves Adam, Eve does.
d) ve does not love anyone.
e) Nothing is furry.
f) Adam, if anyone, is blond.
g) Not all cats are furry.
h) Some cats are not furry.
i) No one is a cat.
j) No cat is a dog.
k) If something purrs, it is a cat.
l) Not everything blond is a cat.
m) A dog is not an animal. (Ambiguous)
n) Not all animals are dogs.
o) Only cats purr.
p) Not only cats are furry.
q) Any dog is not a cat.
r) No blonds love Adam.
s) None but blonds love Adam.
t) Some dog is not a cat.
u) Nothing furry loves anyone.
v) Only cat lovers love dogs. (Ambiguous?)
w) If someone is a son of Adam, he is blond.
x) No son of Adam is a son of Eve.
y) Someone who is a son of Adam is no son of Eve. (Ambiguous)
z) Each cat which loves Adam also loves Eve.
aa) Not everyone who loves Adam also loves Eve.
bb) Anyone who is tickling Eve is tickling Adam.
cc) None but those who love Adam also love Eve.
4-5. Give alternative transcriptions which show the ways in which the following sentences are ambiguous. In this problem you
do not have to eliminate subscripts. (It is sometimes easier to study the ambiguity if we write these sentences in the compact
subscript notation.)
a) Everyone loves someone.
b) Someone loves everyone.
c) Something is a cat if and only if Adam loves it.
d) All cats are not furry. e) Not anyone loves Adam.
4-6. In this section I discussed ambiguities connected with words such as 'a', 'someone', and 'anyone.' In fact, English has a
great many sorts of ambiguity arising from the ways in which words are connected with each other. For example, 'I won't stay
at home to please you.' can mean that if I stay at home, I won't do it in order to please you. But it can also mean that 1 will go
out because going out will please you. 'Eve asked Adam to stay in the house.' can mean that Eve asked Adam to remain in a
certain location, and that location is the house. It can also mean that Eve asked Adam to remain in some unspecified location,
and that she made her request in the house.
For the following English sentences, provide alternative transcripts showing how the sentences are ambiguous. Use the
transcription guides given for each sentence.
a) Flying planes can be dangerous. (Px: x is a plane. Fx: x is flying. Dx: x can be dangerous. Ax: x is an act of flying a
plane.)
b) All wild animal keepers are blond. (Kxy: x keeps y. Wx: x is wild. Ax: x is an animal. Bx: x is blond.)
c) Adam only relaxes on Sundays. (a: Adam. Rxy: x relaxes on day y. Lxy: x relaxes ("is lazy") all day long on day y.
Sx: x is Sunday.)
d) Eve dressed and walked all the dogs. (e: Eve. Cxy: x dressed y. Dx: x is a dog. Wxy: x walked y.)
Linguists use the expression Structural Ambiguity for the kind of ambiguity in these examples. This is because the ambiguities
have to do with alternative ways in which the grammatical structure of the sentences can be correctly analyzed. Structural
ambiguity contrasts with Lexical Ambiguity, which has to do with the ambiguity in the meaning of isolated words. Thus the
most obvious ambiguity of 'I took my brother's picture yesterday.' turns on the ambiguity of the meaning of 'took' (stole vs.
produced a picture). The ambiguity involved with quantifier words such as 'a', 'someone', and 'anyone' is actually structural
ambiguity, not lexical ambiguity. We can see a hint of this in the fact that '(Ǝx)Hx ⊃ J' is logically equivalent to '(Vx)(Hx ⊃ J)
and the fact that '(Vx)Hx ⊃ J' is logically equivalent to (Ǝx)(Hx ⊃ J), as you will prove later on in the course.
3.4.2: Transcribing from English into Logic is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
I'm going to turn now from particularly hard cases to general strategy. If you are transcribing anything but the shortest of sentences,
don't try to do it all at once. Transcribe parts into logic, writing down things which are part logic and part English. Bit by bit,
transcribe the parts still in English into logic until all of the English is gone.
Let's do an example. Suppose we want to transcribe
(18) Any boy who loves Eve is not a furry cat.
(18) says of any boy who loves Eve that he is not a furry cat; that is, it says of all things, x, of which a first thing is true (that x is a
boy who loves Eve) that a second thing is true (x is not a furry cat). So the sentence has the form (Vx)(Px ⊃ Q):
(18a) (Vx)(x is a boy who loves Eve ⊃ x is not a furry cat)
Now all you have to do is to fashion transcriptions of 'x is a boy who loves Eve' and of 'x is not a furry cat' and plug them into
(18a):
(18b) x is a boy who loves Eve: Bx & Lxe
(18c) x is not a furry cat: ~(Fx & Cx)
(Something which is not a furry cat is not both furry and a cat. Such a thing could be furry, or a cat, but not both.) Now we plug
(18b) and (18c) into (18a), getting our final answer:
(18d) (Vx)[(Bx & Lxe) ⊃ ~(Ǝx & Cx)]
Here is another way you could go about the same problem. Think of the open sentence 'Bx & Lxe' as indicating a complex one
place predicate. The open sentence 'Bx & Lxe' presents something which might be true of an object or person such as Adam. For
example, if the complex predicate is true of Adam, we would express that fact by writing in 'a' for 'x' in 'Bx & Lxe', giving 'Ba &
Lae'. Now, thinking of 'Bx & Lxe' as a predi- ' cate, we can use the method of quantifier subscripts which we discussed in section 4-
1. (18) is somewhat like a sentence which asserts that everything is not a furry cat. But (18) asserts this, not about absolutely
everything, but just about all those things which have the complex property Bx & Lxe. So we can write (18) as a universally
quantified sentence with the universal quantifier restricted by the predicate 'Bx & Lxe':
(18e) (Vx)(Bx & Lxe)~(Fx & Cx)
Now you simply use the rule for rewriting subscripts on universal quantifiers, giving (18d).
In yet a third way of working on (18), you could first use the method of subscripting quantifiers before transcribing the complex
predicates into logic. Following this route you would first write.
(18f) (Vx)(x is a boy who loves Eve)(x is not a furry cat)
Now transcribe the English complex predicates as in (18b) and (lac), plug the results into (18f), giving (18e). Then you rewrite the
subscript, giving (18d) as before. You have many alternative ways of proceeding.
Generally, it is very useful to think of complex descriptions as complex predicates. In particular, this enables us to use two place
predicates to construct one place predicates. We really took advantage of this technique in the last example. 'Lxy' is a two place
predicate. By substituting a name for 'y', we form a one place predicate, for example, 'Lxe'. 'Lxe' is a one place predicate which is
true of anything which loves Eve.
Here is another useful way of constructing one place predicates from two place predicates. Suppose we need the one place
predicate 'is married', but our transcription guide only gives us the two place predicate 'Mxy', meaning that x is married to y. To see
how to proceed, consider what it means to say that Adam, for example, is married. This is to say that there is someone to whom
Adam is married. So we can say Adam is married with the sentence '(Ǝy)May'. We could proceed in the same way . to say that Eve,
or anyone else, is married. In short, the open sentence '(Ǝy)Mxy' expresses the predicate 'x is married'.
Here's another strategy point: When 'who' or 'which' comes after a predicate they generally transcribe as 'and'. As you saw in (la),
the complex predicate 'x is a boy who loves Eve' becomes 'Bx & Lxe'. The complex predicate 'x is a dog which is not furry but has
a tail' becomes 'Dx & (~Fx & (Ǝy)Tyx)'.
When 'who' or 'which' comes after, a quantifier word, they indicate a subscript on the quantifier: 'Anything which is not furry but
has a tail' should be rendered as (VX)(~Fx & (Ǝy)Tyx). When the quantifier word itself calls for a subscript, as does 'someone', you
need to combine both these ideas for treating 'who1: 'Someone who loves Eve' is the subscripted quantifier '(Ǝx)Px & Lxe'.
Let's apply these ideas in another example. Before reading on, see if you can use only 'Cx' for 'x is a cat', 'Lxy' for 'x loves y', and
'Oxy' for 'x owns y' and transcribe
(19) Some cat owner loves everyone who loves themselves.
Let's see how you did. (19) says that there is something, taken from among the cat owners, and that thing loves everyone who loves
themselves. Using a subscript and the predicates 'x is a cat owner' and 'x loves everyone who loves themselves', (19) becomes
(19a) (Ǝx) (x is a cat owner)(x loves everyone who loves themselves)
Now we have to fashion transcriptions for the two complex English predicates used in (19a). Someone (or something) is a cat
owner just in case there is a cat which they own:
(19b) x is a cat owner: (Ǝy)(Cy & Oxy)
To say that x loves everyone who loves themselves is to say that x loves, not absolutely everyone, but everyone taken from among
those that are, first of all people, and second, things which love themselves. So we want to say that x loves all y, where y is
restricted to be a person, Py, and restricted to be a self-lover, Lyy:
(19c) x loves everyone who loves themselves: (Vy)(Py & Lyy)Lxy
Putting the results of (19b) and (19c) into (19a), we get
(19d) (Ǝx)(Ǝy)(cy & Oxy)[(Vy)(Py & Lyy)Lxy]
Discharging first the subscript of '(Ǝx)' with an '&' and then the subscript of '(Vy)' with a '⊃', we get
(19e) (Ǝx)(Ǝy)(Cy & Oxy)[(Vy)(Py & Lyy)Lxy]
(19f) (Ǝx)(Ǝy)(Cy & Oxy)[(Vy)[(Py & Lyy)Lxy]}
This looks like a lot of work, but as you practice, you will find that you can do more and more of this in your head and it will start
to go quite quickly.
I'm going to give you one more piece of advice on transcribing. Suppose you start with an English sentence and you have tried to
transcribe it into logic. In many cases you can catch mistakes by transcribing your logic sentence back into English and comparing
your retranscription with the original sentence. This check works best if you are fairly literal minded in retranscribing. Often the
original and the retranscribed English sentences will be worded differently. But look to see if they still seem to say the same thing.
If not, you have almost certainly made a mistake in transcribing from English into logic.
Here is an illustration. Suppose I have transcribed
(20) If something is a cat, it is not a dog.
as
(20a) (Ǝx)(Cx ⊃ ~Dx)
To check, I transcribe back into English, getting
(20b) There is something such that if it is a cat, then it is not a dog.
Now compare (20b) with (20). To make (20b) true it is enough for there to be one thing which, if a cat, is not a dog. The truth of
(20b) is consistent with there being a million cat-dogs. But (20) is not consistent with there being any cat-dogs. I conclude that
(20a) is a wrong transcription. Having seen that (20) is stronger than (20a), I try
(20c) (Vx)(Cx ⊃ ~Dx)
Transcribing back into English this time gives me
(20d) Everything which is a cat is not a dog.
which does indeed seem to say what (20) says. This time I am confident that I have transcribed correctly.
(Is (20) ambiguous in the same way that (5) was? I don't think so!)
Here is another example. Suppose after some work I transcribe
(21) Cats and dogs have tails.
as
(21a) (Vx)[(Cx & Dx) ⊃ (Ǝy)Txy]
To check, I transcribe back into English:
(21b) Everything is such that if it is both a cat and a dog, then it has a tail.
Obviously, something has gone wrong, for nothing is both a cat and a dog. Clearly, (21) is not supposed to be a generalization
about such imaginary cat-dogs. Having noticed this, I see that (21) is saying one thing about cats and then the same thing about
dogs. Thus, without further work, I try the transcription
(21c) (Vx)(Cx ⊃ (Ǝy)Txy) & (Vx)(Dx ⊃ (Ǝy)Txy)
To check (21c), I again transcribe back into English, getting
(21d) If something is a cat, then it has a tail, and if something is a dog, then it has a tail.
which is just a long-winded way of saying that all cats and dogs have tails-in other words, (21). With this check, 1 can be very
confident that (21c) is a correct transcription.
 Exercise
Use this transcription guide for exercises 4-7 and 4-8:

a: Adam Fx: x is furry
e: Eve Px: x is a person
Ax: x is an animal Qx: x purrs
Bx: x is blond Lxy: x loves y
Cx: x is a cat Sxy: x is a son of y
Dx: x is a dog Txy: x is a tail of y
Oxy: x owns y
4-7. Transcribe the following sentences into English:
a) (Ǝx)(Ǝy)(Px & Py & Sxy)
b) ~(Ǝx)(Px & Ax)
c) ~(Vx)[Qx ⊃ (Fx & Cx)]
d) (Ǝx)[Qx & ~(Fx & Cx)]
e) (Vx)~[Pk & (Lxa & Lxe)]
f) (Vx)[Px ⊃ ~(Lxa & Lxe)]
g) (Vx)(Vy)[(Dx & Cy) ⊃ Lxy]
h) (Vx)(Vy)[Dx ⊃ (Cy ⊃ Lxy)]
i) (Ǝx)[Pk & (Ǝy)(Ǝz)(Py & Szy & Lxz)]
j) (Ǝx)[Px & (Ǝy)(Ǝz)(Pz & Syz & Lxz)]
k) (Vx)([Bx & (Ǝy)(Fy & Txy)] ⊃ (Ǝz)(Cz & Txz))
l) (Vx)(Ǝy)Sxy ⊃ [(Ǝz)(Cz & Lxz) ≡ (Ǝz)(Dz & Lxz)])
4-8. Transcribe the following sentences into predicate logic. I have included some easy problems as a review of previous
sections along with some real brain twisters. I have marked the sentences which seem to me clearly ambiguous, and you should
give different transcriptions for these showing the different ways of understanding the English. Do you think any of the
sentences I haven't marked are also ambiguous? You should have fun arguing out your intuitions about ambiguous cases with
your classmates and instructor.
a) All furry cats purr.
b) Any furry cat purrs.
c) No furry cats purr.
d) None of the furry cats purr.
e) None but the furry cats purr. (Ambiguous?)
f) Some furry cats purr.
g) Some furry cats do not purr.
h) Some cats and dogs love Adam.
i) Except for the furry ones, all cats purr.
j) Not all furry cats purr.
k) If a cat is furry, it purrs. A furry cat purrs. (Ambiguous)
m) Only furry cats purr.
n) Adam is not a dog or a cat.
o) Someone is a son.
p) Some sons are blond.
q) Adam loves a blond cat, and Eve loves one too.
r) Adam loves a blond cat and so does Eve. (Ambiguous)
s) Eve does not love everyone.
t) Some but not all cats are furry.
u) Cats love neither Adam nor Eve.
v) Something furry loves Eve.
w) Only people love someone.
x) Some people have sons.
y) Any son of Adam is a son of Eve.
z) Adam is a son and everybody loves him.
aa) No animal is furry, but some have tails.
bb) Any furry animal has a tail.
cc) No one has a son.
dd) Not everyone has a son.
ee) Some blonds love Eve, some do not.
ff) Adam loves any furry cat.
gg) All blonds who love themselves love Eve.
hh) Eve loves someone who loves themself.
ii) Anyone who loves no cats loves no dogs.
jj) Cats love Eve if they love anyone. (Ambiguous)
kk) If anyone has a son, Eve loves Adam. (Ambiguous)
ll) If anyone has a son, that person loves Adam.
mm) Anyone who has a son loves Eve.
nn) If someone has a son, Adam loves Eve.
oo) If someone has a son, that person loves Adam.
pp) Someone who has a son loves Adam. (Ambiguous)
qq) All the cats with sons, except the furry ones, love Eve.
rr) Anyone who loves a cat loves an animal.
ss) Anyone who loves a person loves no animal.
tt) Adam has a son who is not furry.
uu) If Adam's son has a furry son, so does Adam.
vv) A son of Adam is a son of Eve. (Ambiguous)
ww) If the only people who love Eve are blond, then nobody loves Eve.
xx) No one loves anyone. (Ambiguous)
yy) No one loves someone. (Ambiguous)
zz) Everyone loves no one.
aaa) Everyone doesn't love everyone. (Ambiguous!)
bbb) Nobody loves nobody. (Ambiguous?)
ccc) Except for the furry ones, every animal loves Adam.
ddd) Everyone loves a lover. (Ambiguous)
eee) None but those blonds who love Adam own cats and dogs.
fff) No one who loves no son of Adam loves no son of Eve.
ggg) Only owners of dogs with tails own cats which do not love Adam.
hhh) None of Adam's sons are owners of furry animals with no tails.
iii) Anyone who loves nothing without a tail owns nothing which is loved by an animal.
jjj) Only those who love neither Adam nor Eve are sons of those who own none of the animals without tails.
kkk) Anyone who loves all who Eve loves loves someone who is loved by all who love Eve.
4-9. Transcribe the following sentences into predicate logic, making up your own transcription guide for each sentence. Be sure
to show as much of the logical form as possible.
a) No one likes Professor Snarf.
b) Any dog can hear better than any person.
c) Neither all Republicans nor all Democrats are honest.
d) Some movie stars are better looking than others.
e) None of the students who read A Modem Formal Logic Primer failed the logic course.
f) Only people who eat carrots can see well in the dark.
g) Not only people who eat carrots can see as well as people who eat strawberries.
h) Peter likes all movies except for scary ones.
i) Some large members of the cat family can run faster than any horse.
j) Not all people with red hair are more temperamental than those with blond hair.
k) Some penny on the table was minted before any dime on the table.
l) No pickle tastes better than any strawberry.
m) John is not as tall as anyone on the basketball team.
n) None of the pumpkins at Smith's fruit stand are as large as any of those on MacGreggor's farm.
o) Professors who don't prepare their lectures confuse their students.
p) Professor Snarf either teaches Larry or teaches someone who teaches Larry.
q) Not only logic teachers teach students at Up State U.
r) Anyone who lives in Boston likes clams more than anyone who lives in Denver. (Ambiguous)
s) Except for garage mechanics who fix cars, no one has greasy pants.
t) Only movies shown on channel 32 are older than movies shown on channel 42.
u) No logic text explains logic as well as some professors do.
v) The people who eat, drink, and are merry are more fun than those who neither smile nor laugh.
In reviewing this chapter make a short summary of the following to ensure your grasp of these ideas:
a) Restricted Quantifiers
b) Rule ~VS,
c) Rule ~ƎS,
d)Transcription Guide
e) Words that generally transcribe with a universal quantifier
f) Word that generally transcribe with an existential quantifier
g) Negative Quantifier Words
h) Ambiguity
i) Give a summary of important transcription strategies
3.4.3: Transcription Strategies is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
3.5: Natural Deduction for Predicate Logic - Fundamentals

3.5.1: Review and Overview
3.5.4: The Existential Elimination and Universal Introduction Rules- Background in lnformal Argument

3.5: Natural Deduction for Predicate Logic - Fundamentals is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
5.1: Review and Overview
Let's get back to the problem of demonstrating argument validity. You know how to construct derivations which demonstrate the
validity of valid sentence logic arguments. Now that you have a basic understanding of quantified sentences and what they mean,
you are ready to extend the system of sentence logic derivations to deal with quantified sentences.
Let's start with a short review of the fundamental concepts of natural deduction: To say that an argument is valid is to say that in
every possible case in which the premises are true, the conclusion is true also. The natural deduction technique works by applying
truth preserving rules. That is, we use rules which, when applied to one or two sentences, license us to draw certain conclusions.
The rules are constructed so that in any case in which the first sentence or sentences are true, the conclusion drawn is guaranteed to
be true also. Certain rules apply, not to sentences, but to subderivations. In the case of these rules, a conclusion which they license
is guaranteed to be true if all the sentences reiterated into the subderivation are true.
A derivation begins with no premises or one or more premises. It may include subderivations, and any subderivation may itself
include a subderivation. A new sentence, or conclusion, may be added to a derivation if one of the rules of inference licenses us to
draw the conclusion from previous premises, assumptions, conclusions, or subderivations. Because these rules are truth preserving,
if the original premises are true in a case, the first conclusion drawn will be true in that case also. And if this first conclusion is true,
then so will the next. And so on. Thus, altogether, in - any case in which the premises are all true, the final conclusion will be true.
The only further thing you need to remember to be able to write sentence logic derivations are the rules themselves. If you are
feeling rusty, please refresh your memory by glancing at the inside front cover, and review chapters 5 and 7 of Volume I, if you
need to.
Now we are ready to extend our system of natural deduction for sentence logic to the quantified sentences of predicate logic.
Everything you have already learned will still apply without change. Indeed, the only fundamental conceptual change is that we
now must think in terms of an expanded idea of what constitutes a case. For sentence logic derivations, truth preserving rules
guarantee that if the premises are true for an assignment of truth values to sentence letters, then conclusions drawn will be true for
the same assignment. In predicate logic we use the same overall idea, except that for a "case" we use the more general idea of an
interpretation instead of an assignment of truth values to sentence letters. Now we must say that if the premises are true in an
interpretation, the conclusions drawn will be true in the same interpretation.
Since interpretations include assignment of truth values to any sentence letters that might occur in a sentence, everything from
sentence logic applies as before. But our thinking for quantified sentences now has to extend to include the idea of interpretations
as representations of the case in which quantified sentences have a truth value.
You will remember each of our new rules more easily if you understand why they work. You should understand why they are truth
preserving by thinking in terms of interpretations. That is, you should try to understand why, if the premises are true in a given
interpretation, the conclusion licensed by the rule will inevitably also be true in that interpretation.
Predicate logic adds two new connectives to sentence logic: the universal and existential quantifiers. So we will have four new
rules, an introduction and elimination rule for each quantifier. Two of these rules are easy and two are hard. Yes, you guessed it! I'm
going to introduce the easy rules first.
5.1: Review and Overview is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Consider the argument
Everyone is blond. (Vx)Bx
Adam is blond. Ba
Intuitively, if everyone is blond, this must include Adam. So if the premise is true, the conclusion is going to have to be true also. In
terms of interpretations, let's consider any interpretation you like which is an interpretation of the argument's sentences and in
which the premise, '(Vx)Bx', is true. The definition of truth of a universally quantified sentence tells us that '(Vx)Bx' is true in an
interpretation just in case all of its substitution instances are true in the interpretation. Observe that 'Ba' is a substitution instance of
'(Vx)Bx'. So in our arbitrarily chosen interpretation in which '(Vx)Bx' is true, 'Ba' will be true also. Since 'Ba' is true in any
interpretation in which '(Vx)Bx' is true, the argument is valid.
(In this and succeeding chapters I am going to pass over the distinction between someone and something, as this complication is
irrelevant to the material we now need to learn. I could give examples of things instead of people, but that makes learning very
dull.)
The reasoning works perfectly generally:
Universal Elimination Rub: If X is a universally quantified sentence, then you are licensed to conclude any of its substitution
instances below it. Expressed with a diagram, for any name, s, and any variable, u,
| (Vu)(. . . u . . .)
|.
|.
|.
| (. . . s . . .) VE
Remember what the box and the circle mean: If on a derivation you encounter something with the form of what you find in the box,
the rule licenses you to conclude something of the form of what you find in the circle.
Everyone loves Eve. (Vx)Lxe 1 | (Vx)Lxe P
Adam loves Eve. Lae 2 | Lae 1, VE
In forming the substitution instance of a universally quantified sentence, you must be careful always to put the same name
everywhere for the substituted variable. Substituting 'a' for 'x' in '(Vx)Lxx', we get 'Laa', not 'Lxa'. Also, be sure that you substitute
your name only for the occurrences of the variable which are free after deleting the initial quantifier. Using the name 'a' again, the
substitution instance of '(Vx)(Bx ⊃ (Vx)Lxe)' is 'Ba ⊃ (Vx)Lxe'. The occurrence of 'x' in 'Lxe' is bound by the second '(Vx)', and
so is still bound after we drop the first '(Vx)'. If you don't understand this example, you need to review bound and free variables and
substitution instances, discussed in chapter 3.
When you feel confident that you understand the last example, look at one more:
(Vx)(Gx ⊃ Kx) 1 | (Vx)(Gx ⊃ Kx) P
Gf 2 | Gf P
Kf 3 | Gf ⊃Kf 1, VE
4 | Kf 2, 3, ⊃E
 Exercise
5-1. Provide derivations which demonstrate the validity of these arguments. Remember to work from the conclusion backward,
seeing what you will need to get your final conclusions, as well as from the premises forward. In problem (d) be sure you
recognize that the premise is a universal quantification of a conditional, while the conclusion is the very different conditional
with a universally quantified antecedent.
a) (Vx)(Px & Dx) b) (Vx)(Px & Dx) c) (Vx)(Dx Kx)
Pk Pd & Dk (Vx)Dx
ka
d) (Vx)(Mx ⊃ A) e) (Vx)(Fx v Hx) f) (Vx)(~Bx v Lcx)
(Vx)Mx ⊃ A (Vx)(Fx ⊃ Dx) (Vx)Bx ⊃ Lcd
(Vx)(Hx ⊃ Dx)
Dp & Db
g) (Vx)(Lxx ⊃ Lxh) h) (Vx)(Rxx v Rxk)
~Lmh (Vy)~Ryk
~(Vx)Lxx Rcc & Rff
3.5.2: The Universal Elimination Rule is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Adam is blond. Ba
Someone is blond. (Ǝx)Bx
Intuitively, this argument is valid. If Adam is blond, there is no help for it: Someone is blond. Thinking in terms of interpretations,
we see that this argument is valid according to our new way of making the idea of validity precise. Remember how we defined the
truth of an existentially quantified sentence in an interpretation: '(Ǝx)Bx' is true in an interpretation if and only if at least one of its
substitution instances is true in the interpretation. But 'Ba' is a substitution instance of '(Ǝx)Bx'. So, in any interpretation in which
'Ba' is true, '(Ǝx)Bx' is true also, which is just what we mean by saying that the argument "Ba. Therefore (Ǝx)Bx." is valid.
You can probably see the form of reasoning which is at play here: From a sentence with a name we can infer what we will call an
Existential Generalization of that sentence. '(Ǝx)Bx' is an existential generalization of 'Ba'. We do have to be a little careful in
making this notion precise because we can get tripped up again by problems with free and bound variables. What would you say is
a correct existential generalization of '(Vx)Lax'? In English: If Adam loves everyone, then we know that someone loves everyone.
But we have to use two different variables to transcribe 'Someone loves everyone': '(Ǝy)(Vx)Lyx'. If I start with '(Vx)Lax', and
replace the 'a' with 'x', my new occurrence of 'x' is bound by that universal quantifier. I will have failed to generalize existentially
on 'a'.
Here is another example for you to try: Existentially generalize
(i) Ba ⊃ (Vx) Lax
2 3 45
If I drop the 'a' at 2 and 4, write in 'x', and preface the whole with '(Ǝx)', I get
(ii) (Ǝx)(Bx ⊃ (Vx)Lxx) Wrong
1 2 3 45
The 'x' at 4, which replaced one of the 'a's, is bound by the universally quantified 'x' at 3, not by the existentially quantified 'x' at 1,
as we intend in forming an existential generalization. We have to use a new variable. A correct existential generalization of 'Ba ⊃
(Vx)Lax' is
(iii) (Ǝy)(By ⊃ (Vx) Lyx)
1 2 3 45
as are
(iv) (Ǝy)(By ⊃ (Vx) Lax)
1 2 3 45
and
(iii) (Ǝy)(Ba⊃(Vx) Lyx)
1 2 3 45
Here is how you should think about this problem: Starting with a closed sentence, (. . . s . . .), which uses a name, s, take out one or
more of the occurrences of the name s. For example, take out the 'a' at 4 in (i). Then look to see if the vacated spot is already in the
scope of one (or more) quantifiers. In (i) to (v), the place marked by 4 is in the scope of the '(Vx)' at 3. So you can't use 'x'. You
must perform your existential generalization with some variable which is not already bound at the places at which you replace the
name. After taking out one or more occurrences of the name, s, in (. . . s . . . ), replace the vacated spots with a variable (the same
variable at each spot) which is not bound by some quantifier already in the sentence.
Continuing our example, at this point you will have turned (i) into
(vi) Ba ⊃ (Vx)Lya
You will have something of the form (. . . u . . .) in which u is free: 'y' is free in (vi). At this point you must have an open sentence.
Now, at last, you can apply your existential quantifier to the resulting open sentence to get the closed sentence (Ǝu)(. . . u . . .).
To summarize more compactly:
(Ǝu)(. . . u . . .) is an Existential Generalization of (. . . s . . .) with respect to the name s if and only if (Ǝu)(. . . u . . .) results
from (. . . s . . .) by
a) Deleting any number of occurrences of s in (. . . s . . .),
b) Replacing these occurrences with a variable, u, which is free at these occurrences, and
c) Applying (Ǝu) to the result.
(In practice you should read (a) in this definition as "Deleting one or more occurrences of s in (. . . s . . .)." I have expressed (a) with
"any number of' so that it will correctly treat the odd case of vacuous quantifiers, which in practice you will not need to worry
about. But if you are interested, you can figure out what is going on by studying exercise 3-3.)
It has taken quite a few words to set this matter straight, but once you see the point you will no longer need the words.
With the idea of an existential generalization, we can accurately state the rule for existential introduction:
Existential Introduction Rule: From any sentence, X, you are licensed to conclude any existential generalization of X
anywhere below. Expressed with a diagram,
| (. . . s . . .)
|.
|.
|.
| (Ǝu)(. . . u . . .) ƎI
When (Ǝu)(. . . u . . .) is an existential generalization of (. . . s . . .)
Let's look at a new example, complicated only by the feature that it involves a second name which occurs in both the premise and
the conclusion:
Adam loves Eve. Lae
Adam loves someone. (Ǝx)Lax
'(Ǝx)Lax' is an existential generalizaton of 'Lae'. So ƎI applies to make the following a correct derivation:
1 | Lae P
2 | (Ǝx)Lax 1, ƎI
To make sure you have the hang of rule ƎI, we'll do one more example. Notice that in this example, the second premise has an
atomic sentence letter as its consequent. Remember that predicate logic is perfectly free to use atomic sentence letters as
components in building up sentences.
ka 1 | ka P
(Ǝx)Kx ⊃ P 2 | (Ǝx)Kx ⊃ P P
P 3 | (Ǝx)Kx 1, ƎI
4 | P 2, 3, ⊃E
In line 4 I applied 3E to lines 2 and 3. ⊃E applies here in exactly the same way as it did in sentence logic. In particular ⊃E and the
other sentence logic rules apply to sentences the components of which may be quantified sentences as well as sentence logic
sentences.
Now let's try an example which applies both our new rules:
(Vx)Lxx 1 | (Vx)Lxx P
(Ǝx)Lxx 2 | Laa 1, VE
3 | (Ǝx)Lxx 2, ƎI
In addition to illustrating both new rules working together, this example illustrates something else we have not yet seen. In past
examples, when I applied VE I instantiated a universally quantified sentence with a name which already occurred somewhere in the
argument. In this case no name occurs in the argument. But if a universally quantified sentence is true in an interpretation, all of its
substitution instances must be true in - the interpretation. And every interpretation must have at least one object in it. So a
universally quantified sentence must always have at least one substitution instance true in an interpretation. Since a universally
quantified sentence always has at least one substitution instance, I can introduce a name into the situation with which to write that
substitution instance, if no name already occurs.
To put the point another way, because every interpretation always has at least one object in it, I can always introduce a name to
refer to some object in an interpretation and then use this name to form my substitution instance of the universally quantified
sentence.
Good. Let's try yet another example:
(Vx)(Cx ⊃ Mx) 1 | (Vx)(Cx Mx) P
Cd 2 | Cd P
(Ǝx)Mx 3 | Cd ⊃ Md 1, VE
4 | Md 2, 3, ⊃E
5 | (Ǝx)Mx 4, ƎI
Notice that although the rules permit me to apply Ǝ1 to line 2, doing so would not have gotten me anywhere. To see how I came up
with this derivation, look at the final conclusion. You know that it is an existentially quantified sentence, and you know that Ǝ1
permits you to derive such a sentence from an instance, such as 'Md'. So you must ask yourself: Can I derive such an instance from
the premises? Yes, because the first premise says about everything that if it is C, then it is M. And the second premise says that d,
in particular, is C. So applying VE to 1 you can get 3, which, together with 2, gives 4 by ⊃E.
 Exercise
5-2. Provide derivations which demonstrate the validity of the following arguments:
a) Na b) (Vx)(Kx & Px) c) (Vx)(Hx ⊃ ~Dx)
(Ǝx)(Nx v Gx) (Ǝx)Kx & (Ǝx) Px Dg
(Ǝx)~Hx
d) (Vx)Ax & (Vx)Txd e) Fa v Nh f) (Vx) Sx V Jx)
(Ǝx)(Ax & Txd (Ǝx)Fx v (Ǝx)Nx (Ǝx) Sx v (Ǝx)Jx
g) (Ǝx)Rxa ⊃ (Vx)Rax h) Lae v Lea i) (Ǝx)Jx ⊃ Q
Rea (Ǝx)Lax ⊃ A (Vx)Jx
(Ǝx)Rax (Ǝx)Lxa ⊃ A Q
A
j) (Vx)(Max v Mex) k) (Vx)(Kxx ≡ Px) l) (Vx)(~Oxx v Ix)
~(Ǝx)Max v Bg (Vx){Kjx & (Px ⊃Sx)} (Vx)(Ix ⊃ Rxm)
~(Ǝx)Mex v Bg (Ǝx)Sx (Ǝx)Oxx ⊃ (Ǝx)Rxm
(Ǝx)Bx
3.5.3: The Existential Introduction Rule is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
3.5.4: The Existential Elimination and Universal Introduction Rules- Background in
lnformal Argument
Now let's go to work on the two harder rules. To understand these rules, it is especially important to see how they are motivated.
Let us begin by looking at some examples of informal deductive arguments which present the kind of reasoning which our new
rules will make exact. Let's start with this argument:
Everyone likes either rock music or country-western.
Someone does not like rock.
Someone likes country-western.
Perhaps this example is not quite as trivial as our previous examples. How can we see that the conclusion follows from the
premises? We commonly argue in the following way. We are given the premise that someone does not like rock. To facilitate our
argument, let us suppose that this person (or one of them if there are more than one) is called Doe. (Since I don't know this person's
name, I'm using 'Doe' as the police do when they book a man with an unknown name as 'John Doe.') Now, since according to the
first premise, everyone likes either rock or country-western, this must be true, in particular, of Doe. That is, either Doe likes rock,
or he or she likes country-western. But we had already agreed that Doe does not like rock. So Doe must like country-western.
Finally, since Doe likes country-western, we see that someone likes country-western. But that was just the conclusion we were
trying to derive.
What you need to focus on in this example is how I used the name 'Doe'. The second premise gives me the assumption that
someone does not like rock. So that I can talk about this someone, I give him or her a name: 'Doe'. I don't know anything more that
applies to just this person, but I do have a fact, the first premise, which applies to everyone. So I can use this fact in arguing about
Doe, even though I really don't know who Doe is. I use this general fact to conclude that Doe, whoever he or she' might be, does
like country-western. Finally, before I am done, I acknowledge that I really don't know who Doe is, in essence by saying: Whoever
this person Doe might be, I know that he or she likes country-western. That is, what I really can conclude is that there is someone
who likes country-western.
Now let's compare this argument with another:
(1) Everyone either likes rock or country-western.
(2) Anyone who likes country-western likes soft music.
(3) Anyone who doesn't like rock likes soft music.
This time I have deliberately chosen an example which might not be completely obvious so that you can see the pattern of
reasoning doing its work.
The two premises say something about absolutely everyone. But it's hard to argue about 'everyone'. So let us think of an arbitrary
example of a person, named 'Arb', to whom these premises will then apply. My strategy is to carry the argument forward in
application to this arbitrarily chosen individual. I have made up the name 'Arb' to emphasize the fact that I have chosen this person
(and likewise the name) perfectly arbitrarily. We could just as well have chosen any person named by any name.
To begin the argument, the first premise tells us that
(4) Either Arb likes rock, or Arb likes country-western.
The second premise tells us that
(5) If Arb does like country-western, then Arb likes soft music.
Now, let us make a further assumption about Arb:
(6) (Further Assumption): Arb doesn't like rock.
From (6) and (4), it follows that
(7) Arb likes country-western.
And from (7) and (5), it follows that
(8) Arb likes soft music.
Altogether we see that Arb's liking soft music, (8), follows from the further assumption, (6), with the help of the original premises
(1) and (2) (as applied through this application to Arb, in (4) and (5)). Consequently, from the original premises it follows that
(9) If Arb doesn't like rock, then Arb likes soft music.
All this is old hat. Now comes the new step. The whole argument to this point has been conducted in terms of the person, Arb. But
Arb could have been anyone, or equally, we could have conducted the argument with the name of anyone at all. So the argument is
perfectly general. What (9) says about Arb will be true of anyone. That is, we can legitimately conclude that
(3) Anyone who doesn't like rock likes soft music.
which is exactly the conclusion we were trying to reach.
We have now seen two arguments which use "stand-in" names, that is, names that are somehow doing the work of "someone" or of
"anyone". Insofar as both arguments use stand-in names, they seem to be similar. But they are importantly different, and
understanding our new rules turns on understanding how the two arguments are different. In the second argument, Arb could be
anyone-absolutely anyone at all. But in the first argument, Doe could not be anyone. Doe could only be the person, or one of the
people, who does not like rock. 'Doe' is "partially arbitrary" because we are careful not to assume anything we don't know about
Doe. But we do know that Doe is a rock hater and so is not just anyone at all. Arb, however, could have been anyone.
We must be very careful not to conflate these two ways of using stand in names in arguments. Watch what happens if you do
conflate the ways:
Someone does not like rock.
Everyone does not like rock. (Invalid)
The argument is just silly. But confusing the two functions of stand-in names could seem to legitimate the argument, if one were to
argue as follows: Someone does not like rock. Let's call this person 'Arb'. So Arb does not like rock. But Arb could be anyone, so
everyone does not like rock. In such a simple case, no one is going to blunder in this way. But in more complicated arguments it
can happen easily.
To avoid this kind of mistake, we must find some way to clearly mark the difference between the two kinds of argument. I have
tried to bring out the distinction by using one kind of stand-in name, 'Doe', when we are talking about the existence of some
particular person, and another kind of stand-in name, 'Arb', when we are talking about absolutely any arbitrary individual. This
device works well in explaining that a stand-in name can function in two very different ways. Unfortunately, we cannot incorporate
this device in natural deduction in a straightforward way simply by using two different kinds of names to do the two different jobs.
Let me try to explain the problem. (You don't need to understand the problem in detail right now; detailed understanding will come
later. All you need at this point is just a glimmer of what the problem is.) At the beginning of a derivation a name can be arbitrary.
But then we might start a subderivation in which the name occurs, and although arbitrary from the point of view of the outer
derivation, the name might not be arbitrary from the point of view of the subderivation. This can happen because in the original
derivation nothing special, such as hating rock, is assumed about the individual. But inside the subderivation we might make such a
further assumption about the individual. While the further assumption is in effect, the name is not arbitrary, although it can
become arbitrary again when we discharge the further assumption of the subderivation. In fact, exactly these things happened in
our last example. If, while the further assumption (6) was in effect, I had tried to generalize on statements about Arb, saying that
what was true of Arb was true of anyone, I could have drawn all sorts of crazy conclusions. Look back at the example and see if
you can figure out for yourself what some of these conclusions might be.
Natural deduction has the job of accurately representing valid reasoning which uses stand-in names, but in a way which won't allow
the sort of mistake or confusion I have been pointing out. Because the confusion can be subtle, the natural deduction rules are a
little complicated. The better you understand what I have said in this section, the quicker you will grasp the natural deduction rules
which set all this straight.
 Exercise
5-3. For each of the two different uses of stand-in names discussed in this section, give a valid argument of your own,
expressed in English, which illustrates the use.
3.5.4: The Existential Elimination and Universal Introduction Rules- Background in lnformal Argument is shared under a not declared license and
was authored, remixed, and/or curated by LibreTexts.
Here is the intuitive idea for universal introduction, as I used this rule in the soft music example: If a name, as it occurs in a
sentence, is completely arbitrary, you can Universally Generalize on the name. This means that you rewrite the sentence with a
variable written in for all occurrences of the arbitrary name, and you put a universal quantifier, written with the same variable, in
front. To make this intuition exact, we have to say exactly when a name is arbitrary and what is involved in universal
generalization. We must take special care because universal generalization differs importantly from existential generalization.
Let's tackle arbitrariness first. When does a name not occur arbitrarily? Certainly not if some assumption is made about (the object
referred to by) the name. If some assumption is made using a name, then the name can't refer to absolutely anything. If a name
occurs in a premise or assumption, the name can refer only to things which satisfy that premise or assumption. So a name does not
occur arbitrarily when the name appears in a premise or an assumption, and it does not occur arbitrarily as long as such a premise
or assumption is in effect.
The soft music example shows these facts at work. I'll use 'Rx' for 'x likes rock.', 'Cx' for 'x likes country/western.', and 'Sx' for 'x
likes soft music.' Here are the formalized argument and derivation which I am going to use to explain these ideas:
(Vx)(Rx v Cx) 1 | (Vx)(Rx v Cx) P
(Vx)(Cx ⊃ Sx) 2 | (Vx)(Cx ⊃ Sx) P
(Vx)(~Rx ⊃ Sx) 3 | Ra v Ca 1, VE
4 | Ca ⊃ Sa 2, VE
5 | | ~Ra A
6 | | Ra v Ca 3, R
7 | | Ca 5, 6, vE
8 | | Ca ⊃ Sa 4, R
9 | | Sa 7, 8, ⊃E
10 | ~Ra ⊃ Sa 5-9, ⊃I
11 | (Vx)(~Rx ⊃ Sx) 10, VI
Where does 'a' occur arbitrarily in this example? It occurs arbitrarily in lines 3 and 4, because at these lines no premise or
assumption using 'a' is in effect. We say that these lines are Not Governed by any premise or assumption in which 'a' occurs. In
lines 5 through 9, however, 'a' does not occur arbitrarily. Line 5 is an assumption using 'a'. In lines 5 through 9, the assumption of
line 5 is in effect, so these lines are governed by the assumption of line 5. (We are going to need to say that a premise or assumption
always governs itself.) In all these lines something special is being assumed about the thing named by 'a', namely, that it has the
property named by '-R'. So in these lines the thing named by 'a' is not just any old thing. However, in line 10 we discharge the
assumption of line 5. So in line 10 'a' again occurs arbitrarily. Line 10 is only governed by the premises 1 and 2, in which 'a' does
not occur. Line 10 is not governed by the assumption of line 5.
I am going to introduce a device to mark the arbitrary occurrences of a name. If a name occurs arbitrarily we will put a hat on it, so
it looks like this: I. Marking all the arbitrary occurrences of 'a' in the last derivation ' makes the derivation look like this:
1 | (Vx)(Rx v Cx) P
2 | (Vx)(Cx ⊃ Sx) P
3 | Râ v Câ 1, VE
4 | Câ ⊃ Sâ 2, VE
5 | | ~Ra A
6 | | Ra v Ca 3, R
7 | | Ca 5, 6, vE
8 | | Ca ⊃ Sa 4, R
9 | | Sa 7, 8, ⊃E
10 | ~Râ ⊃ Sâ 5-9, ⊃I
11 | (Vx)(~Rx ⊃ Sx) 10, VI
Read through this copy of the derivation and make sure you understand why the hat occurs where it does and why it does not occur
where it doesn't. If you have a question, reread the previous paragraph, remembering that a hat on a name just means that the name
occurs arbitrarily at that place.
I want to be sure that you do not misunderstand what the hat means. A name with a hat on it is not a new kind of name. A name is a
name is a name, and two occurrences of the same name, one with and one without a hat, are two occurrences of the same name. A
hat on a name is a kind of flag to remind us that at that point the name is occurring arbitrarily. Whether or not a name occurs
arbitrarily is not really a fact just about the name. It is a fact about the relation of the name to the derivation in which it occurs. If, at
an occurrence of a name, the name is governed by a premise or assumption which uses the same name, the name does not occur
there arbitrarily. It is not arbitrary there because the thing it refers to has to satisfy the premise or assumption. Only if a name is not
governed by any premise or assumption using the same name is the name arbitrary, in which case we mark it by dressing it with a
hat.
Before continuing, let's summarize the discussion of arbitrary occurrence with an exact statement:
Suppose that a sentence, X, occurs in a derivation or subderivation. That occurrence of X is Governed by a premise or
assumption, Y, if and only if Y is a premise or assumption of X's derivation, or of any outer derivation of X's derivation (an
outer derivation, or outer-outer derivation, and so on). In particular, a premise or assumption is always governed by itself.
A name Occurs Arbitrarily in a sentence of a derivation if that occurrence of the sentence is not governed by any premise or
assumption in which the name occurs. To help us remember, we mark an arbitrary occurrence of a name by writing it with a
hat.
The idea for the universal introduction rule was that we would Universally Generalize on a name that occurs arbitrarily. We have
discussed arbitrary occurrence. Now on to universal generalization.
The idea of a universal generalization differs in one important respect from the idea of an existential generalization. To see the
difference, you must be clear about what we want out of a generalization: We want a new quantified sentence which follows from a
sentence with a name.
For the existential quantifier, '(Ǝx)Lxx', '(Ǝx)Lax', and '(Ǝx)Lxa' all follow from 'Laa'. From the fact that Adam loves himself, it
follows that Adam loves someone, someone loves Adam, and someone loves themself.
Now suppose that the name 'â' occurs arbitrarily in 'Lââ'. We know that "Adam" loves himself, where Adam now could be just
anybody at all. What universal fact follows? Only that '(Vx)Lxx', that everyone loves themself. It does not follow that '(Vx)Lâx' or
'(Vx)Lxâ'. That is, it does not follow that Adam loves everyone or everyone loves Adam. Even though 'Adam' occurs arbitrarily,
'(Vx)Lâx' and '(Vx)Lxâ' make it sound as if someone ("Adam") loves everyone and as if someone ("Adam") is loved by everyone.
These surely do not follow from 'Lââ'. But ƎI would license us to infer these sentences, respectively, from '(Vx)Lâx' and from
'(Vx)Lxâ'.
Worse, â is still arbitrary in '(Vx)Lâx'. So if we could infer '(Vx)Lâx' from 'Lââ', we could then argue that in '(Vx)Lâx', 'â' could be
anyone. We would then be able to infer '(Vy)(Vx)Lyx', that everyone loves everyone! But from 'Lââ' we should only be able to
infer '(Vx)Lxx', that everyone loves themself, not '(Vy)(Vx)Lyx', that everyone loves everyone.
We want to use the idea of existential and universal generalizations to express valid rules of inference. The last example shows that,
to achieve this goal, we have to be a little careful with sentences in which the same name occurs more than once. If s occurs more
than once in (. . . s . . .), we may form an existential generalization by generalizing on any number of the occurrences of s. But, to
avoid the problem I have just described and to get a valid rule of inference, we must insist that a universal generalization of (. . . s .
. .), with respect to the name, s, must leave no instance of s in (. . . s . . .).
In other respects the idea of universal generalization works just like existential generalization. In particular, we must carefully
avoid the trap of trying to replace a name by a variable already bound by a quantifier. This idea works exactly as before, so I will
proceed immediately to an exact statement:
The sentence (Vu)(. . . u . . .) results by Universally Generalizing on the name s in (. . . s . . .) if and only if one obtains (Vu)
(. . . u . . .) from (. . .s. ..)by
a) Deleting all occurrences of s in (. . . s . . .),
b) Replacing these occurrences with a variable, u, which is free at these occurrences, and
c) Applying (Vu) to the result.
(Vu)(. . . u . . .) is then said to be the Universal Generalization of (. . . s . . .) with Respect to the Name s.
With these definitions, we are at last ready for an exact statement of the universal introduction rule:
Universal Introduction Rule: If a sentence, X, appears in a derivation, and if at the place where it appears a name, ŝ, occurs
arbitrarily in X, then you are licensed to conclude, anywhere below, the sentence which results by universally generalizing on
the name ŝ in X. Expressed with a diagram:
Where j occurs arbitrarily in (. . . . . .) and (Vu)(. . ;u . . .) is the univers?l generalization . 1 of (. . . s . . .) with respect to s.
| (. . . ŝ . . .)
| (Vu)(. . . u . . .) VI
Where ŝ occurs arbitrarily in (. . . ŝ . . .) and (Vu)(. . . u. . .) is the universal generalization of (. . . ŝ . . .) with respect to ŝ.
Let's look at two simple examples to illustrate what can go wrong if you do not follow the rule correctly. The first example is the
one we used to illustrate the difference between existential and universal generalization:
Everyone loves themself.
Everyone loves Adam. (Invalid!)
1 | (Vx)Lxx P
2 | Lââ 1,VE
3 | (Vx)Lxâ
Mistaken attempt to apply VI to 2. 3 is not a universal generalization of 2
The second example will make sure you understand the requirement that VI applies only to an arbitrary occurrence of a name:
Adam is blond. 1 | Ba P
Everyone is blond. (Invalid!) 2 | (Vx)Bx
Mistaken attempt to apply VI to 1. 'a' is not arbitrary at 1.
The problem here is that the premise assumes something special about the thing referred to by 'a', that it has the property referred to
by 'B'. We can universally generalize on a name-that is, apply VI-only when nothing special is assumed in this way, that is, when
the name is arbitrary. You will see this even more clearly if you go back to our last formalization of the soft music example and see
what sorts of crazy conclusions you could draw if you were to allow yourself to generalize on occurrences of names without hats.
Let's consolidate our understanding of VI by working through one more example. Before reading on, try your own hand at
providing a derivation for
(Vx)(Lax & Lxa)
(Vx)(Lax ≡ Lxa)
If you don't see how to begin, use the same overall strategy we developed in chapter 6 of volume 1. Write a skeleton derivation
with its premise and final conclusion and ask what you need in order to get the final, or target, conclusion.
1 | (Vx)(Lax & Lxa) P
|?
|?
| (Vx)(Lax ≡ Lxa)
We could get our target conclusion by VI if we had a sentence of the form 'Lab ≡ Lba'. Let's write that in to see if we can make
headway in this manner:
|?
|?
| Lab ≡Lba
| (Vx)(Lax ≡ Lxa) VI
'Lab ≡ Lba' is now our target conclusion. As a biconditional, our best bet is to get it by ≡I from 'Lab ⊃ Lba' and 'Lba ⊃ Lab'. (I
didn't write hats on any names because, as I haven't written the sentences as part oE the derivation, I am not yet sure which
sentences will govern these two conditionals.) The conditionals, in turn, I hope to get from two subderivations, one each starting
from one of the antecedents of the two conditionals:
|?
| | Lab A
||?
| | Lba
| Lab ⊃ Lba ⊃I
|?
| | Lba A
||?
| | Lab
|
| Lba ⊃ Lab ⊃I
| Lab ≡ Lba
| (Vx)(Lax ≡ Lxa) VI
Notice that 'b' gets a hat wherever it appears in the main derivation. There, 'b' is not governed by any assumption in which 'b'
occurs. But 'b' occurs in the assumptions of both subderivations. So in the subderivations 'b' gets no hat. Finally, 'a' occurs in the
original premise. That by itself rules out putting a hat on 'a' anywhere in the whole derivation, which includes all of its
subderivations.
Back to the question of how we will fill in the subderivations. We need to derive 'Lba' in the first and 'Lab' in the second. Notice
that if we apply VE to the premise, using 'b' to instantiate 'x', we get a conjunction with exactly the two new target sentences as
conjuncts. We will be able to apply &E to the conjunction and then simply reiterate the conjuncts in the subderivations. Our
completed derivation will look like this:
2 | Lab & Lba 1, VE
3 | Lab 2, &E
4 | Lbs 2, &E
5 | | Lab A
6 | | Lba 4, R
7 | Lab ⊃ Lba 5-6. ⊃I
8 | | Lba A
9 | | Lab 3, R
10 | Lba ⊃ Lab 8-9, ⊃I
11 | Lab ≡ Lba 7, 10, ≡I
12 | (Vx)(Lax ≡ Lxa) 11, VI
Once more, notice that 'b' gets a hat in lines 2, 3, and 4. In these lines no premise or assumption using 'b' is operative. But in lines 5,
6, 8, and 9, 'b' gets no hat, even though exactly the same sentences appeared earlier (lines 3 and 4) with hats on 'b'. This is because
when we move into the subderivations an assumption goes into effect which says something special about 'b'. So in the
subderivations, off comes the hat. As soon as this special assumption about 'b' is discharged, and we move back out of the
subderivation, no special assumption using 'b' is in effect, and the hat goes back on 'b'.
You may well wonder why I bother with the hats in lines like 2, 3, 4, 7, and 10, on which I am never going to universally
generalize. The point is that, so far as the rules go, I am permitted to universally generalize on 'b' in these lines. In this problem I
don't bother, because applying VI to these lines will not help me get my target conclusion. But you need to develop awareness of
just when the formal statement of the VI rule allows you to apply it. Hence you need to learn to mark those places at which the rule
legitimately could apply.
Students often have two more questions about hats. First, VI permits you to universally generalize on a name with a hat. But you
can also apply ∃1 to a name with a hat. Now that I have introduced the hats, the last example in section 5-3 should really look like
this:
1 | (Vx)(Lxx P
2 | Lââ 1, VE
3 | (∃I)Lxx 2, ∃I
If everyone loves themself, then Arb loves him or herself, whoever Arb may be. But then someone loves themself. When a name
occurs arbitrarily, the name can refer to anything. But then it also refers to something.- You can apply either VI or ∃1 to a hatted
name.
It is also easy to be puzzled by the fact that a name which is introduced in the assumption of a subderivation, and thus does not
occur arbitrarily there, can occur arbitrarily after the assumption of the subderivation has been discharged. Consider this example:
1 | (Vx)Px ⊃ (Vx)Qx P
2 | | Lab A
3 | | (∃I)Px 2, ∃I
4 | | (∃I)Px ⊃ (Vx)Qx 1, R
5 | | (Vx)Qx 3, 4, ⊃E
6 | | Qa 5, VE
7 | Pâ ⊃ Qâ 2-6, ⊃I
8 | (Vx)(Px ⊃ Qx) 7, VI
In the subderivation something is assumed about 'a', namely, that it has the property P. So, from the point of view of the
subderivation, 'a' is not arbitrary. As long as the assumption of the subderivation is in effect, 'a' cannot refer to just anything. It can
only refer to something which is P. But after the subderivation's assumption has been discharged, 'a' is arbitrary. Why? The rules
tell us that 'a' is arbitrary in line 7 because line 7 is not governed by any premises or assumptions in which 'a' occurs. But to make
this more intuitive, notice that I could have just as well constructed th~ sam: subderivation using the name 'b' instead of 'a', using
⊃E to write 'Pb ⊃ Qb' on line 7. Or I could have used 'c', 'd', or any other name. This is why 'a' is arbitrary in line 7. I could have
arrived at a conditional in line 7 using any name I liked instead of using 'a'.
Some students get annoyed and frustrated by having to learn when to put a hat on a name and when to leave it off. But it's worth the
effort to learn. Once you master the hat trick, VI is simple: You can apply VI whenever you have a name with a hat. Not otherwise.
 Exercise
5-4. There is a mistake in the following derivation. Put on hats where they belong, and write in the justification for those steps
which are justified. Identify and explain the mistake.
1 | (Vx)(Bx ⊃ Cx) P
2 | Be ⊃ Ce
3 | | Be A
4 | | Be ⊃ Ce
5 | | Ce
6 | | (Vx)Ce
7 | Be ⊃(Vx)Cx
5-5. Provide derivations which establish the validity of the following arguments. Be sure you don't mix up sentences which are
a quantification of a sentence formed with a '&', a 'v', or a '⊃' with compounds formed with a '&', a 'v', or a '⊃', the components
of which are quantified sentences. For example, '(Vx)(Px & Qa)' is a universally quantified sentence to which you may apply
VE.'(Vx)Px & Qa' is a conjunction to which you may apply &E but not VE.
a) (Vx)(Fx & Gx) b) (Vx)(Mx ⊃ Nx) c) A
(Vx)Fx (Vx)Mx (Vx)(A v Nx)
(Vx)Nx
d) (Vx)Hx & (Vx)Qx e) (Vx)(Kxm & Kmx) f) (Vx)(Fx v Gx)
(Vx)(Hx & Qx) (Vx)Kxm & (Vx)Kmx (Vx)(Fx ⊃ Gx)
(Vx)Gx
g) (Vx)~Px v C h) (Vx)(Rxb ⊃ Rax) i) (Vx)(Gxh ⊃ Gxm)
(Vx)(~-Px v C) (Vx)Rxb ⊃ (Vx)Rax (Vx)(~Gxm ⊃ -Gxh)
j) (Vx)(Mx ⊃ Nx) k) T ⊃ (Vx)Mdx l) (Vx)(Hff ⊃ Lxx)
(Vx)(Nx ⊃ Ox) (Vx)(T ⊃ Mdx) Hff ⊃ (Vx)Lxx
(Vx)(Mx ⊃ Ox)
m) (Vx)Px v (Vx)Qx n) (Vx)Hx o) (Vx)(Sx ≡ Ox)
(Vx)(Px v Qx) (3x)Hx ⊃ (Vx)(Hx ⊃ Jx) (Vx)Sx ≡ (Vx)Ox
(Vx)x
p) (∃x)Px ⊃ A q) ~(∃x)Px r) ~(Vx)Px s) (Vx)Px ⊃ A
(Vx)(Px ⊃ A) (Vx)~Px (∃x)~Px (∃x)(Px ⊃ A)
t) ~(Vx)(Jx ⊃ ~Kx) u) ~(∃x)Qx v H v) ~(∃x)Dx
(∃x)(Jx & Kx) (Vx)(~Qx v H) (Vx)(Dx ⊃ Kx)
3.5.5: The Universal Introduction Rule is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
VI and ∃E are difficult rules. Many of you will have to work patiently - over this material a number of times before you understand
them clearly. But if you have at least a fair understanding of VI, we can proceed to ∃E because ultimately these two rules need to
be understood together.
Let's go back to the first example in Section 5.4: Everyone likes either rock music or country-western. Someone does not like rock.
So someone likes country-western. I will symbolize this as
(Vx)(Rx v Cx)
(∃x)~Rx
(∃x)Cx
In informally showing this argument's validity, I used 'Doe', which I will now write just as 'd', as a stand-in name for the unknown
"someone" who does not like rock. But I must be careful in at least two respects:
i. I must not allow myself to apply VI to the stand-in name, 'd'. Otherwise, I could argue from '(∃x)~Rx' to '~Rd' to
'(Vx)~Rx'. In short, I have to make sure that such a name never gets a hat.
ii. When I introduce the stand-in name, 'd', I must not be assuming anything else about the thing to which 'd' refers other
than that '-R' is true of it.
It's going to take a few paragraphs to explain how we will meet these two requirements. To help you follow these paragraphs, I'll
begin by writing down our example's derivation, which you should not expect to understand until you have read the explanation.
Refer back to this example as you read:
(Vx)(Rx v Cx) 1 | (Vx)(Rx v Cx) P
(∃x)~Rx 2 | (∃x)~Rx P
(∃x)Cx 3 | d | ~Rd A
4 | | (Vx)(Rx v Cx) 1, R
5 | | Rd v Cd 4, VE
6 | | Cd 3, 5, vE
7 | | (∃x)Cx 6, ∃I
8 | (∃x)Cx 2, 3-7, ∃E
I propose to argue from the premise, '(∃x)~Rx', by using the stand-in name, 'd'. I will say about the thing named by 'd' what
'(∃x)~Rx' says about "someone". But I must be sure that 'd' never gets a hat. How can I guarantee that? Well, names that occur in
assumptions can't get hats anywhere in the subderivation governed by the assumption. So we can guarantee that 'd' won't get a hat
by introducing it as an assumption of a subderivation and insisting that 'd' never occur outside that subderivation. This is what I
did in line 3. '~Rd' appears as the subderivation's assumption, and the 'd' written just to the left of the scope line signals the
requirement that 'd' be an Isolated Name. That is to say, 'd' is isolated in the subderivation the scope line of which is marked with
the 'd'. An isolated name may never appear outside its subderivation.
Introducing 'd' in the assumption of a subderivation might seem a little strange. I encounter the sentence, '(∃x)~Rx', on a derivation.
I reason: Let's assume that this thing of which '~R' is true is called 'd', and let's record this assumption by starting a subderivation
with '-Rd' as its assumption, and see what we can derive. Why could this seem strange? Because if I already know '(∃x)~Rx', no
further assumption is involved in assuming that there is something of which '-R' is true. But, in a sense, I do make a new
assumption in assuming that this thing is called 'd'. It turns out that this sense of making a special assumption is just what we need.
By making 'd' occur in the assumption of a subderivation, and insisting that 'd' be isolated, that it appear only in the subderivation, I
guarantee that 'd' never gets a hat. But this move also accomplishes our other requirement: If 'd' occurs only in the subderivation, 'd'
cannot occur in any outer premise or assumption.
Now let's see how the overall strategy works. Look at the argument's subderivation, steps 3-7. You see that, with the help of
reiterated premise 1, from '~Rd' I have derived '(∃x)Cx'. But neither 1 nor the conclusion '(∃x)Cx' uses the name 'd'. Thus, in this
subderivation, the fact that I used the name 'd' was immaterial. I could have used any other name not appearing in the outer
derivation. The real force of the assumption '~Rd' is that there exists something of which '~R' is true (there is someone who does
not like rock). But that there exists something of which '~R' is true has already been given to me in line 2! Since the real force of
the assumption of line 3 is that there exists something of which '~R' is true, and since I am already given this fact in line 2, I don't
really need the assumption 3. I can discharge it. In other words, if I am given the truth of lines 1 and 2, I know that the conclusion
of the subderivation, 7, must also be true, and I can enter 7 as a further conclusion of the outer derivation.
It is essential, however, that 'd' not appear in line 7. If 'd' appeared in the final conclusion of the subderivation, then I would not be
allowed to discharge the assumption and enter this final conclusion in the outer derivation. For if 'd' appeared in the subderivation's
final conclusion, I would be relying, not just on the assumption that '~R' was true of something, but on the assumption that this
thing was named by 'd'.
The example's pattern of reasoning works perfectly generally. Here is how we make it precise:
A name is Isolated in a Subderivation if it does not occur outside the subderivation. We mark the isolation of a name by
writing the name at the top left of the scope line of its subderivation. In applying this definition, remember that a sub-sub-
derivation of a subderivation counts as part of the subderivation.
Existential Elimination Rule: Suppose a sentence of the form (∃u)(. . . u. . .) appears in a derivation, as does a subderivation
with assumption (. . . s . . .), a substitution instance of (∃u)(. . . u . . .). Also suppose that s is isolated in this subderivation. If
X is any of the subderivation's conclusions in which s does not occur, you are licensed to draw X as a further conclusion in
the outer derivation, anywhere below the sentence (∃u)(. . . u . . .) and below the subderivation. Expressed with a diagram:
Where (. . . s . . .) is a substitution instance of (∃u) (. . . u . . .) and s is isolated in the derivation.
| (∃u)(. . . u . . .)
| s | (. . . s . . .)
||.
||.
||.
||x
| X ∃E
When you annotate your application of the ∃E rule, cite the line number of the existentially quantified sentence and the inclusive
line numbers of the subderivation to which you appeal in applying the rule.
You should be absolutely clear about three facets of this rule. I will illustrate all three.
Suppose the ∃E rule has been applied, licensing the new conclusion, X, by appeal to a sentence of the form (∃u)(. . . u . . .)
and a subderivation beginning with assumption (. . . s . . .):
1) s cannot occur in any premise or prior assumption governing the
subderivation,
2) s cannot occur in (∃u)(. . . u . . .), and
3) s cannot occur in X.
All three restrictions are automatically enforced by requiring s to be isolated in the subderivation. (Make sure you understand why
this is correct.) Some texts formulate the ∃E rule by imposing these three requirements separately instead of requiring that s be
isolated. If you reach chapter 15, you will learn that these three restrictions are really all the work that the isolation requirement
needs to do. But, since it is always easy to pick a name which is unique to a subderivation, I think it is easier simply to require that
s be isolated in the subderivation.
Let us see how things go wrong if we violate the isolation requirement in any of these three ways. For the first, consider:
Ca 1 | Ca P
(∃x)Bx (Invalid!) 2 | (∃x)Bx P
(∃x)(Cx &Bx) 3 | a| Ba A
4 | | Ca 1, R
5 | | Ca & Ba 3, 4, &I
6 | | (∃x)(Ca & Bx) 5,∃I
7 | (∃x)(Cx &Bx)
Mistaken attempt to apply ∃E to 2 and 3-6. 'a' occurs in premise 1 and is not isolated in the sub derivation.
From the fact that Adam is clever and someone (it may well not be Adam) is blond, it does not follow that any one is both clever
and blond.
Now let's see what happen if one violates the isolation requirement in the second way:
(Vx)(∃y)Lxy 1 | (Vx)(∃y)Lxy P
(∃x)Lxx (Invalid!) 2 | (∃y)Lây 1, VE
3 | a| Laa A
4 | | (∃x)Lxx 3, ∃I
5 | (∃x)Lxx
Mistaken attempt to apply ∃E to 2 and 3-4, 'a' occurs in 2 and is not isolated in the subderivation.
From the fact that everyone loves someone, it certainly does not follow that someone loves themself.
And, for violation of the isolation requirement in the third way:
(∃y)Bx 1 | (∃x)Bx P
(Vx)Bx (Invalid!) 2 | a| Ba A
3 | | Ba 2, R
4 | Bâ
5 | (Vx)Bx
Mistaken attempt to apply ∃E to 1 and 2-3, 'a' occurs in 4 and is not isolated in the subderivation.
From the fact that someone is blond, it will never follow that everyone is blond.
One more example will illustrate the point about a sub-sub-derivation being part of a subderivation. The following derivation is
completely correct:
1 | (Vx)(Cx ⊃ ~Bx) P
2 | (∃x)Bx P
3 | d |Bd A
4 | | | Cd A
5 | | | (Vx)(Cx ~Bx) 1, R
6 | | | Cd ⊃~Bd 5, VE
7 | | | ~Bd 4, 6, ⊃E
8 | | 3| Bd 3, R
9 | | ~Cd 4-8, ~I
0 | 2| (∃x)~Cx 9, ∃I
11 1| (∃x)~Cx 2, 3-10, ∃E
You might worry about this derivation: If 'd' is supposed to be isolated in subderivation 2, how can it legitimately get into sub-sub-
derivation 3?
A subderivation is always part of the derivation in which it occurs, and the same holds between a sub-sub-derivation and the
subderivation in which it occurs. We have already encountered this fact in noting that the premises and assumptions of a derivation
or subderivation always apply to the derivation's subderivations, its sub-sub-derivations, and so on.
Now apply this idea about parts to the occurrence of 'd' in sub-subderivation 3 above: When I say that a name is isolated in a
subderivation I mean that the name can occur in the subderivation and all its parts, but the name cannot occur outside the
subderivation.
Here is another way to think about this issue: The 'd' at the scope line of the second derivation means that 'd' occurs to the right of
the scope line and not to the left. But the scope line of subderivation 3 is not marked by any name. So the notation permits you to
use 'd' to the right of this line also.
I hope that you are now beginning to understand the rules for quantifiers. If your grasp still feels shaky, the best way to understand
the rules better is to go back and forth between reading the explanations and practicing with the problems. As you do so, try to keep
in mind why the rules are supposed to work. Struggle to see why the rules are truth preserving. By striving to understand the rules,
as opposed to merely learning them as cookbook recipes, you will learn them better, and you will also have more fun.
 Exercise
5-6. There is one or more mistakes in the following derivation. Write the hats where they belong, justify the steps that can be
justified, and identify and explain the mistake, or mistakes.
1 | (Vy)(∃y)Lxy P
2 | (∃y)Lxb
3 | | Lab A
4 | | (Vx)Lay
5 | | (∃x)(Vy)Lxy
6 | (∃x)(Vy)Lxy
5-7. Provide derivations which establish the validity of the following arguments:
a) (∃x)Ix b) (∃x)(A ⊃ Px) c) (∃x)Hmx
(Vx)(Ix ⊃ Jx) A ⊃ (∃x)Px (Vx)(~Hmx v Gxn)
(∃x)Jx (∃x)Gxn
d) (∃x)(Cfx &Cxf) e) (∃x)(Px vQx) f) (∃x)Px v (∃x)Qx
(∃x)Cfx & (∃x)Cxf (∃x)Px v (∃x)Qx (∃x)(Px v Qx)
g) (∃x)(Px ⊃ A) h) (V)(Px ⊃ A) i) (∃x)(Lxa ≡ Lex)
(Vx)Px ⊃ A (∃x)Px ⊃ A (VVx)Lxa
(Vx)Lex
j) (Vx)(Gsx ⊃ ~Gxs)
(∃x)Gxs ⊃ (∃x)~Gsx
k) (∃x)(Px v Qx) l) (∃x)(~Mxt v Mtx) m) (∃x)Hxg v (∃x)Nxf
(Vx)(Px ⊃ kx) (∃x)(Mtx ⊃ Axx) (Vx)(Hxg ⊃ Cx)
(Vx)(Qx ⊃ Kx) (∃x)(~Mxt v Axx) (Vx)(Nxf ⊃ Cx)
(∃x)kx (∃x)Cx
n) (Vx)[(Fx v Gx) ⊃ Lxx] o) (Vx){Fx ⊃ (Rxa v Rax]
(∃x)~(Lxx) (∃x)~Rxa
(∃x)~Fx & (∃x)~Gx (Vx)~Rax ⊃ (∃x)~Fx
p) (∃x)Qxj q) (Vx)~Fx r) (∃x)~Fx
(∃x)(Qxj v Dgx) ⊃ (Vx)Dgx ~(∃x)Fx ~(Vx)Fx
(Vx)(Dgx v Qjx)
s) (Vx)(Jxx ⊃ ~Jxf) t) (∃x)Px v Qa u) A ⊃ (∃x)Px
~(∃x)(Jxx & Jxf) (Vx)~Px (∃x)(A ⊃ Px)
(∃x)Qx
5-8. Are you bothered by the fact that ∃E requires use of a subderivation with an instance of the existentially quantified
sentence as its assumption? Good news! Here is an alternate version of ∃E which does not require starting a subderivation:
| (∃u)(. . . u . . .)
| (Vu)(. . . u . . .)
| X ∃E
Show that, in the presence of the other rules, this version is exchangeable with the ∃E rule given in the text. That is, show that
the above is a derived rule if we start with the rules given in the text. And show that if we start with all the rules in the text
except for ∃E, and if we use the above rule for ∃E, then the ∃E of the text is a derived rule.
Here is a list of important terms from this chapter. Explain them briefly and record your explanations in your notebook:
a) Truth Preserving Rule of Inference
b) Sound
c) Complete
d) Stand-in Name
e) Govern
f) Arbitrary Occurrence
g) Existential Generalization
h) Universal Generalization
i) Isolated Name
j) Existential Introduction Rule
k) Existential Elimination Rule
l) Universal Introduction Rule
m) Universal Elimination Rule
3.5.6: The Existential Elimination Rule is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
3.6: More on Natural Deduction for Predicate Logic


3.6: More on Natural Deduction for Predicate Logic is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
In chapter 5 I wanted you to focus on understanding the basic rules for quantifiers. So there I avoided the complications that arise
when we have sentences, such as '(Vx)(Vy)(Px & Py)', which stack one quantifier on top of another. Such sentences involve no
new principles. It's just a matter of keeping track of the main connective. For example, '(Vx)(Vy)(Px & Qy)' is a universally
quantified sentence, with '(Vx)' as the main connective. You practiced forming substitution instances of such sentences in chapter 3.
The substitution instance of '(Vx)(Vy)(Px & Qy)' formed with 'a' (a sentence you could write when applying VE) is '(Vy)(Pa &
Qy)'.
You will see how to deal with such sentences most quickly by just looking at a few examples. So let's write a derivation to establish
the validity of
(Vx)(Vy)(Px & Qy) 1 | (Vx)(Vy)(Px & Qy) P
(Vx)Px & (Vx)Qx 2 | (Vx)(Pâ & Qy) 1, VE
3 | Pâ & Qb 1, VE
4 | Pâ 1, VE
5 | Qb 1, VE
6 | (Vx)Px 1, VE
7 | (Vx)Qx 1, VE
8 | (Vx)Px & (Vx)Qx 1, VE
In line 2 I applied VE by forming the substitution instance of 1 using the name 'a'. Then in line 3 I formed a substitution instance of
the universally quantified line 2.
Let's look at an example of multiple existential quantification. The basic ideas are the same. But observe that in order to treat the
second existential quantifier, we must start a sub-sub-derivation:
(∃x)(∃y)(Px & Qy)
(∃x)Px & (∃x)Qx
1 | (∃x)(∃y)(Px & Qy) P
2 | a| (∃y)(Pa & Qy) A
3 | | b| Pa & Qb A
4 | | | Pa 3, &E
5 | | | Qb 3, &E
6 | | | (∃x)Px 4, ∃I
7 | | | (∃x)Qx 5, ∃I
8 | | 3| (∃x)Px & (∃x)Qx 6, 7, &I
9 | 2| (∃x)Px & (∃x)Qx 2, 3-8, ∃E
10 1| ∃x)Px & (∃x)Qx 1, 2-9, ∃E
In line 2 I wrote down '(∃y)(Pa & Qy)', a substitution instance of line 1, formed with 'a', substituted for 'x', which is the variable in
the main connective, '(∃x)', of line 1. Since I plan to appeal to 3E in application to line 1, I make '(∃y)(Pa & Qy)' the assumption of
a subderivation with 'a' an isolated name. I then do the same thing with '(∃y)(Pa & Qy)', but because this is again an existentially
quantified sentence to which I will want to apply 3E, I must make my new substitution instance, 'Pa & Qb', the assumption of a
sub-sub-derivation, this time with 'b' the isolated name.
In the previous example, I would have been allowed to use 'a' for the second as well as the first substitution instance, since I was
applying VE. But, in the present example, when setting up to use two applications of ∃E, I must use a new name in each
assumption. To see why, let's review what conditions must be satisfied to correctly apply ∃E to get line 9. I must have an
existentially quantified sentence (line 2) and a subderivation (sub-sub-derivation 3), the assumption of which is a substitution
instance of the existentially quantified sentence. Furthermore, the name used in forming the substitution instance must be isolated
to the subderivation. Thus, in forming line 3 as a substitution instance of line 2, I can't use 'a'. I use the name 'b' instead. The 'a'
following 'P' in line 3 does not violate the requirement. 'a' got into the picture when we formed line 2, the substitution instance of
line 1, and you will note that 'a' is indeed isolated to subderivation 2, as required, since sub-sub-derivation 3 is part of
subderivation 2.
Here's another way to see the point. I write line 3 as a substitution instance of line 2. Since I will want to apply ∃E, the name I use
must be isolated to subderivation 3. If I tried to use 'a' in forming the substitution instance of line 2, I would have had to put an 'a'
(the "isolation flag") to the left of scope line 3. I would then immediately see that I had made a mistake. 'a' as an isolation flag
means that 'a' can occur only to the right. But 'a' already occurs to the left, in line 2. Since I use 'b' as my new name in subderivation
3, I use 'b' as the isolation flag there. Then the 'a' in line 3 causes no problem: All occurrences of 'a' are to the right of scope line 2,
which is the line flagged by 'a'.
All this is not really as hard to keep track of as it might seem. The scope lines with the names written at the top to the left (the
isolation flags) do all the work for you. 'a' can only appear to the right of the scope line on which it occurs as an isolation flag. 'b'
can only occur to the right of the scope line on which it occurs as an isolation flag. That's all you need to check.
Make sure you clearly understand the last two examples before continuing. They fully illustrate, in a simple setting, what you need
to understand about applying the quantifier rules to multiply quantified sentences.
Once you have digested these examples, let's try a hard problem. The new example also differs from the last two in that it requires
repeated use of a quantifier introduction rule instead of repeated use of a quantifier elimination rule. In reading over my derivation
you might well be baffled as to how I figured out what to do at each step. Below the problem I explain the informal thinking I used
in constructing this derivation, so that you will start to learn how to work such a problem for yourself.
(Vx)Px ⊃ (∃x)Qx
(∃x)(∃y)(Px ⊃ Qy)
1 | (Vx)Px ⊃ (∃x)Qx P
2 | | ~(∃x)(∃y)(Px ⊃ Qy) A
3 | | | ~Pa A
4 | | | ~Qb ⊃ ~Pa ⊃, W
5 | | | Pa ⊃ Qb 4, CP
6 | | | (∃y)(Pa ⊃ Qy) 5, ∃I
7 | | | (∃x)(∃y)(Px ⊃ Qy) 6, ∃I
8 | | | ~(∃x)(∃y)(Px ⊃ Qy) 2, R
9 | | Pâ 3-8, RD
10 | | (Vx)Px 9, VI
11 | | | (∃x)Qx A
12 | | | b| Qb A
13 | | | | Pa ⊃ Qb 12, W
14 | | | | (∃y)(Pa ⊃ Qy) 13, ∃I
15 | | | | (∃x)(∃y)(Px ⊃ Qy) 14, ∃I
16 | | | (∃x)(∃y)(Px ⊃ Qy) 11, 12-15, ∃E
17 | | | ~(∃x)(∃y)(Px ⊃ Qy) 2, R
18 | | ~(∃x)Qx 11-17, ~I
19 | | (Vx)Px ⊃ (∃x)Qx 1, R
20 | | (∃x)Qx 10, 19, ⊃E
21 | (∃x)(∃y)(Px ⊃ Qy) 2-20, RD
My basic strategy is reductio, to assume the opposite of what I want to prove. From this I must get a contraction with the premise.
The premise is a conditional, and a conditional is false only if its antecedent is true and its consequent is false. So I set out to
contradict the original premise by deriving its antecedent and the negation of its consequent from my new assumption.
To derive (Vx)Px (line 10), the premise's antecedent, I need to derive Pâ. I do this by assuming ~Pa from which I derive line 7,
which contradicts line 2. To derive ~(∃x)Qx (line 18), the negation of the premise's consequent, I assume (∃x)Qx (line 11), and
derive a contradiction, so that I can use ~I. This proceeds by using ∃E, as you can see in lines 11 to 16.
Now it's your turn to try your hand at the following exercises. The problems start out with ones much easier than the last example-
and gradually get harder!
 Exercise
6-1. Provide derivations to establish the validity of the following argument (a):
(a) (∃x)Lxx (a-1) (Vx)Lxx
(∃x)(∃y)Lxy (Vx)(Vy)Lyx
Note that the argument, (a-1) is invalid. Prove that this argument is invalid by giving a counterexample to it (that is, an
interpretation in which the premise is true and the conclusion is false). Explain why you can't get from (Vx)Lxx to (Vx)
(Vy)Lxy by using VE and VI as you can get from (∃x)Lxx to (∃x)(∃y)Lxy by using ∃E and ∃I.
b) (Vx)(Vy)Lxy b-1) (∃x)(∃y)Lxy
(Vx)Lxx (∃x)Lxx
Note that the argument, (b-1) is invalid. Prove that this argument is invalid by giving a counterexample to it. Explain why you
can't get from (∃x)(∃y)Lxy to (∃x)Lxx by using ∃E and ∃I as you can get from (Vx)(Vy)Lxy to (Vx)Lxx by using VE and VI.
c) (Vx)(Vy)Lxy d) (∃x)(∃y)Lxy e) (∃x)(Vy)Lxy e-1) (Vy)(∃x)Lxy
(Vy)(Vx)Lxy (∃y)(∃x)Lxy (Vy)(∃x)Lxy (∃x)(Vy)Lxy
Note that the converse argument, (e-1) is invalid. Prove this by providing a counterexample.
f) (Vx)Px & (Vx)Qx g) (∃x)Px & (∃x)Qx h) (Vx)Px v (Vx)Qx
(Vx)(Vy)(Px & Qy) (∃x)(∃y)(Px & Qy) (Vx)(Vy)(Px v Qy)
i) (∃x)Px v (∃x)Qx j) (∃x)(∃y)(Px v Qy) k) (Vx)(Vy)(Lxy ⊃ ~Lxy)
(∃x)(∃y)(Px v Qy) (∃x)Px v (∃x)Qx (Vy)~Lxx
l) (Vx)(Vy)(Px ⊃ Qy) m) (∃x)(∃y)(Px ⊃ Qy) n) (∃x)(Vy)(Px ⊃ Qy)
(∃x)Px ⊃ (Vx)Qx (Vx)Px ⊃ (∃x)Qx (Vx)Px ⊃ (Vx)Qx
o) (Vx)(∃y)(Px ⊃ Qy) p) (Vx)Px ⊃ (Vx)Qx q) (Vx)(Vy)(Px v Qy)
(∃x)Px ⊃ (∃x)Qx (∃x)(Vy)(Px ⊃ Qy) (Vx)Px v (Vx)Qy
r) (∃x)(Vy)Jxy
(∃y)(∃z)(Hzy & ~Py)
(Vz)(Vw)[(Jzw & ~Pw) ⊃ Gz]
(∃z)Gz
3.6.1: Multiple Quantification and Harder Problems is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
Problem 5-7(q) posed a special difficulty:
(Vx)~Fx 1 | (Vx)~Fx P
~(∃x)Fx 2 | | (∃x)Fx A
3 | | a| Fa A
4 | | | (Vx)~Fx 1, R
5 1| 2| 3| ~Fa 4, VE
???
We would like to apply ~I to derive ~(∃x)Fx. To do this, we need to get a contradiction in subderivation 2. But we can use the
assumption of subderivation 2 only by using ∃E, which requires starting subderivation 3, which uses 'a' as an isolated name. We do
get a sentence and its negation in subderivation 3, but these sentences use the isolated name 'a', so that we are not allowed to use
∃E to bring them out into subderivation 2 where we need the contradiction. What can we do?
We break this impasse by using the fact_that from a contradiction you can prove anything. Be sure you understand this general fact
before we apply it to resolving our special problem. Suppose that in a derivation you have already derived X and ~X. Let Y be any
sentence you like. You can then derive Y:
|.
|.
|.
|X
| ~X
| |~Y A
||XR
| | ~X R
| Y RD
We can use this general fact to resolve our difficulty in the following way. Since anything follows from the contradiction of 'Pa' and
'~Pa', we can use this contradiction to derive a new contradiction, 'A & ~A', which does not use the name 'a'. ∃E then licenses us to
write 'A & ~A' in derivation 2 where we need the contradiction.
To streamline our work, we will introduce several new derived rules. The first is the one I have just proved, that any sentence, Y,
follows from a contradiction:
Contradiction
|X
| ~X
| Y CD
In practice, I will always use a standard contradiction, 'A & ~A', for Y. I will also use a trivial reformulation of the rules ~I and Rd
expressed in terms of a conjunction of a sentence and its negation where up to now these rules have, strictly speaking, been
expressed only in terms of a sentence and the negation of the sentence on separate lines:
Negation Introduction Reductio
| | X | | ~X
| | Y & ~Y | | Y & ~Y
| ~X ~| | X RD
These derived rules enable us to deal efficiently with problem 5-7(q) and ones like it:
(Vx)~Fx 1 | (Vx)~Fx P
~(∃x)Fx 2 | | (∃x)Fx A
3 | | a| Fa A
4 | | | (Vx)~Fx 1, R
5 | | | ~Fa 4, VE
6 | | | A & ~A 3, 4, CD
7 | | A & ~A 2, 3-6 ∃E
8 | ~(∃x)Fx 2-7, ~I
Let's turn now to four more derived rules, ones which express the rules of logical equivalence, ~V and ~∃, which we discussed in
chapter 3. There we proved that they are correct rules of logical equivalence. Formulated as derived rules, you have really done the
work of proving them in problems 54q) and (r) and 5-7(q) and (r). To prove these rules, all you need do is to copy the derivations
you provided for those problems, using an arbitrary open sentence (. . . u . . .), with the free variable u, instead of the special case
with the open sentence 'Px' or 'Fx' with the free variable 'x'.
Negated Quantifier Rules
| ~(Vu)(. . . u . . . ) | (∃u)~(. . . u . . .)
| (∃u)~(. . . u . . .) ~V | ~(Vu)(. . . u . . . ) ∃~
| ~(Vu)(. . . u . . . ) | (Vu)~(. . . u . . .)
| (∃u)~(. . . u . . .) ~∃ | ~(∃u)(. . . u . . . ) V~
A word of caution in using these negated quantifier rules: Students often rush to apply them whenever they see the opportunity. In
many cases you may more easily see how to get a correct derivation by using these rules than if you try to make do without the
rules. But often, if you work hard and are ingenious, you can produce more elegant derivations without using the quantifier
negation rules. In the following exercises, use the rules so that you have some practice with them. But in later exercises, ' be on the
lookout for clever ways to produce derivations without the quantifier negation rules. Instructors who are particularly keen on their
students learning to do derivations ingeniously may require you to do later problems without the quantifier negation rules. (These
comments do not apply to the derived contradiction rule and derived forms of ~I and RD rules. These rules iust save work which is
invariably boring, so you should use them whenever they will shorten your derivations.)
 Exercise
6-2.
a) (Vx)Px b) (Vx)(Fx ⊃ G) c) ~(Vx)(Vy)Lxy
(Vx)~Qx (Vx)(Gx ⊃ Hx) (∃x)(∃y)~Lxy
~(∃x)(Px = Qx) ~(∃x)HX
~(∃x)Fx
d) ~(∃x)(∃y)Lxy e) ~(∃x)(Px v Qx) f) ~(Vx)(Px & Qx)
(Vx)(Vy)~Lxy (Vx)~PX & (Vx)~Qx (∃x)~Px v (∃x)~Qx
g) (Vx)[~(∃y)Rxy & ~(∃y)Ryx] h) (∃x)[Px ⊃ (Vy)(Py ⊃ Qy)]
(Vx)(Vy)~Rxy ~(∃x)QX
~(Vx)Px
i) (∃y)(∃z)[(Vx)~Rxy v (Vx)~Rxz]
~(Vy)(Vz)(∃x)(Rxy & Rxz)
3.6.2: Some Derived Rules is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
This section straightforwardly applies concepts you have already learned for sentence logic. We said that a sentence of sentence
logic is a logical truth if and only if it is true in all cases, that is, if and only if it comes out true for all assignments of truth values to
sentence letters. The concept of logical truth is the same in predicate logic if we take our cases to be interpretations of a sentence:
A closed predicate logic sentence is a Logical Truth if and only if it is true in all its interpretations.
Proof of logical truth also works just as it did for sentence logic, as we discussed in section 7-3 of Volume I. A derivation with no
premises shows all its conclusions to be true in all cases (all assignments of truth values to sentence letters in sentence logic, all
interpretations in predicate logic). A brief reminder of the reason: If we have a derivation with no premises we can always tack on
unused premises at the beginning of the derivation. But any case which makes the premises of a derivation true makes all the
derivation's conclusions true. For any case you like, tack on a premise in which that case is true. Then the derivation's conclusions
will be true in that case also:
A derivation with no premises shows all its conclusions to be logical truths.
Contradictions in predicate logic also follow the same story as in sentence logic. The whole discussion is the same as for logical
truth, except that we replace "true" with "false":
A closed predicate logic sentence is a Contradiction if and only if it is false in all its interpretations.
To demonstrate a sentence, X, to be a contradiction, demonstrate its negation, ~X, to be a logical truth. That is, construct a
derivation with no premises, with ~X as the final conclusion.
If you did exercise 7-5 (in volume I), you learned an alternative test for contradictions, which also works in exactly the same way
in predicate logic:
A derivation with a sentence, X, as its only premise and two sentences, Y and ~Y, as conclusions shows X to be a
contradiction.
Exercise 7-8 (volume I) dealt with the concept of inconsistency. Once more, the idea carries directly over to predicate logic. I state
it here, together with several related ideas which are important in more advanced work in logic:
A collection of closed predicate logic sentences is Consistent if there is at least one interpretation which makes all of them
true. Such an interpretation is called a Model for the consistent collection of sentences. If there is no interpretation which
makes all of the sentences in the collection true (if there is no model), the collection is Inconsistent.
A finite collection of sentences is inconsistent if and only if their conjunction' is a contradiction.
To demonstrate that a finite collection of sentences is inconsistent, demonstrate their conjunction to be a contradiction.
Equivalently, provide a derivation with all of the sentences in the collection as premises and a contradiction as the final
conclusion.
Finally, in predicate logic, the idea of logical equivalence of closed sentences works just as it did in sentence logic. We have
already discussed this in section 3-4:
Two closed predicate logic sentences are Logically Equivalent if and only if in each of their interpretations the two sentences
are either both true or both false.
Exercise 4-3 (volume I) provides the key to showing logical equivalence, as you already saw if you did exercise 7-9 (volume I).
Two sentences are logically equivalent if in any interpretation in which the first is true the second is true, and in any interpretation
in which the second is true the first is true. (Be sure you understand why this characterization comes to the same thing as the
definition of logical equivalence I just gave.) Consequently
To demonstrate that two sentences, X and Y, are logically equivalent, show that the two arguments, "X. Therefore Y." and
"Y. Therefore X." are both valid. That is, provide two derivations, one with X as premise and Y as final conclusion and one
with Y as premise and X as final conclusion.
 Exercise
6-3. Provide derivations which show that the following sentences are logical truths:
a) (Vx)(Vy)Lxy ⊃ (∃x)(∃y)Lxy
b) (Vx)(Gx v ~Gx)
c) (Vx)(∃y)(Ax & By) ⊃ (∃x)(Ax & Bx)
d) (∃y)[ky & (Vx)(Dx ⊃ Rxy)] ⊃ (Vx)[Dx ⊃ (∃y)(ky & Rxy)]
e) (∃x)(Vy)(Fy ⊃ Fx)
f) (Vx)(∃y)(Fy ⊃ Fx)
g) (∃x)(Vy)(Fx ⊃ Fy)
6-4. Provide derivations which show that the following sentences are contradictions:.
a) (Vx)(Ax ⊃ Bx) & (∃x)(~Bx & (Vy)Ay)
b) (Vx)(Rxb ⊃ ~Rxb) & (∃x)Rxb
c) (Vx)(Vy)Lxy & (∃y)~Lyx]
d) (Vx)(∃y)(Mx & ~My)
e) (Vx)(∃x)(Vw)(∃z)(Lxw & ~Lyz)
6-5. Provide derivations which show that the following collections of sentences are inconsistent:
a) (Vx)Kx, (Vy)~(ky v Lya)
b) (Vy)(∃y)Rxy, (∃x)(Vy)~Rxy
c) (∃x)Dx, (Vx)(Dx ⊃ (Vy)(Vz)Ryz), (∃x)(∃y)~Rxy
d) (∃x)(∃y)(Rxx & ~Ryy & Rxy), (Vx)(Vy)(Rxy ⊃ Ryx), (Vx)(Vy)(Vz)[(Rxy & Ryz) ⊃ Rxz)]
6-6. a) List the pairs of sentences which are shown to be logically equivalent by the examples in this chapter and any of the
derivations in exercises 6-1 and 6-8.
b) Write derivations which show the following three arguments to be valid. (You will see in the next part of this exercise that
there is a point to your doing these trivial derivations.)
(Vx)Rxa (Vx)Rxx (Vx)Px
(∃x)Rxa (∃x)Rxx (∃x)Px
c) Note that the three derivations you provided in your answer to (b) are essentially the same. From the point of view of these
derivations, 'Rxa' and 'Rxx' are both open sentences which we could have just as well have written as P(u), an arbitrary
(perhaps very complex) open sentence with u as its only free variable. In many of the problems in 5-5 and 5-7, I threw in
names and repeated variables which played no real role in the problem, just as in the first two derivations in (b) above. (I did so
to keep you on your toes in applying the new rules.) Find the problems which, when recast in the manner illustrated in (b)
above, do the work of proving the following logical equivalences. Here, P(u) and Q(u) are arbitrary open sentences with u as
their only free variable. A is an arbitrary closed sentence.
(Vu)(P(u) & Q(u)) is logically equivalent to (Vu)P(u) & (Vu)Q(u)
(∃u)(P(u) v Q(u)) is logically equivalent to (∃u)P(u) v (∃u)Q(u)
A ⊃ (Vu)P(u) is logically equivalent to (Vu)(A ⊃ P(u))
A ⊃ (∃u)P(u) is logically equivalent to (∃u)(A ⊃ P(u))
(Vu)P(u) ⊃ A is logically equivalent to (∃u)(P(u) ⊃ A)
(∃u)P(u) ⊃ A is logically equivalent to (Vu)(P(u) ⊃ A)
d) Prove, by providing a counterexample, that the following two pairs of sentences are not logically equivalent. (A
counterexample is an interpretation in which one of the two sentences is true and the other is false.)
(Vx)(Px v Qx) is not logically equivalent to (Vx)Px v (Vx)Qx
(∃x)(Px & Qx) is not logically equivalent to (∃x)Px & (∃x)Qx
e) Complete the work done in 6-l(c) and (d) to show that the following pairs of sentences are logically equivalent. (R is an
arbitrary open sentence with u and v as its only two free variables.)
(Vu)(Vv)R(u, v) is logically equivalent to (Vv)(Vu)R(u, v)
(∃u)(∃v)R(u, v) is logically equivalent to (∃v)(∃u)R(u, v)
6-7. Here are some harder arguments to prove valid by providing derivations. In some cases it is easier to find solutions by
using the derived rules for negated quantifiers. But in every case you should look for elegant solutions which do not use these
rules.
a) (Vx)[(∃y)(Lxy v Lyx) ⊃ Lxx]
(∃x)(∃y)Lxy
(∃x)Lxx
(Everyone who loves or is loved by someone loves themself. Someone loves someone. Therefore, someone loves
themself.)
b) (Vx)(Hx ⊃ Ax)
(Vx)[(∃y)(Hy & Txy) ⊃ (∃y)(Ay & Txy)]
(Horse are animals. Therefore horses' tails are animal's tails)
c) (Vx)(Vy)[(∃z)Lyz ⊃ Lxy]
(∃x)(∃Y)Lxy
(Vx)(Vy)Lxy
(Everyone loves a lover. Someone loves someone. Therefore, everyone loves everyone.)
d) (Vx)(Vy)([(∃x)(Rxy & ~Rxz) ⊃ Lxy]
~(∃x)Lxx
(Vx)(Vy)(~Ryx ⊃ ~Rxy)
e) (Vx){(∃x)Lxy 3 (∃x)[(Vz)Lyz & Lxy]}
(∃x)(∃y)Lxy
(∃x)(Vy)Lxy
(Everyone who loves someone loves someone who loves everyone. Someone loves someone. Therefore, someone loves
everyone.)
f) (Vx)[Px ⊃ (Vy)(Hy ⊃ Rxy)]
(∃x)(Px & (∃y)~Rxy)
~(Vx)Hx
g) (Vx)[(Ex ⊃ (Vy)(Hy ⊃ Wxy)]
(∃x)[Hx & (VY)(Dy ⊃ Wxy)]
(Vx)(Vy)(Vz)[(Wxy & Wyz) ⊃ Wxz
(Vx)[∃x ⊃ (Vy)(Dy ⊃ Wxy)]
(Any elephant weighs more than a horse. Some horse weighs more than any donkey. If a first thing weighs more than a
second, and the second weighs more than a third, the first weighs more than the third. Therefore, any elephant weighs
more than any donkey.)
h) (Vx)(∃y)(Py ⊃ Qx)
(∃y)(Vx)(Py ⊃ Qx)
Note that in general a sentence of the form (Vx)(∃y)X does not imply a sentence of the form (∃y)(Vx)X (See problem 6
-1(e)). However, in this case, the special form of the conditional makes the argument valid.
i) (∃x)Px ⊃ (∃x)Qx
(Vx)(∃y)(Px ⊃ Qy)
j) (∃y)(Px ⊃ (Vx)Qx
(Vx)(Vy)(Px ⊃ Qy)
k) (Vx){Bx ⊃ [(∃y)Lxy ⊃ (∃y)Lyz]}
(Vx)[(∃y)Lyx ⊃ Lxx]
~(∃x)Lxx
(Vx)(Bx ⊃ (Vy)~Lxy)
(All blond lovers are loved. All those who are loved love themselves. No one loves themself. Therefore, all blonds love
no one.)
l) (Vx){Fx ⊃ [Hx & (~Cx & ~Kx)]}
(Vx)[(Hx & ~(∃y)Nxy) ⊃ Dx]
(Vx)(Fx ⊃ (∃y)Nxy)
m) (Vy)(Cy ⊃ Dy)
(Vx)(∃y)[(Hx & Cx) & (Gy & Ryx)]
(∃x)Dx ⊃ (Vy)(Vz)(Ryz ⊃ Dy)
(∃x)(Gx & Cx)
n) (Vx)(Vy){(Rdy & Rxd) ⊃ Rxy]
(Vx)(Bx ⊃ Rdx)
(∃x)(Bx & Rxd)
(∃x)[Bx & (Vy)(By ⊃ Rxy)]
 chapter review Exercise
Write short explanations in your notebook for each of the following.

a) Contradiction Rule
b) Quantifier Negation Rules
c) Logical Truth of Predicate Logic
d) Test for a Logical Truth
e) Contradiction of Predicate Logic
f) Test for a Contradiction
g) Consistent Set of Sentences
h) Inconsistent Set of Sentences
i) Test for a Finite Set of Inconsistent Sentences
j) Logical Equivalence of Predicate Logic Sentences
k) Test for Logical Equivalence
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SECTION OVERVIEW
3.7: Truth Trees for Predicate Logic - Fundamentals

3.7.4: Negated Quantified Sentences

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LibreTexts.
You have already learned the truth tree method for sentence logic. And now that you have a basic understanding of predicate logic
sentences, you are ready to extend the truth tree method to predicate logic.
Let's go back to the basics of testing arguments for validity: To say that an argument is valid is to say that in every possible case in
which the premises are true, the conclusion is true also. We reexpress this by saying that an argument is valid if and only if it has no
counterexamples, that is, no possible cases in which the premises are true and the conclusion false. When we were doing sentence
logic, our possible cases were the lines of a truth table, and in any one problem there were only finitely many lines. In principle, we
could always check all the truth table lines to see if any of them were counterexamples. Often the truth tree method shortened our
work. But the trees were really just a labor-saving device. We could always go back and check through all the truth table lines.
Predicate logic changes everything. In predicate logic our cases are interpretations, and there are always infinitely many of these.
Thus we could never check through them all to be sure that there are no counterexamples. Now truth trees become much more than
a convenience. They provide the only systematic means we have for searching for counterexamples.
Everything we learned about truth trees in sentence logic carries over to predicate logic. Someone gives us an argument and asks us
whether it is valid. We proceed by searching for a counterexample. We begin by listing the premises and the denial of the
conclusion as the beginning of a tree. Just as before, if we can make these true we will have a case in which the premises are true
and the conclusion false, which is a counterexample and which shows the argument to be invalid. If we can establish that the
method does not turn up a counterexample, we conclude that there is none and that the argument is valid.
We have boiled our job down to the task of systematically looking for a case which will make true the initial sentence on a tree. In
sentence logic we did this by applying the rules for the connectives '&', 'v', '~', '⊃', and '='. These rules broke down longer
sentences into shorter ones in all the minimally sufficient possible ways which would make the longer sentences true by making the
shorter ones true. Since, in sentence logic, this process terminates in sentence letters and negated sentence letters, we got branches
which (if they do not close) make everything true by making the sentence letters and negated sentence letters along them true. In
this way you should think of each branch as a systematic way of developing a line of a truth table which will make all the sentences
along the branch true.
The tree method for predicate logic works in exactly the same way, with just one change: Each branch is no longer a way of
developing a line of a truth table which will make all the sentences along the branch true. Instead, a branch is a way of developing
an interpretation which will make all the sentences along the branch true. All you have to do is to stop thinking in terms of
building a line of a truth table (an assignment of truth values to sentence letters). Instead, start thinking in terms of building an
interpretation.
Let's see this strategy in action. Consider the example that got us started on predicate logic, way back in chapter 1:
Everybody loves Eve. (Vx)Lxe
Adam loves Eve. Lae
We begin our search for an interpretation in which the premise is true and the conclusion is false by listing the premise and the
denial of the conclusion as the initial lines of a tree:
1 (Vx)Lxe P
2 ~Lae ~C
We already know quite a bit about any interpretation of these two sentences which makes them both true. The interpretation will
have to have something called 'a' and something called 'e', and '~Lae' will have to be true in the interpretation. '~Lae' is already a
negated atomic sentence. We cannot make it true by making some shorter sentence true.
But we can make '(Vx)Lxe' true by making some shorter sentences true. Intuitively, '(Vx)Lxe' says that everybody loves Eve. In our
interpretation we have a (Adam) and e (Eve). In this interpretation, in this little novel or story of the way the world might be, we
can make it true that everybody loves Eve by making it true that Adam loves Eve and making it true that Eve loves Eve. So we
extend the branch representing our interpretation with the sentences 'Lae' and 'Lee':
a, e 1 (Vx)Lxe P
2 ~Lae ~C
3 Lae 1, v
4 Lee 1, v
X
And the branch closes! The branch includes both '~Lae' and 'Lae', where the first is the negation of the second. They cannot both be
true in an interpretation. We had to include '~Lae' to get an interpretation which makes the conclusion of the argument false. We
had to include 'Lae' to get an interpretation which makes '(Vx)Lxe' true. But no interpretation can make the same sentence both true
and false. So there is no interpretation which makes lines 1 and 2 true-there is no counterexample to the argument. And so the
argument is valid.
Let's talk more generally about how I got lines 3 and 4 out of line 1. Already, when we have just lines 1 and 2, we know that our
branch will represent an interpretation with something called 'a' and something called 'e'. We know this because our interpretation
must be an interpretation of all the sentences already appearing, and these sentences include the names 'a' and 'e'. Our immediate
objective is to make '(Vx)Lxe' true in this interpretation. But we know that a universally quantified sentence is true in an
interpretation just in case all its substitution instances are true in the interpretation. So to make '(Vx)Lxe' true in the interpretation
we must make 'Lae' and 'Lee' true in the interpretation. This is because 'Lae' and 'Lee' are the substitution instances of '(Vx)Lxel
formed with the interpretation's names, 'a' and 'e'.
Notice that I did something more complicated than simply checking line 1 after working on it and putting the annotation '1,V' after
lines 3 and 4. The rule for the universal quantifier differs in this respect from all the other rules. The other rules, when applied to a
"target" sentence, tell us to write something at the bottom of every open branch on which the target sentence appears. When this is
done, we have guaranteed that we have made the target sentence true in all possible minimally sufficient ways. We thus will never
have to worry about the target sentence again. To note the fact that we are done with the sentence, we check it.
But the rule for the universal quantifier is not like this. First, in applying the rule to a universally quantified sentence, we have to
search the branch on which the target sentence appears for names. Then, at the bottom of every open branch on which the target
sentence appears, we must instantiate the target sentence with each name which occurs along that branch. To help keep track of
which names have already been used to instantiate the target sentence, we list them as we use them.
You might think that when we have thus accounted for all the names on the branch we are done with the target sentence and can
check it. But you will see that new names can arise after first working on a universally quantified target sentence. In such a case
we must come back and work on the universally quantified sentence again. Because we must recognize the possibility of having to
return to a universally quantified sentence, we never check the sentence as a whole. Instead, we list the names which we have thus
far used in the sentence, because once a universally quantified sentence has been instantiated with a given name, we never have to
instantiate it with the same name again.
Here is a summary statement of our rule:
Rule V: If a universally quantified sentence (Vu)(. . . u . . .) appears as the entire sentence at a point on a tree, do the
following to each open branch on which (Vu)(. . . u . . .) appears. First, collect all the names s1, s2, s3, . . . that appear along
the branch. (If no name appears on the branch, introduce a name so that you have at least one name.) Then write the
substitution instances (. . . s1 . . .), (. . . s2 . . .), (. . . s3 . . .), . . . at the bottom of the branch, and write the names s1, s2, s3, . . .
to the left of (Vu)(. . . u . . .). Do not put a check by (Vu)(. . . u . . .).
Several facets of this rule bear further comment. First, in working om a universally quantified sentence on a given branch, you only
need to instantiate it with the names along that branch. If the same universally quantified sentence occurs along a second branch,
that second branch calls for use of the names that occur along that second branch. This is because each branch is going to represent
its own interpretation. Also, when instructed to write a substitution instance, (. . . s . . .), at the bottom of an open branch, you do
not need to write it a second time if it already appears.
Next, the rule instructs you to write all of the substitution instances, (. . . s1 . . .), (. . . s2 . . .), (. . . s3 . . .), . . . at the bottom of every
open path. But if the path closes before you are done, of course you can stop. Once a path closes, it cannot represent a
counterexample, and further additions will make no difference. Thus, in the last example, I could have correctly marked the path as
closed after line 3, omitting line 4. You can make good use of this fact to shorten your work by choosing to write down first the
substitution instances which will get a branch to close. But don't forget that if the branch does not close, you must list all the
substitution instances.
Finally, listing the names used to the left of (Vu)(. . . u . . .) is a practical reminder of which names you have already used to
instantiate (Vu)(. . . u . . .). But this reminder is not foolproof because it does not contain the information about which branch the
substitution instance appears on. In practice, this isn't a difficulty because in almost every problem your substitution instance will
appear on all branches. Indeed, when a universally quantified sentence appears on a branch it never hurts to introduce a substitution
instance formed with a name which had not otherwise appeared on that branch.
Let me illustrate how the rule applies when no name appears on a path. At the same time, I will show you how we will write down
counterexamples:
A1AP
~(Vx)Bx 2 ~~(Vx)Bx ~C
a3 (Vx)Bx 2, ~~
4 Ba 3, V
Invalid. Counterexample: D = {a}; Ba & A
'A' is an atomic sentence letter which we make true in the counterexample. 'A' is not a name. So when we get to line 3 and need to
instantiate '(Vx)Bx' with all the names in the interpretation we are building, we find that we don't have any names. What do we do?
Every interpretation must have at least one thing in it. So when applying the rule V to a universally quantified sentence on a branch
which has no names, we have to introduce a name to use. This is the only circumstance in which the V rule tells us to introduce a
new name. Any name will do. In this example I used 'a'.
Notice how I indicated the counterexample to the argument provided by the open branch. The counterexample is an interpretation
which makes everything along the branch true. You read the counterexample off the open branch by listing the names which occur
on the branch and the atomic and negated atomic sentences which occur along the branch. The rules have been designed so that
these shortest sentences make true the longer sentences from which they came, which in turn make true the still longer sentences
from which they came, and so on, until finally everything along the open branch is true.
 Exercise
7-1. Use the truth tree method to test the following arguments for validity. In each problem, state whether or not the argument
is valid; if invalid, give a counterexample.
a) (Vx)(kx & Jx) b) (Vx)(Fx ⊃ Gx) c) (Vx)(Cx ⊃ Ix)
Ka ~Ga Ch v Ih
Fa
d) A ⊃ (Vx)Mx e) (Vx)(Bx ⊃ Cx) f) (Vx)(Ne ≡ Px)
A (Vx)Bx Pg
Mg & Mi Ca & Cb
g) (Vx)(Kx v Ax) h) (Vx)(Dx v Gx) i) (Vx)(Sx ≡ Tx)
~kj (Vx)(Dx ⊃ Jx) Sb v ~Ta
Ad (Vx)(Gx ⊃ Jx)
Ja
j) ~Tfg v (Vx)Px
Ph ⊃ (Vx)Qx
Tfg ⊃ Qh
3.7.1: The Rule for Universal Quantification is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Somebody is blond. (∃x)Bx
Adam is blond. Ba
As we noted in chapter 2, this argument is obviously invalid. If somebody is blond, it does not follow that Adam is blond. The
blond might well be somebody else. We will have to keep the clear invalidity of this argument in mind while formulating the rule
for existentially quantified sentences to make sure we get the rule right.
Begin by listing the premise and the negation of the conclusion:
1 (∃x)Bx P
2 ~Ba ~C
As in the last example, we already know that we have an interpretation started, this time with one object named 'a'. We also know
that '~Ba' will have to be true in this interpretation. Can we extend the interpretation so as also to make '(∃x)Bx' true?
We have to be very careful here. We may be tempted to think along the following lines: An existentially quantified sentence is true
in an interpretation just in case at least one of its substitution instances is true in the interpretation. We have one name, 'a', in the
interpretation, so we could make '(3x)Bx' true by making its substitution instance, 'Ba', true. But if we do that, we add 'Ba' to a
branch which already has '-Ba' on it, so that the branch would close. This would tell us that there are no counterexamples,
indicating that the inference is valid. But we know the inference is invalid. Something has gone wrong.
As I pointed out in introducing the example, the key to the problem is that the blond might well be someone other than Adam. How
do we reflect this fact in our rules? Remember that in extending a branch downward we are building an interpretation. In so doing,
'we are always free to add new objects to the interpretation's domain, which we do by bringing in new names in sentences on the
tree. Since there is a possibility that the blond might be somebody else, we indicate this by instantiating our existentially quantified
sentence with a new name. That is, we make '(∃x)Bx' true by writing 'Bb' at the bottom of the branch, with 'b' a new name. We
bring the new name, 'b', into the interpretation to make sure that there is no conflict with things that are true of the objects which
were in the interpretation beforehand.
The completed tree looks like this:
1 (∃x)Bx P
2 ~Ba ~C
3 Bb 1, ∃, New name
New name Invalid. Counterexample: D = {a,b); ~Ba & Bb
The open branch represents a counterexample. The counterexample is an interpretation with domain D = {a,b), formed with the
names which appear on the open branch. The open branch tells us what is true about a and b in this interpretation, namely, that ~Bb
& Bb.
You may be a little annoyed that I keep stressing 'new name'. I do this because the new name requirement is a very important
aspect of the rule for existentially quantified sentences-an aspect which students have a very hard time remembering. When I don't
make such a big fuss about it, at least 50 percent of a class forgets to use a new name on the next test. By making this fuss I can
sometimes get the percentage down to 25 percent.
Here is the reason for the new name requirement. Suppose we are working on a sentence of the form (∃u)(. . . u . . .) such as
'(∃x)Bx' in our example. And suppose we try to make it true along each open branch on which it appears by writing a substitution
instance, (. . . t . . .), at the bottom of each of these branches. Now imagine, as happened in our example, that ~(. . . t . . .)-or
something which logically implies ~(. . . t . . .)-already appears along one of these branches. In the example we already had '~Ba'.
This would lead to the branch closing when in fact we can make a consistent interpretation out of the branch. We can always do this
by instantiating (∃u)(. . . u . . .) with a new name, say, s, a name which does not appear anywhere along the branch. We use this
new name in the instantiation (. . . s . . .). Then (. . . s . . .) can't conflict with a sentence already on the branch, and we are
guaranteed not to have the kind of trouble we have been considering.
Not infrequently you get the right answer to a problem even if you don't use a new name when instantiating an existentially
quantified sentence. But this is just luck, or perhaps insight into the particular problem, but insight which cannot be guaranteed to
work with every problem. We want the rules to be guaranteed to find a counterexample if there is one. The only way to guarantee
this is to write the rule for existentially quantified sentences with the new name requirement. This guarantees that we will not get
into the kind of difficulty which we illustrated with our example:
Rule ∃: If an existentially quantified sentence (∃u)(. . . u. . .) appears as the entire sentence at some point on a tree, do the
following to each open branch on which (∃u)(. . . u . . .) appears: First pick a new name, s, that is, a name which does not
appear anywhere on the branch. Then write the one substitution instance (. . . s . . .) at the bottom of the branch. Put a check
by (∃u)(. . . u . . .).
Why do we always need a new name for an' existentially quantified sentence but no new name for a universally quantified sentence
(unless there happens to be no names)? In making a universally quantified sentence true, we must make it true for all things (all
substitution instances) in the interpretation we are building. To make it true for more things, to add to the interpretation, does no
harm. But it also does no good. If a conflict is going to come up with what is already true in the interpretation, we cannot avoid the
conflict by bringing in new objects. This is because the universally quantified sentence has to be true for all the objects in the
interpretation anyway.
We have an entirely different situation with existentially quantified sentences. They don't have to be true for all things in the
interpretation. So they present the possibility of avoiding conflict with what is already true in the interpretation by extending the
interpretation, by making each existentially quantified sentence true of something new. Finally, since the rules have the job of
finding a consistent interpretation if there is one, the rule for existentially quantified sentences must incorporate this conflict-
avoiding device.
 Exercise
7-2. Test the following arguments for validity. State whether each argument is valid or invalid; when invalid, give the
counterexamples shown by the open paths.
a) A ⊃ (∃x)Gx b) (∃x)Dx c) (∃x)(Px & Qx)
A (∃x)~Dx Pa v Qb
Gb A
d) (∃x)Px e) A v B
(∃x)Qx A ⊃ (∃x)Nx
Pm ≡ Qm B ⊃ (∃x)Nx
Ng
3.7.2: The Rule for Existential Quantification is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Now let's apply our rules to some more involved examples. Let's try the argument
(Vx)Lxe v (Vx)~Lxa
~Lae
~(∃x)Lxa
I am going to write out the completed tree so that you can follow it as I explain each step. Don't try to understand the tree before I
explain it. Skip over it, and start reading the explanation, referring back to the tree in following the explanation of each step.
We begin by listing the premises and the negation of the conclusion. Our first move is to apply the rule for double negation to line
3, giving line 4. Next we work on line 1. Notice that even though '(Vx)' is the first symbol to appear on line 1, the sentence is not a
universally quantified sentence. Ask yourself (As in chapters 8 and 9 in volume I): What is the last thing I do in building this
sentence up from its parts? You take '(Vx)Lxe' and '(Vx)~Lxa' and form a disjunction out of them. So the main connective is a
disjunction, and to make this sentence true in the interpretation we are building, we must apply the rule for disjunction, just as we
used it in sentence logic. This gives line 5.
In lines 1 through 4 our tree has one path. Line 5 splits this path into two branches. Each branch has its own universally quantified
sentence which we must make true along the branch. Each branch also has '(∃x)Lxa', which is common to both branches and so
must be made true along both branches. What should we do first?
When we work on '(∃x)Lxa' we will have to introduce a new name. It is usually better to get out all the new names which we will
have to introduce before working on universally quantified sentences. To see why, look at what would have happened if I had
worked on line 5 before line 4. Looking at the right branch I would have instantiated '(Vx)~Lxa' with 'a' and 'e'. Then I would have
returned to work on line 4, which would have introduced the new name 'c'. But now with a new name 'c' on the branch I must go
back and instantiate (Vx)~Lxa' with 'c'. To make this sentence true, I must make it true for all instances. If a new name comes up in
midstream, I must be sure to include its instance. Your work on a tree is more clearly organized if you don't have to return in this
way to work on a universally quantified sentence a second time.
We will see in the next chapter that in some problems we cannot avoid returning to work on a universally quantified sentence a
second time. (It is because sometimes we cannot avoid this situation that we must never check a universally quantified sentence.)
But in the present problem we keep things much better organized by following this practical guide:
Practical Guide: Whenever possible, work on existentially quantified sentences before universally quantified sentences.
Now we can complete the problem. I work on line 4 before line 5. Line 4 is an existentially quantified sentence. The rule ∃ tells me
to pick a new name, to use this new name in forming a substitution instance for the existentially quantified sentence, and to write
this instance at the bottom of every open path on which the existentially quantified sentence appears. Accordingly, I pick 'c' as my
new name and write the instance 'Lca' on each branch as line 6. Having done this, I check line 4, since I have now ensured that it
will be made true along each open path on which it appears.
Finally, I can work on line 5. On the left branch I must write substitution instances for '(Vx)Lxe' for all the names that appear along
that branch. So below '(Vx)Lxe' I write 'Lae', 'Lee', and 'Lce', and I write the names 'a', 'e', and 'c' to the left of the target sentence
'(Vx)Lxe' to note the fact that this sentence has been instantiated with these three names. The branch closes because 'Lae' of line 7
conflicts with '~Lae' on line 2. On the right branch I have '(Vx)~Lxa'. At the bottom of the branch I write its substitution instances
for all names on the branch, giving '~Laa', '~Lea', and '-La'. Again, I write the names used to the left of the target sentence. '~Lca' is
the negation of 'Lca' on line 6. So the right branch closes also, and the argument is valid.
One more comment about this example: The new name requirement did not actually avoid any trouble in this particular case. If I
had used either of the old names 'a' or 'e', I would in this case have gotten the right answer. Moreover, the tree would have been
shorter. You may be thinking: What a bother this new name requirement is! Why should I waste my time with it in a case like this?
But you must follow the new name requirement scrupulously if you want to be sure that the tree method works. When problems get
more complicated, it is, for all practical purposes, impossible to tell whether you can safely get away without using it. The only
way to be sure of always getting the right answer is to use the new name requirement every time you instantiate an existentially
quantified sentence.
Now let's try an example which results by slightly changing the first premises of the last case:
(Vx)(Lxe v ~Lxa)
~Lae
~(∃x)Lxa
Instead of starting with a disjunction of two universally quantified sentences, we start with a universal quantification of a
disjunction:
Lines 1, 2, and 3 list the premises and the negation of the conclusion. Line 4 gives the result of applying ~~ to line 3. Looking at
line 1, we ask, - What was the very last step performed in writing this sentence? The answer: applying a universal quantifier. So it
is a universally quantified sentence. But line 4 is an existentially quantified sentence. Our practical guide tells us to work on the
existentially before the universally quantified sentence. Accordingly, I pick a new name, 'c', and use it to instantiate '(∃x)Lxa',
giving me line 5. Now I can return to line 1 and apply the rule V. At this point, the names on the branch are 'a', 'e', and 'c'. So I get
the three instances of 1 written on lines 6, 7, and 8, and I record the names used to the left of line 1. Lines 9, 10, and 11 apply the v
rules to lines 6, 7, and 8. Notice that I chose to work on line 8 before line 7. I am free to do this, and I chose to do it, because I
noticed that the disjunction of line 8 would close on one branch, while the disjunction of line 7 would not close on any branches.
We have applied the rules as far as they can be applied. No sentence can be made true by making shorter sentences true. We are left
with two open branches, each of which represents a counterexample to the original argument. Let's write these counterexamples
down.
The branch labeled (i) at the bottom has the names 'e', 'c', and 'a'. (In principle, the order in which you list information on a branch
makes no difference. But it's easiest to read off the information from the bottom of the branch up.) So I indicate the domain of
branch (i)'s interpretation by writing D = {e,c,a}. What is true of e, c, and a in this interpretation? The branch tells us that e bears L
to itself, that c bears L to e, that a does not bear L to itself, that c bears L to a and that a does not bear L to e. In short, the
interpretation is
D = {e,c,a); Lee & Lce & ~Laa & Lca & ~Lae
To read an interpretation of an open branch, you need only make a list of the branch's names and the atomic and negated atomic
sentences which appear along the branch. We use the format I have just indicated to make clear that the names are names of the
objects of the domain, and the atomic and negated atomic sentences describe what is true of these objects. Check your
understanding by reading the counterexample off the branch labeled (ii). You should get
D = {e,a,c); ~Lea & Lce & ~Laa & Lca & ~Lae
Notice that neither of these counterexamples as read off the branches constitutes complete interpretations. The branches fail to
specify some of the atomic facts that can be expressed with 'L', 'a', 'c', and 'e'. For example, neither branch tells us whether 'LC' is
true or false. We have seen the same situation in sentence logic when sometimes we had a relevant sentence letter and an open
branch on which neither the sentence letter nor its negation appeared as the entire sentence at a point along the branch. Here, as in
sentence logic, this happens when a branch succeeds in making everything along it true before completing a selection of truth
values for all relevant atomic sentences. In effect, the branch represents a whole group of interpretations, one for each way of
completing the specification of truth values for atomic sentences which the branch does not mention. But for our purposes it will be
sufficient to read off the interpretation as the branch presents it and call it our counterexample even though it may not yet be a
complete interpretation.
 Exercise
7-3. Test the following arguments for validity. State whether each argument is valid or invalid, when invalid, give the
counterexamples shown by the open paths.
a) (∃x)(Px 3 Qx) b) (∃x)Cx c) (∃x)Jx v (∃x)Kx
~(Vx)(Px & ~Qx) ~(∃x)~Cx (Vx)~Jx
~(Vx)~kx
3.7.3: Applying the Rules is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
7.4: Negated Quantified Sentences
7-4. NEGATED QUANTIFIED SENTENCES
To complete the rules for quantification, we still need the rules for the negation of quantified sentences. As always, we illustrate
with a simple example:
Lae I Lae P (3x)Lxe 2 -(3x)Lxe -C
What will make line 2 true? All we need do is to make use of the equivalence rule for a negated existential quantifier, which we
proved in section 34. (Remember: "Not one is" comes to the same thing as "All are not." If you have forgotten that material, reread
the first few paragraphs of section 34you will very quickly remember how it works.) '-(3x)Lxe' is true in an interpretation if and
only -if '(Vx)-Lxe' is true in the interpretation. So we can make '-(3x)Lxe' true by making '(Vx)-Lxe' true. This strategy obviously
will work for any negated existentially quantified sentence. So we reformulate the -3 rule for logical equivalence as a rule for
working on a negated existentially quantified sentence on a tree:
Rule -3: If a sentence of the form -(3u)(. . . u . . .) appears as the full sentence at any point on a tree, write (Vu)-(. . . u . . .)at the
bottom of every open branch on which -(3u)(. . . u . . .) appears. Put a check by -(3u)(. . . u . . .).
With this rule, we complete our problem as follows:
1 Lae P 42 -(3x)Lxe -C a, e, 3 (Vx)-Lxe 2, -3 4 -Lae 3, v 5 -Lee 3, v X Valid
Note that I did not use a new name when I worked on line 2. The new name rule does not enter anywhere in this problem because
the problem has no existentially quantified sentence as the full sentence at some point on the tree. Consequently, the rule 3 never
applies to any sentence in this problem. Line 2 is the negation of an existentially quantified sentence, and we deal with such a
sentence with our new rule, -3. Line 2 gives line 3 to which the rule V applies.
The story for the negation of a universally quantified sentence goes the same way as for a negation of an existentially quantified
sentence. Just as "not one is" comes to the same thing as "all are not," "not all are" comes to the same thing as "some are not." In
other words, we appeal to the rule -V for logical equivalence in exactly the same way as we appealed to the rule -3 for logical
equivalence. Reformulating for application to a negated universally quantified sentence on a tree, we have
Rule -V: If a sentence of the form -(Vu)(. . . u . . .) appears as the full sentence at any point on a wee, write (31)-(. . . u . . .) at the
bottom of
140 Truth Trees for Predicate Logic: Fundamentab 7-4. Negated Quantijkd Sentences 121
every open branch on which -(Vu)(. . . u . . .) appears. Put a check by -(Vu)(. . . u . . .).
Once again, this rule works because -(Vu)(. . . u . . .) is equivalent to (3u)-(. . . u . . .), as we noted in section 3-4 and as you proved
in exercise 3-5.
Here is an example to illustrate the -V rule:
(3x)Lxe J1 (3x)Lxe P J2 --(Vx)Lxe -C J3 (3x)-Lxe 2, -V 4 Lce 1, 3, New name 5 -Lde 3, 3, New name
Invalid. Counterexample: D = {d,e,c); -Lde & Lce
In this example, note that failure to follow the new name rule at step 5 would have incorrectly closed the branch. Also note that we
do not instantiate line 2 with any of the names. Line 2 is not a universally quantified sentence. Rather, it is the negation of a
universally quantified sentence which we treat with the new rule -V.
Now you have all the rules for quantifiers. It's time to practice them.
Before you go to work, let me remind you of the three most common mistakes students make when working on trees. First of all,
you must be sure you are applying the right rule to the sentence you are working on. The key is to determine the sentence's main
connective. You then apply the rule for that connective (or for the first and second connectives in case of negated sentences). You
should be especially careful with sentences which begin with a quantifier. Some are quantified sentences, some are not; it depends
on the parentheses. '(Vx)(Px > A)' is a universally quantified sentence, so the rule V applies to it. It is a universally quantified
sentence because the initial universal quantifier applies to the whole following sentence as indicated by the parentheses. By way of
contrast, '(Vx)(Lxa 3 Ba) & Lba' is not a universally quantified sentence. It is a conjunction, and the rule & is the rule to apply to it.
The second mistake to watch for especially carefully is failure to instantiate a universally quantified sentence with all the names
that appear on the sentence's branch. When a universally quantified sentence appears on a branch, you are not finally done with the
sentence until either the branch closes or you have instantiated it with all the names that appear on the branch. - - 1
Finally, please don't forget the new name requirement when you 1 work on an existentially quantified sentence. When instantiating
an existentially quantified sentence,. you use only one name, but that I name must not yet appear anywhere on the branch. 1
7-4. Use the truth tree method to test the following arguments for validity. In each problem, state whether or not the argument is
valid; if invalid give a counterexample. al) (Vx)Px & (Vx)Qx a2) (Vx)(Px & Qx) (Vx)(Px & Qx) (Vx)Px & (Vx)Qx bl ) (3x)Px v
(3x)Qx b2) (3x)(Px v Qx) (~X)(PX v Qx) (3x)Px v (3x)Qx el) A > (Vx)Px e2) (Vx)(A > Px) (Vx)(A > Px) A 3 (Vx)Px fl ) A 3
(3x)Px f2) (3x)(A > Px) (3x)(A > Px) A 3 (3x)Px 144 Tmth TWW for PndicaU Logic: FI
CHAPTER SUMMARY EXERCISES Here are the new ideas from this chapter. Make sure you understand them, and record your
summaries in your notebook. a) Rule V d) Rule -V b) Rule 3 e) Rule -3 c) New Name Requirement f) Reading an Interpretation
Off an Open Branch
7.4: Negated Quantified Sentences is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
SECTION OVERVIEW
3.8: More on Truth Trees for Predicate Logic

3.8.1: Contradictions, Logical Truth, Logical Equivalence, and Consistency
3.8.2: Truth Trees with Multiple Quantifiers
3.8.3: Three Shortcuts
3.8.4: Infinite Trees

3.8: More on Truth Trees for Predicate Logic is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
8.1: Contradictions, Logical Truth, Logical Equivalence, and Consistency
In this section we are going to see how to apply the truth tree method to test predicate logic sentences for some familiar properties.
This will be little more than a review of what you learned for sentence logic. The ideas are all the same. All we have to do is to
switch to talking about interpretations where before we talked about lines of a truth table.
Let's start with logical contradiction. In sentence logic we say that a sentence is a contradiction if and only if it is false in all
possible cases, where by "possible cases" we mean assignments of truth values to sentence letters-in other words, lines of the
sentence's truth table. Recharaaerizing the idea of a possible case as an interpretation, we have
A dosed predicate logic sentence is a CmrtradicCion if and only if it is false in all of its interpretations.
The truth tree test for being a contradiction also carries over directly from sentence logic. The truth tree method is guaranteed to
lind an interpretation in which the initial sentence or sentences on the tree 'are true, if there is such an interpretation. Consequently
To test a sentence, X, for being a contradiction make X the first he of a truth tree. If there is an interpretation which makes X true,
the tree method will find such an interpretation, which will provide a counterexample to X being a contradiction. If all branches
close, there is no interpretation in which X is true. In this case, X is false in all of its interpretations; that is, X is a contradiction.
Here is a very simple example. We test '(3x)(Bx & -Bx)' to see whether it is a contradiction:
J1 (3x)(Bx & -Bx) S (The sentence being tested) 42 (Ba & -Ba) 1, 3, New name 3 Ba 2, 4 - Ba 2, X
The sentence is a contradiction.
The idea of a logical truth carries over from sentence logic in exactly the same way. In sentence logic a sentence is a logical truth if
it is true for all possible cases, understood as all truth value assignments. Now, taking possible cases to be interpretations, we say
A closed predicate logic sentence is a Logccal Truth if and only if it is true in all of its interpretations.
To determine whether a sentence is a logical truth, we must, just as we do in sentence logic, look for a counterexample-that is, a
case in which the sentence is false. Consequently To test a predicate logic sentence, X, for being a logical truth, make -X the first
line of a tree. If there is an interpretation which makes -X true, the tree method will find such an interpretation. In such an
interpretation, X is false, so that such an interpretation provides a counterexample to X being a logical truth. If all branches close,
there is no interpretation in which -X is true, and so no interpretation in which X is false. In this event, X is true in all of its
interpretations; that is, X is a logical truth. Again, let's illustrate with a simple example: Test '(3x)Bx v (3x)-Bx' to see if it is a
logical truth: -[(gx)~x v (~x)-Bx] -S (The negation of the sentence being tested) J 2 -(3x)Bx 1, -v 43 -(ax)-Bx 1, -v a4 (Vx)-Bx 2, -3
a5 Wx)--Bx 3, -3 6 - Ba 4, v 7 --Ba 5, v X The sentence is a logical truth. The tree shows that there are no interpretations in which
line 1 is true. Consequently, there are no interpretations in which the original sentence (the one which we negated to get line 1) is
false. So this original sentence is a logical truth. Notice that I had to introduce a name when I worked on line 4. Line 4 is a
universally quantified sentence, and having no name at that point I h'ad to introduce one to start my try at an interpretation. Line 5
is another universally quantified sentence, and when I worked on it, I already had the name 'a'. So I instantiated line 5'with 'a'. At
no place on this tree did the new name requirement of the rule 3 apply. This is because at no place on the tree is the entire sentence
an existentially quantified sentence. In particular, the sentences of lines 2 and 3 are negated existentially quantified sentences, not
existentially quantified sentences, so the rule 3 and the new name requirement do not apply to them. It's time to talk about logical
equivalence. We already discussed this subject in section 3-4, which you may want to review at this point. For completeness, let's
restate the definition: Two closed predicate logic sentences are Logically Equivalent if and only if in each of their interpretations
the two sentences are either both true or both false. Do you remember how we tested for logical equivalence of sentence logic
sentences? Once again, everything works the same way in predicate logic. Two closed predicate logic sentences have the same
truth value in one of their interpretations if and only if their biconditional is true in the interpretation. So the two sentences will
agree in truth value in all of their interpretations if and only if their biconditional is true in all of their interpretations-that is, if and
only if their biconditional is a logical truth. So to test for logical equivalence we just test for the logical truth of the biconditional:
To determine whether the closed predicate logic sentences, X and Y, are logically equivalent, test their biconditional, X=Y, for
logical truth. That is, make -(X=Y) the first line of a tree. If all branches close, -(X=Y) is a logical truth, so that X and Y are
logically equivalent. If there is an open branch, X and Y are not logically equivalent. An open branch will be an interpretation in
which one of the two sentences is true and the other false, so that such an open branch provides a counterexample to X and Y being
logically equivalent.
Here is another way in which you can test two sentences, X and Y, for logical equivalence. Consider the argument "X. Therefore
Y." with X as premise and Y as conclusion. If this argument is invalid, there is a counterexample, an interpretation in which X is
true and Y is false. Thus if "X. Therefore Y." is invalid, X and Y are not logically equivalent, and a counterexample to the argument
is also a counterexample which shows X and Y not to be logically equivalent. The same goes for the argument "Y. Therefore X.",
this time taking the second sentence, Y, as premise and the first sentence, X, as conclusion. If this argument is invalid there is a
counterexample, that is, an interpretation in which Y is true and X is false, and hence again a counterexample to X and Y being
logically equivalent.
Now, what happens if both the arguments "X. Therefore Y." and "Y. Therefore X." are valid? In this event every interpretation in
which X is ' true is an interpretation in which Y is true (the validity of "X. Therefore Y."), and every interpretation in which Y is
true is an interpretation in which X is true (the validity of "Y. Therefore X."). But that is just another way of saying that in each
interpretation X and Y have the same truth value. If whenever X is true Y is true and whenever Y is true X is true, we can't have a
situation (an interpretation) in which one is true and the other is false. Thus, if "X. Therefore Y." and "Y. Therefore X." are both
valid, X and Y are logically equivalent:
To determine whether the closed predicate logic sentences, X and Y, are logically equivalent, test the two arguments "X. Therefore
Y." and "Y. Therefore X." for validity. If either argument is invalid, X and Y are not logically equivalent. A counterexample to
either argument is a counterexample to the logical equivalence of X and Y. If both arguments are valid, X and Y are logically
equivalent.
In fact, the two tests for logical equivalence really come to the same thing. To see this, suppose we start out to determine whether X
and Y are logically equivalent by using the first test. We begin a tree with -(X=Y) and apply the rule -=: A 2 X-X I,-= 3 -Y Y I,-=
Now notice that the left-hand branch, with X followed by -Y, is just the way we start a tree which tests the validity of the argument
"X. Therefore Y.". And, except for the order of -X and Y, the right-hand branch looks just like the tree which we would use to test
the validity of the argument "Y. Therefore X.". So far as the right-hand branch goes, this order makes no difference. Because we are
free to work on the lines in any order, what follows on the right-hand branch is going to look the same whether we start it with -X
followed by Y or Y follow by -X.
In sum, lines 2 and 3 in our tree are just the beginning of trees which test the validity of "X. Therefore Y." and "Y. Therefore X.".
Thus the completed tree will contain the trees which test the arguments "X. Therefore Y." and "Y. Therefore X.". And, conversely,
if we do the two trees which test the arguments "X. Therefore Y." and "Y. Therefore X." we will have done all the work which
appears in the tree we started above, the tree which tests X=Y for logical truth. So the two ways of determining whether X and Y
are logically equivalent really involve the same work.
If you did all of exercise 7-4 you have already tested 11 pairs of sentences for logical equivalence! In each of these pairs you tested
two arguments, of the form "X. Therefore Y." and "Y. Therefore X.". Using our new test for logical equivalence, you can use your
work to determine in each of these problems whether or not the pair of sentences is logically equivalent.
Truth trees also apply to test sets of sentences for consistency. Recall from section 9-2 in volume I that a set of sentence logic
sentences is consistent if and only if there is at least one case which makes all of the sentences in the set true. Interpreting cases as
interpretations, we have
A Model of a set of one or more predicate logic sentences is an interpretation in which all of the sentences in the set are true.
A set of one or more predicate logic sentences is consistent just in case it has at least one model, that is, an interpretation in which
all of the sentences in the set are true.
To test a finite set of predicate logic sentences for consistency, make the sentence or sentences in the set the initial sentences of a
tree. If the tree closes, there is no interpretation which makes all of the sentences true together (no model) and the set is
inconsistent. An open branch gives a model and shows the set to be consistent.
Every truth tree test of an argument is also a test of the consistency of the argument's premises with the negation of the argument's
conclusion. An argument is valid if and only if its premises are inconsistent with the negation of the argument's conclusion. In other
words, an argument is invalid if and only if its premises are consistent with the negation of its conclusion. Thus one can view the
truth tree test for argument validity as a special application of the truth tree test for consistency of sets of sentences. (If you have
any trouble understanding this paragraph, review exercise 9-7 in volume I. Everything in that exercise applies to predicate logic in
exactly the same way as it does to sentence logic.)
EXERCISES
8-1. Test the following sentences to determine which are logical truths, which are contradictions, and which are neither. Show your
work and state your conclusion about the sentence. Whenever you find a counterexample to a sentence being a logical truth or a
contradiction, give the counterexample and state explicitly what it is a counterexample to. (Vx)Dx v (3x)-Dx b)(Vx)Kx & (3x)-Kx
(Vx)Nx v (Vx)-Nx d)(Vx)Jx & (Vx)-Jx (3x)Bx v (3x)-Bx f )(3x)Px & (3x)-Px [(Vx)Gx v (Vx)Hx] & -(Vx)(Gx v Hx) (Vx)(Kx v
Jx) 3 [(3x)-Kx 3 (3x)Jxl [(Vx)Mx 3 (Vx)-Nx] & (3x)(-Mx & Nx) [(3x)Hx 3 (Vx)(Ox 3 Nx)] 3 [(3x)(Hx & Ox) > (Vx)Nx] (3x)[-Sx
& (Gx v Kx)] v [(Vx)Gx 3 (Vx)(Sx v Kx)] [(Vx)Fx v (Vx)Gx] = [(3x)-Fx & -(Vx)Gx]
8-2. Use the truth tree method to test the following sets of sentences for consistency. In each case, state your conclusion about the
set of sentences, and if the set of sentences is consistent, give a model. a) (3x)Px. (3x)-Px b) (Vx)Px, (Vx)-Px c) (Vx)Px, (3x)-Px d)
(Vx)-Fx, (Vx)Sx, (3x)[(-Fx 3 Sx) 3 Fx] e) (3x)Gx & (3x)Qx, -(3x)(Gx & Qx) f) (Vx)(Gx v Qx), -[(Vx)Gx v (Vx)Qx] g) (3x)Ux v
Dx), (Vx)Ux 3 -Hx), (Vx)(Dx 3 Hx), (Vx)Ux = (Dx v Hx)]
8-3. Explain the connections among consistency, logical truth, and logical contradiction.
8-4. By examining your results from exercise 7-4(a) through (k), determine which pairs of sentences are logically equivalent and
which are not. This is more than an exercise in mechanically applying the test for logical equivalence. For each pair of sentences,
see if you can understand intuitively why the pair is or is not logically equivalent. See if you can spot any regularities. logical truth.
To determine this, we must look for a counterexample to its being a logical truth, that is, an interpretation in which it is false. So we
- make the negation of the sentence we are testing the first line of a tree. Here are the first six lines: J1 -(3x)[Lxa 3 WyILyal -S a 2
pix)-[Lxa 3 Wy)Lya] 1, -3 J3 -[Laa 3 Wy)Lyal 2, V 4 Laa 3, -> J5 -(Vy)Lya 3, -3 6 (3~)-Lya 5, -V We begin with the negation of
the sentence to be tested. Line 2 applies the rule for a negated quantifier, and line 3 instantiates the resulting universally quantified
sentence with the one name on the branch. Lines 4, 5, and 6 are straightforward, first applying the rule -3 to line 3 and then the rule
-V to line 5. But now the rules we have been using all along are going to force on us something we have not seen before. Applying
the rule 3 to line 6 forces us to introduce a new name, say, 'b', giving '-Lba' as line 7. This has repercussions for line 2. When we
worked on line 2 we instantiated it for all the names we had on that branch at that time. But when we worked on line 6 we got a
new name, 'b'. For the universally quantified line 2 to be true in the interpretation we are building, it must be true for all the names
in the interpretation, and we now have a name which we did not have when we worked on line 2 the first time. So we must return
to line 2 and instantiate it again with the new name, 'b'. This gives line 8. Here, with the final two easy steps, is the way the whole
tree looks: -(3x)ILxa 3 (Vy)Lyal Wx)-[Lxa 3 (Vy)Lyal -[Laa 3 (Vy)Lya] Laa -(Vy)Lya (3~)-Lya -Lba -[Lba 3 (Vy)Lya] Lba -
(Vy)Lya X -S 1, -3 2, v 3, -3 3, -3 5, -v 6, 3, New name 2, v 8, -3 8, -3
8.1: Contradictions, Logical Truth, Logical Equivalence, and Consistency is shared under a not declared license and was authored, remixed,
8.2: Truth Trees with Multiple Quantifiers
8-2. TRUTH TREES WITH MULTIPLE QUANTIFIERS
The sentence is a logical truth. In the last chapter I tried to keep the basics in the limelight by avoiding the complication of multiple
quantifiers. Multiple quantifiers involve no new rules. But they do illustrate some circumstances which you have not yet seen.
Suppose I asked you to determine whether '(3x)[Lxa 3 (Vy)Lya]' is a logical truth. To determine this, we must look for a
counterexample to its being a logical truth, that is, an interpretation in which it is false. So we - make the negation of the sentence
we are testing the first line of a tree. Here are the first six lines: J1 -(3x)[Lxa 3 WyILyal -S a 2 pix)-[Lxa 3 Wy)Lya] 1, -3 J3 -[Laa 3
Wy)Lyal 2, V 4 Laa 3, -> J5 -(Vy)Lya 3, -3 6 (3~)-Lya 5, -V We begin with the negation of the sentence to be tested. Line 2 applies
the rule for a negated quantifier, and line 3 instantiates the resulting universally quantified sentence with the one name on the
branch. Lines 4, 5, and 6 are straightforward, first applying the rule -3 to line 3 and then the rule -V to line 5. But now the rules we
have been using all along are going to force on us something we have not seen before. Applying the rule 3 to line 6 forces us to
introduce a new name, say, 'b', giving '-Lba' as line 7. This has repercussions for line 2. When we worked on line 2 we instantiated
it for all the names we had on that branch at that time. But when we worked on line 6 we got a new name, 'b'. For the universally
quantified line 2 to be true in the interpretation we are building, it must be true for all the names in the interpretation, and we now
have a name which we did not have when we worked on line 2 the first time. So we must return to line 2 and instantiate it again
with the new name, 'b'. This gives line 8. Here, with the final two easy steps, is the way the whole tree looks:
The sentence is a logical truth.
We do not need to work on line 10 because line 7 is the negation of line 9, and the branch thus closes. Indeed, I could have omitted
line 10. There was no way for us to avoid going back and working on line 2 a second time. There was no way in which we could
have worked on the
130 More on Truth Trees for Predicate Logic 8-3. Three Shortc& 13 1
existentially quantified sentence of line 6 before working on line 2 the first time. The sentence of line 6 came from inside line 2.
Thus we could get line 6 only by instantiating line 2 first. You should always be on the watch for this circumstance. In multiple
quantified sentences it is always possible that an existentially quantified sentence will turn up from inside some larger sentence.
The existential quantifier will then produce a new name which will force us to go back and instantiate all our universally quantified
sentences with the new name. This is why we never check a universally quantified sentence.
Here is another example. We will test the following argument for validity: Everyone loves someone. (V~Y~Y Anyone who loves
someone (Vx)[(3y)Lxy 3 Lxxl loves themself. Wx)Lxx Everyone loves themselves. (Vx)(3y)Lxy (Vx)[(3y)Lxy 3 Lxxl -(Vx)Lxx
(3x)-Lxx -Laa (3y)Lay Lab (3y)Lay 3 Laa A -(3y)Lay Laa (Vy)-Lay x -Lab X Valid P P - C 3, -v 4, 3 1, v 6, 3, New name 2, v
Getting this tree to come out as short as I did requires some care in choosing at each stage what to do next. For example, 1 got line
8 by instantiating the universally quantified line 2 with just the name 'a'. The rule V requires me to instantiate a universally
quantified sentence with all the names on the branch. But it does not tell me when I have to do this. I am free to do my instantiating
of a universally quantified sentence in any order I like, and if I can get all branches to close before I have used all available names,
so much the better. In the same way, the rule V requires that I return to instantiate lines 1 and 2 with the new name 'b', which arose
on line 7. But the rule doesn't tell me when I have to do that. With a combination of luck and skill in deciding what to do first, I can
get all branches to close before returning to line 1 or line 2 to instantiate with 'b'. In this circumstance I can get away without using
'b' in these sentences.
However, in any problem, if I instantiate with fewer than all the names and I have failed to close all the branches, then I must return
and put - the names not yet used into all universally quantified sentences which appear on open branches.
8.2: Truth Trees with Multiple Quantifiers is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
8.3: Three Shortcuts
In general, it is very dangerous to do two or more steps at the same time, omitting explicitly to write down one or more steps which
the rules require. When you fail to write down all the steps, it becomes too easy to make mistakes and too hard to find mistakes
once you do make them. Also, omitting steps makes it extremely hard for anyone to correct your papers. However, there are three
step-skipping shortcuts which are sufficiently clear-cut that, once you are proficient, you may safely use. You should begin to use
these shortcuts only if and when your instructor says it is alright to do so.
Suppose you encounter the sentence -(Vx)(Vy)Lxy on a tree. The rules as I have given them require you to proceed as follows: J1 -
(Vx)(Vy)Lxy J2 (3x1-(Vy)Lxy 1, -V J3 -(Vy)Lay 2, 3 J4 Qy)-Lay 3, -V 5 -Lab 4, 3, New name
Now look at line 2. You may be tempted to apply the rule -V inside the sentence of line 2. In most cases, applying a rule inside a
sentence is disastrous. For example, if you should try to instantiate '(Vx)Bx' inside ' -[(Vx)Bx v A]', you will make hash of your
answer. But in the special case of the rule for negated quantifiers, one can justify such internal application of the rule.
In the example we have started, an internal application of the rule -V gives the following first three lines of the tree:
In fact, we can sensibly skip line 2 and simply "push" the negation sign through both quantifiers, changing them both. Our tree now
looks like this:
We can do the same with two consecutive existential quantifiers or a mixture of quantifiers:
Indeed, if a sentence is the negation of a triply quantified sentence, you could push the negation sign through all three quantifiers,
changing each quantifier as you go.
Why is this shortcut justified? To give the reason in a very sketchy way, the subsentences to which we apply the negated quantifier
rule are logically equivalent to the sentences which result from applying the rule. In short, we are appealing to the substitution of
logical equivalents. To make all this rigorous actually takes a little bit of work (for the reasons explained in exercise 34, and I will
leave such niceties to your instructor or to your work on logic in a future class.
Here is another shortcut: Suppose you have a multiple universally quantified sentence, such as '(Vx)(Vy)Lxy', on a tree that already
has several names, say, 'a' and 'b'. Following the rules explicitly and instantiating with all the names is going to take a lot of writing:
a, b, 1 (Vx)(tly)Lxy a, b, 2 Wy)Lay 1, V 3 Laa 2,V 4 Lab 2, V a, b, 5 Wy)Lby 1, V 6 Lba 5, V 7 Lbb 5, V
In general, it is not a good idea to skip steps, because if we need to look for mistakes it is often hard to reconstruct what steps we
skipped. But we won't get into trouble if we skip steps 2 and 5 in the above tree:
1 Wx)Wy)Lxy (a, a), (a, b), (b, a), (a, b) 2 Laa 1, V, V 3 Lab 1, V, V 4 Lba 1, V, V 5 Lbb 1, V, V
(In noting on line 1 what names I have used in instantiating the doubly universally quantified sentence '(Vx)(Vy)Lxy', I have
written down the pain of names I have used, being careful to distinguish the order in which they occurred, and I wrote them to the
right of the line simply because I did not have room on the left.)
In fact, if you think you can get all branches to close by writing down just - some of the lines 2 to 5, write down only what you
think you will need. But if in doing so you do not get all branches to close, you must be sure to come back and write down the
instances you did not include on all open branches on which line 1 occurs.
What about using the same shortcut for a doubly existentially quantified sentence? That is, is it all right to proceed as in this mini-
tree?
1 (3x)(3y)Lxy 2 Lab 1, 3, 3, New names
This is acceptable if you are very sure that the names you use to instantiate both existential quantifiers are both new names, that is,
names that have not yet appeared anywhere on the branch.
Our last shortcut does not really save much work, but everyone is tempted by it, and it is perfectly legitimate: You may drop double
negations anywhere they occur, as main connectives or within sentences. This step is fully justified by the law of substitution of
logical equivalents from sentence logic.
A final reminder: You may use these shortcuts only if and when your instructor judges that your proficiency is sufficient to allow
you to use them safely. Also, do not try to omit other steps. Other shortcuts are too likely to lead you into errors.
8.3: Three Shortcuts is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
8.4: Infinite Trees
8-4. INFINITE TREES
So far, truth trees have provided a mechanical means for testing arguments for validity and sentences for consistency, logical truth,
logical contradiction, or logical equivalence. But if logic were a purely mechanical procedure it could not have enough interest to
absorb the attention of hundreds of logicians, mathematicians, and philosophers. However, in one way, the truth tree method is not
purely mechanical.
Let's test the sentence '(Vx)(3y)Lxy' for consistency. That is, let's look for an interpretation which makes it true:
d, c, b, a, 1 Wx)(3y)Lxy S 42 (3y)Lay 1, V 3 Lab 2, 3, New name J4 (3y)Lby 1, V 5 Lbc 4, 3, New name J6 (3x)Lcy 1, V 7 Lcd 6,
3, New name 1
34 More on Truth Trees for Predicate Logic 8-4. Injinite Trees 135
The tree starts with the universally quantified sentence '(Vx)(3y)Lxy'. At this point the tree has no names, so I pick a name, 'a', and
use it to instantiate 1, giving 2. Line 2 is an existentially quantified sentence, so I must instantiate it with a new name, 'b', giving
line 3. But having the new name, 'b', I must go back and make 1 true for 'b'. This produces 4, again an existentially quantified
sentence, which calls up the new name, 'c'. Now I must go back once more to 1 and instantiate it with 'c', producing the existentially
quantified 6 and the new name, 'd', in 7. I am going to have to return to 1 with 'd'. By this time you can see the handwriting on the
wall. This procedure is never going to end! The tree is just going to keep on growing. What does this mean? What has gone wrong?
Your immediate reaction may be that the troublesome new name requirement has clearly gummed up the works. The tree keeps on
growing without end only because we keep needing to use a new name each time the existentially quantified sentence comes up.
It's the new name from the existentiall; quantified sentence which has to be used to instantiate the universally quantified sentence
which produces a new existentially quantified sentence which . . . and so on.
On the other hand, we know that without the new name requirement, the method will not always do its job. So what should we
make of this situation?
First, let us understand what this infinite tree represents. It represents an interpretation with infinitely many names. The tree goes
on forever, and corresponding to it we have a domain, D = {a,b,c,d, . . .}, and a specification that Lab & Lbc & Lcd & . . . . In other
words, each object bears the relation L to the next.
Perhaps you have noticed that we can simplify the interpretation by supposing that there really is only one object to which all of the
infinitely many names refer. This gives an interpretation in which there is only one thing, a, such that 'Laa' is true. In this
interpretation it is true that for each thing (there is only one, namely a) there is something (namely a itself) such that Laa.
This is the last straw! you may well be saying to yourself. The new name requirement horribly complicates things, in this case by
unnecessarily making the tree infinite. In this case the requirement prevents the method from finding the simplest interpretation
imaginable which makes the original sentence true!
In fact we could rewrite the rules so that they would find this simple interpretation in the present case. But then the new rules
would have some analogue of the new name requirement, an analogue which would provide a similar difficulty in some other
problem. Let us say a bit more specifically what the difficulty comes to. In the infinite tree we have seen, it is very easy to tell that
the tree will go on forever. And it is easy to figure out what infinite interpretation the infinite tree will provide. But in more
complicated problems it will not be so easy. The rub is that there can be no mechanical way which will apply to all cases to tell us,
in some limited number of steps, whether or not the tree will eventually close. - There will always be cases in which, even after
thousands of pages, we will still not know whether the tree will close in just a few more steps or whether it will go on forever.
One can show that this problem, or some analogue of it, will come up no matter how we write the rules of logic. Indeed, this is one
of the many exciting things about logic. The rules can be mechanically applied. But logic will always leave room for insight and
ingenuity. For in difficult problems the mechanical rules may never tell you whether the tree will eventually close. In these cases
you can find out only by being clever.
Unfortunately, we must stop at this point. But I hope that all of you have been able to get a glimpse of one of the ways in which
logic can be exciting. If you continue your study of logic beyond this course, you will come to understand why and how the
problem of infinite trees is really a very general fact about all formulations of predicate logic, and you will understand the essential
limitations of predicate logic in a much more thorough way.
EXERCISES
8-5. Test the following sentences to determine which are logical truths, which are contradictions, and which are neither. Show your
work and state your conclusion about the sentence. Whenever you find a counterexample to a sentence being a logical truth or a
contradiction, give the counterexample and state explicitly what it is a counterexample to.
8-6. Use the truth tree method to test the following sets of sentences for consistency. In each case, state your conclusion about the
sets of sentences, and if the set of sentences is consistent, give a model. 136 More on Td Treesfor Pwdhtc Logic
8-7. Use the truth tree method to determine which of the following are logically equivalent. Give counterexamples as appropriate.
and Wx)Px & Wx)Qx and (3x)Px & (3x)Qx and Wx)Px v Wx)Qx and (3x)Px v (3x)Qx and (3x)Px 3 Wx)Qx and (Vx)Px 3 (3x)Qx
and Wx)Px 3 Wx)Qx and (3x)Px 3 (3x)Qx (3y)Wx)ky and (3x)Bx & Wy)Hy
8-8. Are the following arguments valid? If not, give a counterexample. (All cats are animals. Therefore all tails of cats are tails of
animals.)
CHAPTER SUMMARY EXERCISES Here are items from this chapter for you to review and record in summary: a) Contradiction
b) Truth Tree Test for Contradictions c) Logical Truth d) Truth Tree Test for Logical Truth e) Logical Equivalence f) Truth Tree
Test for Logical Equivalence g) Consistency h) Model i) Truth Tree Test for Consistency j) Three Permissible Truth Tree Shortcuts
k) Infinite Trees 8-4. I* Trees 137
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SECTION OVERVIEW
3.9: Identity, Functions, and Definite Descriptions

3.9.1: Identity
3.9.2: Inference Rules for Identity
3.9.3: Functions
3.9.4: Definite Descriptions
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3.9.1: Identity
Identity, Functions, and Definite Descriptions
9-1. IDENTITY
Clark Kent and Superman would seem to be entirely different people. Yet it turns out they are one and the same. We say that they
are Identical. Since identity plays a special role in logic, we give it a permanent relation symbol. We express 'a is identical to b'
with 'a= b', and the negation with either '-(a = b)' or 'a # b'.
'=' is not a connective, which forms longer sentences from shorter sentences. '= ' is a new logical symbol which we use to form
atomic sentences out of names and variables. But as we did with the connectives, we can explain exactly how to understand '=' by
giving truth conditions for closed sentences in interpretations. Just follow the intuitive meaning of identity: To say that s= t is to say
that the thing named by s is identical to the thing named by t; that is, that the names s and t refer to the same object. (Logicians say
that s and t have the same referent, or that they are Co-Referential.) To summarize
'=' flanked by a name or a variable on either side is an atomic sentence. If s and t are names, t= s is true in an interpretation if s and
t name the same thing. s=t is false if s and t name different things. The negation of an identity sentence can be written either as -
(s=t) or as sf t.
Identity is easy to understand, and it is extraordinarily useful in expressing things we could not say before. For example, '(3x)'
means that there is one or more x such that. . . . Let's try to say that there is exactly one x such that . . . , for which we will introduce
the traditional expres- . sion '(3x!)' (read "E shriek"). We could, of course, introduce '(3x!)' as a new connective, saying, for
example, that '(3x!)Bx1 is true in an interpretation just in case exactly one thing in the interpretation is B. But, with the help of
identity, we can get the same effect with the tools at hand, giving a rewriting rule for '(gx!)' much as we did for subscripted
quantifiers in chapter 4.
To say that there is exactly one person (or thing) who is blond is to say, first of all, that someone is blond. But it is further to say
that nothing else is blond, which w'e can reexpress by saying that if anything is blond, it must be (that is, be identical to) that first
blond thing. In symbols, this is '(3x)[Bx & (Vy)(By 3 y = x)r.
Before giving a general statement, I want to introduce a small, new expository device. Previously I have used the expression '(. . . u
. . .)' to stand for an arbitrary sentence with u the only free variable. From now on I am going to use expressions such as P(u) and
Q(u) for the same thing:
Boldface capital letters followed by a variable in parentheses, such as P(u) and Q(u), stand for arbitrary sentences in which u, and
only u, may be free. Similarly, R(u,v) stands for an arbitrary sentence in which at most u and v are free.
In practice P(u), Q(u), and R(u,v) stand for open sentences with the indicated variable or variables the only free variable. However,
for work in Part I1 of this Volume, I have written the definition to accommodate degenerate cases in which u, or u and v, don't
actually occur or don't occur free. If you are not a stickler for detail, don't worry about this complication: Just think of P(u), Q(u),
and R(u,v) as arbitrary open sentences. But if you want to know why I need, to be strictly correct, to cover degenerate cases, you
can get an idea from exercise 13-3. With this notation we can give the E! rewrite rule:
Rule for reuniting 31: For any open formula P(u) with u a free variable, (3u!)P(u) is shorthand for (3u)[P(u) & (v)(P(v) 3 v=u)],
where v is free for u in P(u), that is, where v is free at all the places where u is free in P(u).
Once you understand how we have used '=' to express the idea that exactly one of something exists, you will be able to see how to
use '=' to express many related ideas. Think through the following exemplars until you see why the predicate logic sentences
expresses what the English expresses:
There are at least two x such that Fx: (3x)(3y)[x f y & Fx & Fy]. 1
There are exactly two x such that Fx: (3x)(3y){xf y & Fx & Fy & (Vz)[Fz 3 (z =x v z=y)]).
There are at most two x such that Fx: (Vx)(Vy)(Vz)[(Fx & Fy & Fz) 3 (x= y v x= z v y = z)].
We can also use '=' to say some things more accurately which previously we could not say quite correctly in predicate logic. For
example, when we say that everyone loves Adam, we usually intend to say that everyone other than Adam loves Adam, leaving it
open whether Adam loves himself. But '(Vx)' means absolutely everyone (and thing), and thus won't exempt Adam. Now we can
use '=' explicitly to exempt Adam:
Everyone loves Adam (meaning, everyone except possibly Adam himself): (Vx)(x f a 3 Lxa).
In a similar way we can solve a problem with transcribing 'Adam is the tallest one in the class'. The problem is that no one is taller
than themself, so we can't just use '(Vx)', which means absolutely everyone. We have to say explicitly that Adam is taller than all
class members except Adam.
Adam is the tallest one in the class:
(Vx)[(Cx & xf a) 3 Tax].
To become familiar with what work '=' can do for us in transcribing, make sure you understand the following further examples:
Everyone except Adam loves Eve: (Vx)(xf a 3 Lxe) & -be. Only Adam loves Eve: (Vx)(Lxe = x = a), or Lae & (Vx)(Lxe 3 x = a).
Cid is Eve's only son: (Vx)(Sxe = x=c), or Sce & (Vx)(Sxe 3 x=c).
EXERCISES 9-1. Using Cx: x is a clown, transcribe the following: a) There is at least one clown. b) There is no more than one
clown. c) There are at least three clowns. d) There are exactly three clowns. e) There are at most three clowns.
9-2. Use the following transcription guide: a: Adam Sxy: x is smarter than y e: Eve Qxy: x is a parent of y Px: x is a person
Oxy: x owns y Rx: x is in the classroom Mxy: x is a mother of y Cx: x is a Cat Fx: x is furry Transcribe the following: Three
people love Adam. (Three or more) Three people love Adam. (Exactly three) Eve is the only person in the classroom.
Everyone except Adam is in the classroom. Only Eve is smarter than Adam. Anyone in the classroom is smarter than Adam.
Eve is the smartest person in the classroom. Everyone except Adam is smarter than Eve. Adam's only cat is furry. Everyone has
exactly one maternal grandparent. No one has more than two parents.
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9.2: Inference Rules for Identity
9-2. INFERENCE RULES FOR IDENTITY
You now know what '=' means, and you have practiced using '=' to say various things. You still need to learn how to use '=' in
proofs. In this section I will give the rules for '=' both for derivations and for trees. If you have studied only one of these methods of
proof, just ignore the rules for the one you didn't study.
As always, we must guide ourselves with the requirement that our rules be truth preserving, that is, that when applied to sentences
true in an interpretation they should take us to new sentences also true in that interpretation. And the rules need to be strong enough
to cover all valid arguments.
To understand the rules for both derivations and trees, you need to appreciate two general facts about identity. The first is that
everything is self-identical. In any interpretation which uses the name 'a', 'a=al will be true. Thus we can freely use statements of
self-identity. In particular, selfidentity should always come out as a logical truth.
The second fact about identity which our rules need to reflect is
144 Identity, Functions, ad Dejinitc Descriptionr 9-2. Inference Ruler for Identity 143
just this: If a=b, then anything true about a is true about b, and conversely.
I'm going to digress to discuss a worry about how general this second fact really is. For example, if Adam believes that Clark Kent
is a weakling and if in addition Clark Kent is Superman, does it follow that Adam believes that Superman is a weakling? In at least
one way of understanding these sentences the answer must be "no," since Adam may well be laboring under the false belief that
Clark Kent and Superman are different people.
Adam's believing that Clark Kent is a weakling constitutes an attitude on Adam's part, not just toward a person however named, but
toward a person known under a name (and possibly under a further description as well). At least this is so on one way of
understanding the word 'believe'. On this way of understanding 'believe', Adam's attitude is an attitude not just about Clark Kent
but about Clark Kent under the name 'Clark Kent'. Change the name and we may change what this attitude is about. What is
believed about something under the name 'a' may be different from what is believed about that thing under the name 'b', whether or
not in fact a=b.
This problem, known as the problem of substitutivity into belief, and other so-called "opaque" or "intensional" contexts, provides a
major research topic in the philosophy of language. I mention it here only to make clear that predicate logic puts it aside. An
identity statement, 'a= b', is true in an interpretation just in case 'a' and 'b' are names of the same thing in the interpretation. Other
truths in an interpretation are specified by saying which objects have which properties, which objects stand in which relations to
each other, and so on, irrespective of how the objects are named. In predicate logic all such facts must respect identity.
Thus, in giving an interpretation of a sentence which uses the predicate 'B', one must specify the things in the interpretation, the
names of these things, and then the things of which 'B' is true and the things of which 'B' is false. It is most important that this last
step is independent of which names apply to which objects. Given an object in the interpretation's domain, we say whether or not
'B' is true of that object, however that thing happens to be named. Of course, we may use a name in saying whether or not 'B' is true
of an object-indeed, this is the way I have been writing down interpretations. But since interpretations are really characterized by
saying which predicates apply to which objects, if we use names in listing such facts, we must treat names which refer to the same
thing, so-called Co-Referential Names, in the same way. If 'a' and 'b' are names of the same thing and if we say that 'B' is true of
this thing by saying that 'Ba' is true, then we must also make 'Bb' true in the interpretation.
In short, given the way we have defined truth in an interpretation, if 'a= b' is true in an interpretation, and if something is true of 'a'
in the interpretation, then the same thing is true of 'b' in the interpretation. - Logicians say that interpretations provide an
Extensional Semantics for predicate logic. "Semantics" refers to facts concerning what is true, and facts concerning meaning,
insofar as rules of meaning have to do with what comes out true in one or another circumstance. "Extensional" means that the
Extension of a predicate-the collection of things of which the predicate is true-is independent of what those things are called. Parts
of English (e.g., 'Adam believes Clark Kent is a weakling') are not extensional. Predicate logic deals with the special case of
extensional sentences. Because predicate logic deals with the restricted and special case of extensional sentences, in predicate logic
we can freely substitute one name for another when the names name the same thing.
Now let's apply these two facts to write down introduction and elimination rules for identity in derivations. Since, for any name, s,
s=s is always true in an interpretation, at any place in a derivation which we can simply introduce the identity statement s = s:
I Where r is any name.
If s does not occur in any governing premises or assumptions, it occurs arbitrarily and gets a hat. To illustrate, let's demonstrate that
'(Vx)(x= x)' is a logical truth:
The second fact, that co-referential names can be substituted for each other, results in the following two rules:
The indicated substitutions may be for any number of occurrences of the name substituted for.
To illustrate, let's show that '(Vx)(Vy)[x = y 3 (Fx 3 Fy)]' is a logical truth:
144 Identity, Functions, and Definite Descrif~tMN 9-2. Inference Ruks for Identity 145
Now we'll do the rules for trees. We could pmceed much as we did with derivations and require that we write identities such as
'a=a' wherever this will make a branch close. An equivalent but slightly simpler rule instructs us to close any branch on which there
appears a negated self-identity, such as 'afa'. This rule makes sense because a negated self-identity is a contradiction, and if a
contradiction appears on a branch, the branch cannot represent an interpretation in which all its sentences are true. In an exercise
you will show that this rule has the same effect as writing selfidentities, such as 'a=a', wherever this will make a branch close.
Rule f : For any name, s, if sf s appears on a branch, close the branch.
Let's illustrate by proving '(Vx)(x =x)' to be a logical truth:
J1 -(Vx)(x=x) -S J2 (3x)(xf X) 1, -V 3 afa 2, 3 X
The second rule for trees looks just like the corresponding rules for derivations. Substitute co-referential names:
Rule =: For any names, s and t, if s=t appears on a branch, substitute s for t and t for s in any expression on the branch, and write
the result at the bottom of the branch if that sentence does not already appear on the branch. Cite the line numbers of the equality
and the sentence into which you have substituted. But do not check either sentence. Application of this rule to a branch is not
completed until either the branch closes or until all such substitutions have been made.
Let's illustrate, again by showing '(Vx)(Vy)[x= y 3 (Fx 3 Fy)]' to be a logical truth:
J1 -(Vx)(Vy)lx=y 3 (Fx 3 Fy)l J2 (3x)(3y)-Ix=y 3 (Fx 3 Fy)l J3 -[a=b > (Fa 3 Fb)] 4 a=b 45 -(Fa 3 Fb) 6 Fa 7 -Fb 8 -Fa X
Before closing this discussion of identity, I should mention that identity provides an extreme example of what is called an
Equivalence Relation. Saying that identity is an equivalence relation is to attribute to it the following three characteristics:
Identity is Reflexive. Everything is identical with' itself: (Vx)(x=x). In general, to say that relation R is reflexive is to say that
(Vx)R(x,x).
Identity is Symmetric. If a first thing is identical with a second, the second is identical with the first: (Vx)(Vy)(x= y 3 y = x). In
general, to say that relation R is symmetric is to say that (Vx)(Vy)(R(x,y) 3 R(y,x)).
Identity is Transitive. If a first thing is identical with a second, and the second is identical with a third, then the first is identical
with the third: (Vx)(Vy)(Vz)[(x = y & y = z) 3 x = z]. In general, to say that relation R is transitive is to say that (Vx)(Vy)(Vz)
[(R(x,y) & R(y,z)) 3 R(x,z)].
You can prove that identity is an equivalence relation using either derivations or trees.
Here are some other examples of equivalence relations: being a member of the same family, having (exactly) the same eye color,
being teammates on a soccer team. Items which are related by an equivalence relation can be treated as the same for certain
purposes, depending on the relation. For example, when it comes to color coordination, two items with exactly the same color can
be treated interchangeably. Identity is the extreme case of an equivalence relation because "two" things related by identity can be
treated as the same for all purposes.
Equivalence relations are extremely important in mathematics. For example two numbers are said to be Equul Modulo 9 if they
differ by an exact multiple of 9. Equality modulo 9 is an equivalence relation which is useful in checking your arithmetic (as you
know if you have heard of the "rule of casting out 9s").
9-3. Show that each of the two =E rules can be obtained from the other, with the help of the =I rule.
146 Identify, Fundmas, and Definite Descriptions 9-4.
Show that the rule f is equivalent to requiring one to write, on each branch, self-identities for each name that occurs on the branch.
Do the following three exercises using derivations, trees, or both:
9-5. Show that the following are logical truths:
9-6. Show that (3x)(Vy)(Fy = y = x) and (3x!)Fx are logically equivalent.
9-7. Prove that = is an equivalence relation.
9-8. Show the validity of the following arguments: a) (Vx)(x= a 3 Fx) b) Fa C) (3x)(Fx & x=a) Fa (Vx)(x = a 3 Fx) Fa d) (Vx)(x= a
3 Fx) e) Pa fl a=b (VX)(FX 3 ~b) (3y)(y=a&y=b)
9-9. 1 stated that being teammates on a soccer team is an equivalence relation. This is right, on the assumption that no one belongs
to more than one soccer team. Why can the relation, being teammata on a soccer team, fail to be an equivalence relation if someone
belongs to two teams? Are there any circumstances under which being teammates on a soccer team is an equivalence relation even
though one or more people belong to more than one team?
9.2: Inference Rules for Identity is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
9.3: Functions
9-3. FUNCTIONS
Often formal presentations of functions leave students bewildered. But if you have done any high school algebra you have an
intuitive idea of a function. So let's start with some simple examples from algebra.
For our algebraic examples, the letters 'x', 'y', and 'z' represent variables for numbers. Consider the expression 'y = 2x + 7'. This
means that if you put in the value 3 for x you get the value 2 x 3 + 7 = 13 for y. If you put in the value 5 for x, you get the value 2 x
5 + 7 = 17 for y. Thus the expression 'y = 2x + 7' describes a rule or formula for calculating a value for y if you give it a value for x.
The formula always gives you a definite answer. Given some definite value for x, there is exactly one value for y which the formula
tells you how to calculate.
Mathematicians often use expressions like 'f(x)' for functions. Thus, instead of using the variable y in the last example, I could have
written 'f(x) = 2x + 7' This means exactly what 'y = 2x + 7' means. When you put in a specific number for x, 'f(x)' serves as a name
for the value y, so that we have y = flx). In particular, 'f(3) is a name for the number which results by putting in the value 3 for x in
2x + 7. That is, 'f(3)' is a name for the number 13, the number which results by putting in the value 3 for x in f(x) = 2x + 7.
This is all there is to functions in logic. Consider the name 'a'. Then 'f(a)' acts like another name. To what does 'f(a)' refer? That
depends, of course, on what function f( ) is, which depends on how 'f( )' is interpreted. In specifying an interpretation for a sentence
in which the function symbol 'f( )' occurs, we must give the rule which tells us, for any name s, what object f(s) refers to. When we
deal with interpretations in which there are objects with no names, this must be put a little more abstractly: We must say, for each
object (called the Argument of the function), what new object (called the Value of the function) is picked out by the function f( )
when f( ) is applied to the first object. The function must be well defined, which means that for each object to which it might be
applied, we must specify exactly one object which the function picks out. For each argument there must be a unique value.
So far I have talked only about one place functions. Consider the example of the mathematical formula 'z = 3x + 5y - 8'. which we
can also write as 'z = g(x,y)' or as 'g(x,y) = 3x + 5y - 8'. Here g( , ) has two '
148 Identity, Functions, and Dejnite Descripciorrr 9-3. Functions 149 arguments.
You give the function two input numbers, for example, x = 2 and y = 4, and the function gives you a single, unique output-in this
case, the number 3 x 2 + 5 x 4 - 8 = 18. Again, the idea carries over to logic. If 'g( , )' is a two place function symbol occurring in a
sentence, in giving an interpretation for the sentence we must specify the unique object the function will pick out when you give it
a pair of objects. If our interpretation has a name for each object the same requirement can be expressed in this way: For any two
names, s and t, 'g(s,t)' refers to a unique object, the one picked out by the function g( , ) when g( , ) is applied to the arguments s
and t. We can characterize functions with three, four, or any number of argument places in the same kind of way.
To summarize
The interpretation of a one place function specifies, for each object in the interpretation's domain, what object the function picks
out as its value when the function is applied to the first object as its argument. The interpretation of a two place function similarly
specifies a value for each pair of arguments. Three and more place functions are interpreted similarly.
Incidentally, the value of a function does not have to differ from the argument. Depending on the function, these may be the same
or they may be different. In particular, the trivial identity function defined by (Vx)(f(x) = X) is a perfectly well-defined function.
In the last sentence I applied a function symbol to a variable instead of a name. How should you understand such an application? In
an interpretation, a name such as 'a' refers to some definite object. A variable symbol such as 'x' does not. Similarly, 'f(a)' refers to
some definite object, but Xx)' does not. Nonetheless, expressions such as 'f(x)' can be very useful. The closed sentence '(Vx)Bf(x)'
should be understood as saying that every value of 'f(x)' has the property named by 'B'. For example, let us understand 'Bx' as 'x is
blond' and 'f(x)' as referring to the father of x. That is, for each person, x, f(x) is the father of x, so that Xa)' refers to Adam's father,
Xe)' refers to Eve's father, and so on. Then '(Vx)Bf(x)' says that everyone's father is blond.
In sum, function symbols extend the kind of sentences we can write. Previously we had names, variables, predicate symbols, and
connectives. Now we introduce function symbols as an extension of the category of names and variables. This involves the new
category called Tern:
We extend the vocabulary of predicate logic to include Function Symboh, written with lowercase italicized letters followed by
parentheses with places for writing in one, two, or more arguments.
All names and variables are Tern. A function symbol applied to any term or terms (a one place function symbol applied to one
term, a two place function symbol applied to two terms, etc.) is again a term. Only such expressions are terms.
In forming sentences, terms function exactly as do names and variables. One may be written after a one place predicate, two after a
two place predicate, and so on.
Do not confuse function symbols (lowercase italicized letters followed by parentheses with room for writing in arguments) with
such expressions as P(u) and R(u,v). These latter expressions are really not part of predicate logic at all. They are part of English
which I use to talk about arbitrary open predicate logic sentences.
Notice that these definitions allow us to apply functions to functions: If 'f( )' is a one place function symbol, 'f(f(a))' is a well-
defined term. In practice, we leave out all but the innermost parentheses, writing 'f(f(a))' as tff(a)' What does such multiple
application of a function symbol mean? Well, if f(x) = x2, then ff(x) is the square of the square of x. If x = 3, thenff(3) = (3')' = 9* =
81. In general, you determine the referent ofthat is, the object referred to by -ma)' as follows: Look up the referent of 'a'. Apply the
function f to that object to get the referent of 'f(a)'. Now apply f a second time to this new object. The object you get after the
second application off is the referent of tff(a)'.
Function symbols can be combined to form new terms in all kinds of ways. If 'f( )' is a one place function symbol and 'g( , )' is a
two place function symbol, the following are all terms: 'f(a)', Xy)', 'g(a,x)', 'fg(a,x)'- that is, flg(aA1. 'g[f(a), f(b)l', and gf.f(x),
g(a,b)l'.
We need one more definition:
A term in which no variables occur is called a Consant or a Constant Term.
Only constant terms actually refer to some specific object in an interpretation. But closed sentences which use nonconstant terms
still have truth values. In applying the truth definitions for quantifiers, we form substitution instances, substituting names for
variables within function symbols as well as elsewhere. Thus, in applying the definition for the truth of a universally quantified
sentence in an interpretation to '(Vx)Laf(x)', we look at the substitution instances '4(a)', 'Laf(b)', 'Laf(c)', and so on. We then look to
see if the relation L holds between a and the object f(a), between a and the object f(b), and so on. Only if all these instances hold is
'(Vx)Laf(x)' true in the interpretation.
The rules for functions simply reflect the fact that constant terms formed by applying function symbols to other constant terms have
definite referents, just as names do. However, the generality of these new referring terms may be restricted. For example, the
constant function f defined by (Vx)(f(x) = a) can only refer to one thing, namely, a. Thus, when it is important that nothing be
assumed about a constant term we must use a name and not a function symbol applied to another constant term.
150 Identity, Functions, Md Definite Descriptions
For derivations this means that we should treat constant terms all alike in applying the rules VE and 31. In applying 3E, our
isolated name must still be a name completely isolated to the subderivation to which the 3E rule applies. (Strictly speaking, if you
used an isolated function symbol applied to an isolated name, no difficulty would arise. But it's simpler just to let the isolated name
requirement stand as a requirement to use an isolated name.)
In applying VI only names can occur arbitrarily. For example, we must never put a hat on a term such as 'f(a)'. The hat means that
the term could refer to absolutely anything, but often the value of a function is restricted to only part of an interpretation's domain.
So we can't apply VI to a function symbol. However, if a name appears in no governing premise or assumption and occurs as the
argument of a function symbol, we can apply VI to the name. For example, if 'a' appears in no governing premise or assumption,
we could have 'Bf(2)' as a line on a derivation, to which we could apply VI to get '(x)Bf(x)'. To summarize
In derivations, treat all constant terms alike in applying VE and 31. Apply VI and 3E only to names.
Let's try this out by showing that '(Vx)(3y)(f(x) = y)' is a logical truth. This sentence says that for each argument a function has a
value. The way we treat functions in giving interpretations guarantees that this statement is true in all interpretations. If our rules
are adequate, this fact should be certified by the rules:
Note that this derivation works without any premise or assumption. =I allows us to introduce the identity of line 1. Since 'a' does
not occur in any governing premise or assumption, it occurs arbitrarily, although the larger term 'f(a)' does not occur arbitrarily. 'a'
could refer to absolutely anything-that is, the argument to which the function is applied could be any object at all. However, the
result of applying the function f to this arbitrary object might not be just anything. In line 2 we apply 31 to the whole term 'f(a)', not
just to the argument 'a'. This is all right because we are existentially, not universally, generalizing. If f(5) =f(l), then f(2) is identical
with something. Finally, in line 3, we universally generalize on the remaining arbitrarily occurring instance of 'a'.
Let's try something harder. '(Vx)(3y)lf(x) = y & (Vz)(f(x) = z 3 z = y)]' says that for each argument the function f has a value and
furthermore this value is unique. Again, the way we treat functions in giving interpretations guarantees that this statement is true in
all interpretations. So our rules had better enable us to show that this sentence is a logical truth:
One more example will illustrate 3E and VE as applied to terms using function symbols. Note carefully how in applying VE the
constant term to use in this problem is not a name, but 'f(a)', a function symbol applied to a name:
Similar thinking goes into the rules for trees. All constant terms act as names when it comes to the rule V. But for the rule 3 we
want a name that could refer to anything in the interpretation-that was the reason for requiring that the name be new to the branch.
So for 3 we need a new name, which must be a name, not a function symbol, applied to another constant term:
In trees, instantiate all universally quantified sentences with all constant terms that occur along the branch, unless the branch closes.
Instantiate each existentially quantified sentence with a new name.
Let us illustrate the new rules with the same sentence as before, '(Vx)(3y)lf(x) = y & (Vz)(f(x) = z 3 z = y)]'. As I mentioned, this
sentence says that f has a unique value for each argument. Since the way we treat functions in giving interpretations ensures that
this sentence is true in all interpretations, our rules had better make this sentence come out to be a logical truth:
152 Identity, Funcfimrr, and Definite Desmitk 9-4. Definite Desmitions 153 -
(Vx)(3y){f(x) = y & (Vz)[f(x) = z 3 z = yl} (3x)(Vy)-{f(x) = y & (Vz)[f(x) = z 3 z = yl}
Notice that to get everything to close I used the term 'f(a)' in substituting into line 3. Also, note that the right branch does not close
at line 9. Line 9 is not, strictly speaking, the negation of line 8 since, strictly speaking, 'f(a) = b' and 'b = f(a)' are different
sentences.
The occurrence of functions in trees has an unpleasant feature. Suppose that a universally quantified sentence such as '(Vx)Pf(x)'
appears on a tree. This will be instantiated, at least once, say, with 'a', giving 'Pf(a)'. But now we have a new constant, 'f(a)', which
we must put into '(Vx)Pf(x))' giving 'Pff(a)'. This in turn gives us a further constant, 'ff(a)'-and clearly we are off on an infinite
chase. In general, open trees with function symbols are infinite when, as in '(Vx)Pf(x)', a function symbol occurs as a nonconstant
term inside the scope of a universal quantifier
EXERCISES
9-10. Provide derivations andor trees to establish that the following are logical truths:
9-11. Provide derivations andor trees to establish the validity of the following arguments:
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9.4: Definite Descriptions
9-4. DEFINITE DESCRIPTIONS
Let's transcribe
(1) The one who loves Eve is blond.
We need a predicate logic sentence which is true when (1) is true and false when it is false. If there is exactly one person who loves
Eve and this person is blond, (1) is true. If this person is not blond, (1) clearly is false. But what should we say about (1) if no one
loves Eve, or more than one do?
If no one, or more than one love Eve, we surely can't count (1) as true. If we insist that every sentence is true or false, and since (1)
can't be true if none or more than one love Eve, we will have to count (1) as false under these conditions. Thinking about (1) in this
way results in transcribing it as
(la) (3x!)(Lxe & Bx).
which is true if exactly one person loves Eve and is blond, and is false if such a person exists and is not blond or if there are none or
more than one who love Eve.
From a perspective wider than predicate logic with identity we do not have to take this stand. We could, instead, suggest that there
being exactly one person who loves Eve provides a precondition for, or a Presupposition of, the claim that the one who loves Eve is
blond. This means that the condition that there is exactly one person who loves Eve must hold for (1) to be either true or false. If
the presupposition holds-if there is exactly one person who loves Eve-then (1) is true if this unique person is blond and false if he
or she is not blond. If the presupposition fails-if there is none or more than one who love Eve-then we say that (1) is neither true
151 Idmtity, Functions, and Dejinite Descriptions 9-4. Dejinite Desm'pth 155
nor false. One can design more complex systems of logic in which to formalize this idea, but predicate logic with identity does not
have these resources. Hence, (la) is the best transcription we can provide.
Grammatically, 'the one who loves Eve' functions as a term. It is supposed to refer to something, and we use the expression in a
sentence by attributing some property or relation to the thing purportedly referred to. We can mirror this idea in predicate logic by
introducing a new kind of expression, (The u)P(u), which, when there is a unique u which is P, refers to that object. We would then
like to use (The u)P(u) like a name or other constant term in combination with predicates. Thus we would transcribe
(1) as (lb) B(The x)Lxe.
Read this as the predicate 'B' applied to the "term" '(The x)Lxe'. 'The one who loves Eve' and '(The x)Lxe' are called Dejinite
Descriptions, respectively in English and in logic. Traditionally, the definite description formBut what should we say in predicate
logic about the transcription of (2)? We can see (2) as the negation of (1) in two very different ways. We can see (2) as the definite
description '(The x)Lxe applied to the negated predicate '-B' in which case we have (2a) -B(The x)Lxe, rewritten as (3x!)(Lxe & -
Bx). When we think of (1) and (2) this way, we say that the definite description has Primary Occurrence or Wide Scope. Or we can
see (2) as the negation of the whole transcribed sentence: (2b) -[B(The x)Lxe], rewritten as -(3x!)(Lxe & Bx). Thinking of (1) and
(2) in this second way, we say that the definite description has Secondary Occurrence or Narrow Scope. When transcribing an
English sentence with a definite description into logic, you will always have to make a choice between treating the definite
description as having ing operator, (The u), is written with an.upside-down Greek letter iota, primary or secondary occurrence. like
this:
(LU)P(U).
Here are some examples of definite descriptions transcribed into predicate logic: a) The present king of France: (The x)Kx. b) The
blond son of Eve: (The x)(Bx & Sxe). c) The one who loves all who love themselves:
(The x)(Vy)(Lyy > Lxy).
But we can't treat (The x)P(x) like an ordinary term, because sometimes such "terms" don't refer. Consequently, we need a
rewriting rule, just as we did for subscripted predicates and '(gx!)', to show that expressions like (lb) should be rewritten as (la):
Rule for rewriting Definite Descriptions Using '(The u)': Q[(The u)P(u)] is shorthand for (3u!)[P(u) & Q(u)], where P(u) and Q(u)
are open formulas with u the only free variable.
This treatment of definite descriptions works very smoothly, given the limitations of predicate logic. It does, however, introduce an
oddity about the negations of sentences which use a definite description. How should we understand
(2) The one who loves Eve is not blond.
Anyone who holds a presupposition account will have no trouble with (2): They will say that if the presupposition holds, so that
there is just one person who loves Eve, then (2) is true if the person is not blond and false if he or she is blond. If the presupposition
fails, then (2), just as (I), is neither true nor false.
EXERCISES Transcription Guide a: Adam Dx: x is dark-eyed e: Eve Fxy: x is a father of y c: Cain Sxy: xisasonofy Bx: x is blond
Cxy: x is more clever than y Lxy: x loves y
9-12. Transcribe the following. Expressions of the form (The u) and (gu!) should not appear in your final answers. The son of Eve
is blond. The son of Eve is more clever than Adam. Adam is the father of Cain. Adam loves the son of Eve. Adam loves his son.
Cain loves the blond. The paternal grandfather of Adam is dark-eyed. The son of Eve is the son of Adam. The blond is more clever
than the dark-eyed one. The most clever son of Adam is the father of Eve. The son of the father of Eve is more clever than the
father of the son of Adam.
9-13. Transcribe the negations of the sentences of exercise 9-12,, once with the definite description having primary occurrence and
once with secondary occurrence, indicating which transcription is which. Comment on how you think the notions of primary and
secondary occurrence should work when a sentence has two definite descriptions.
CHAPTER SUMMARY EXERCISES This chapter has introduced the following terms and ideas. Summarize them briefly. Identity
Referent Co-Referential @PI) Self-Identity Extensional Extensional Semantics Rule =I for Derivations Rule = E for Derivations
Rule = for Trees Rules # for Trees Reflexive Relation Symmetric Relation Transitive Relation Equivalence Relation Function One
Place Function Two and Three Place Functions Arguments of a Function Function Symbols Term Constant, or Constant Term
Rules for Function Symbols in Derivations Rules for Function Symbols in Trees Presupposition Definite Description Rewrite Rule
for Definite Descriptions Primary Occurrence (Wide Scope) of a Definite Description Secondary Occurrence (Narrow Scope) of a
Definite Description
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SECTION OVERVIEW
3.10: Metatheory - The Basic Concepts

3.10.1: Object Language and Metalanguage
3.10.3: Soundness and Completeness
3.10.4: Some Further Notation and Definitions

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10.1: Object Language and Metalanguage
Metatheory The Basic Concepts 10-1. OBJECT LANGUAGE AND METALANGUAGE In metatheory, we analyze and prove
facts about logic, as opposed to using logic. To proceed clearly, we must bear in mind that the language in which we do logic is
distinct from the language in which we study logicthat is, that the language of sentence and predicate logic is distinct from English.
The distinction has been implicit throughout the text. It is time to make this distinction explicit and give it a name. Since the
language of sentence and predicate logic is the language we study and talk about, we call it an Olrjct language. An OljcGt
Language is a language we study and talk abut Our object language is the language of sentence and predicate logic. Our object
language has an infinite stock of sentence letters, names, one place predicates, two place predicates, and in general, n-place
predicates. (In section 15-5 we also add function symbols.) We contrast our object language with the language, called a
Mehhnguage, which we use to talk about our object language. Our metalanguage is English, to which we add a few convenient
items. Most of these you have already seen. For example, think about how we have been using boldface capital 'X' and 'Y' to range
over sentences in the object language. In so doing, we are really using 'X' and 'Y as a simple extension of English, as a new
linguistic device which we use in talking about the lan- 158 Metafheoiy: The Basic Concepts 10-1. Object Language ad
Metalanguage guage of sentence and predicate logic. We have used 's', r', 'u', 'v', 'P(u)', and 'R(u,v)' similarly. Since these are
variables used in the metalanguage to range over object language sentences, names, variables, and open sentences, we call them
Metavariables. I will now add three more kinds of metavariables to be used as part of English in talking about our object language.
I will use boldface script capitals 'X', 'Y', and '2' to talk generally about sets of object language sentences. A set of sentences is just
a collection of one, two, or more sentences where the order of the sentences does not matter. I will also include among sets of
sentences infinite sets, with infinitely many sentences in them, and the somewhat funny case of the Empty Set, that is, the
degenerate case in which the set contains nothing. Next, I will use 'I', 'J', . . . as metavariables ranging over interpretations. When,
as in chapter 15, we will be concerned with predicate logic sentences, interpretations will be described by a generalization of the
idea I introduced in chapter 2. For chapters 11 to 14, in which we will be concerned only with sentence logic, interpretations will
just be assignments of truth values to atomic sentence letters, that is, specifications of conditions which determine truth values for
sentence logic sentences. I will use 'I' and 'J' as part of English to talk generally about interpretations, as when I say that two
sentences, X and Y, are logically equivalent if and only if, for each I which is an interpretation of both X and Y, either X and Y are
both true in I or X and Y are both false in I. As a last metavariable I will use 'T' to range over truth trees. I will also add to English
the special symbol 1' as an abbreviation of the word 'Therefore'. Z\X stands for the argument which has sentences in the set Z as its
premises and X as its conclusion. This is exactly what I have previously written as "Z. Therefore X." I did not previously introduce
1' as an abbreviation for 'therefore' because I wanted to be sure that you did not mistakenly think that 1' was part of the object
language. But now that we have made the object languagelmetalanguage distinction explicit, I can introduce 1' as an abbreviation
for 'therefore' and ask you to be careful to understand that 1' is an abbreviation in English, not a connective of the object language
we are studying. To summarize A Metulanguuge is a language, distinct from the object language, which we use to talk about the
object language. Our metalanguage is English, augmented with metavariables as follows: 'X', 'Y', 'Z', . . . range over object
language sentences; 'X, 'I", 'Z', . . . range over sets of object language sen- *Only after typesetting made large-scale changes in type
a practical impossibility, I learned that the compositor's capital boldface italic was almost indistinguishable from the roman
boldface type. However, I have used Z everywhere as my metavariable for sets of sentences, with only two minor exceptions
(where I use W); and Z never occurs as a metavariable for sentences. By remembering that Z ranges over sets of sentences. I hope
that the reader will be able to make the needed contrast. I regret not having provided a truly distinctive typeface. tences*; 's', 't', . . .
range over names in the object language; 'u', 'v', . . . range over variables of the object language; 'P(u)' and 'R(u, v)' . . . range over
sentences of the object language in which u (or u and v) may be free; 'I1, 'J', . . . range over interpretations; and T' ranges over trees.
We also use 9' as an abbreviation for 'therefore' in the metalanguage, so that 'Z\X stands for the argument with premises taken from
the sentences in the set Z and conclusion the sentence X. To understand better the interplay between object and metalanguage, you
also need to understand the distinction between Use and Mention. Let's talk for a moment about Adam: In so doing I mention (that
is, I refer to) this person. I might say about Adam that he is blond. Now, let us talk, not about the person, Adam, but about this
person's name, 'Adam'. For example, I might say that 'Adam' is spelled with four letters. Note how I accomplished this. To talk
about the name, I take the name and enclose it in single quotation marks. If I use the name without quotes, I use the name to
mention (that is, to talk about) the person. If I use the name enclosed in quotes, I use the quoted name-really a name of the name-to
mention (talk about) the name. Throughout this text I have tried hard (but not always successfully!) to observe the distinction
between use and mention. Thus, when in the text I have talked about an object language sentence, such as 'A&B', I have been
careful always to enclose it in quotes. When such a sentence is displayed as an example, like this I omit the quotes. This is because
of the convention, universal in logic and philosophy, that offsetting a formal expression functions just like quoting it, so that you
know that we are talking about what has been displayed rather than using what is displayed to make a statement or reference. In
contrast, when I use a metavariable I do not put quotes around it. Thus I might say that if the sentence X is a conjunction, then X
contains the symbol I&'. Notice that there are no quotes around the boldface letter. This is because I was using it to make a general
statement, not mentioning the letter. In contrast, I do use quotes when I mention (that is, talk about) the boldface letter, as in the
following statement: In the previous example I used the symbol 'X' as an example of how metavariables can be used. Now let's
look at a problematic case. Suppose I say that any sentence of the form X&Y is true just in case X and Y are both true. I have,
writing in the metalanguage, used 'X' and 'Y' to make a general statement. But in so doing I used the expression 'X&Y', which
contains the object language symbol '&'. Furthermore, in some sense I made a statement about the symbol '&'. I didn't assert a
conjunction. Instead, I talked about all sentences which have '&' as their main connective. 160 Metatheory: The Basic Concepts 10-
2. Syntrur and Senurnties 161 Here's the problem. I was tacitly talking about the symbol I&'. But I didn't quote it. I really should
have used quotes around '&'. But it's not clear how I could do that without putting quotes around 'X' and 'Y', which I was using and
not mentioning! Philosophers have invented some fancy notation to make more precise what is going on in such cases. But
introducing this further notation would be to pass the point of diminishing returns for our present needs. Instead, I am simply going
to ask you to understand that such "mixed" cases of use and mention, formed with metalanguage boldface variables and object
language connectives, are a device which I use to talk generally about all sentences of the indicated form. I must mention one
further twist in our conventions. Our object language provides a very precise and compact way of expressing truth functional facts.
It would be a shame not to be able to use this compact notation in our metalanguage and to have to write everything out in
imprecise, long-winded English. So we will occasionally allow ourselves the luxury of using expressions of the object language to
make statements as part of the metalanguage. You can think of the metalanguage, English, as incorporating or being extended by a
copy of the object language. You can always tell when I talk about, or mention, logical notation as part of the object language, for
in these cases I will always quote or display the expressions. When I use, as opposed to mention, logical notation as part of the
metalanguage, the notation will not be quoted. Furthermore, when I use, as opposed to mention, logical notation as part of the
metalanguage, I will use the notation with metalanguage variables. You can spot these metalanguage variables as belonging to the
metalanguage because I always write them in boldface. Strictly speaking, my notation does not distinguish between use of logical
notation in the metalanguage 10-1. For each of the underlined expressions, say whether the expression is being used as part of the
metalanguage, mentioned as part of the metalanguage, used as part of the object language, or mentioned as part of the object
language. a) If there is a proof of a sentence then there is a proof of the sentence XVY. b) The sentence '{Vx)(Bx v -Bx)' is a logical
truth. C) Any sentence of the form (Vu)lP(u) v -P(u)l is a logical truth. d) 'r is a metavariable. and the mixed use-mention cases
which I described two paragraphs back. But in practice this imprecision causes no confusion. 10-2. SYNTAX AND SEMANTICS
Much of metatheory deals with connections between syntax and semantics, another distinction which I have tacitly observed
throughout the text. A fact of Syntux is a fact which concerns symbols or sentences insofar as the fact can be determined from the
form of the symbols or sentences, from the way they are written. The point is that facts of syntax do not depend on what the
symbols mean. A fact of Semantics, on the other hand, concerns the referents, interpretation, or (insofar as we understand this
notion) the meaning of symbols and sentences. In particular, semantics has to do with the referents of expressions, the truth of
sentences, and the relations between various cases of these. Here are some examples: Syntactic facts include the fact that 'A&B' is a
well-formed sentence of sentence logic, that 'AB&' is not a well-formed sentence, and that 'A&B' is a conjunction. Syntactic facts
include more general facts which can be determined from form alone, such as the fact that the derivation rule &E and the truth tree
rule & apply to any sentence of the form X&Y and that any sentence of the form XVY is derivable from (that is, there is a proof
from) a sentence of the form -X 3 Y. One thing to keep in mind is that whether or not a given string of sentences counts as a formal
proof (a derivation or a tree) is a syntactic fact. All the rules of proof have been carefully stated so that they appeal only to facts
about how sentences are written, not about how they are interpreted. Whether or not a string of sentences qualifies as a proof
depends only on the form, not on the content of the sentences. To see this clearly, consider that you could program a computer to
check and construct proofs. The computer need not know anything about how the sentences are interpreted. For example, the
computer need not know that you are supposed to think of '&' as meaning 'and'. It need only be programmed so that if a sentence of
the form X&Y appears on a derivation or tree, then below this sentence it can write both the sentences X and Y. Examples of
semantic facts include the fact that any interpretation which makes 'A3B' true makes '-B3-A' true, that '(Vx)(Px v -Px)' is true in all
interpretations, and that '(Vx)Px' is true in some interpretations and false in others. Semantic facts include more general facts such
as the fact that any existentially quantified sentence is true in an interpretation if one of its substitution instances is true in the
interpretation. To summarize the distinction between syntactic and semantic facts 10-3. Soundness and Completeness 169 Facts of
Sp are facts having to do with the form of expressions. Syntactic facts contrast with facts of Semantics which have to do with the
truth, reference, and the meaning of expressions. EXERCISES 10-2. Which of the following facts are syntactic facts and which
seman tic facts? a) Any interpretation which makes '(Vx)(Ax & Bx)' true makes '(Vx)Axl true b) The expression 'A&BvC' is not a
well-formed sentence, though it would be if parentheses were put around the 'A&B1. c) A sentence of the form --X can be derived
from a sentence of the form X d) In some interpretations 'a' and 'b' have the same referent. In some interpretations they do not. e) If
X and Y are well-formed sentences, then so is their conjunction. f) If the arpment X\Y is valid, then so is the argument -n-X. g) A
model of a set of sentences (that is, an interpretation in which each sentence in the set is true) is a model for any subset of the set
(that is, any smaller set of sentences all the sentences of which are contained in the original set). h) If there is a proof of the
sentence X from the sentences in the set Z, then there is a proof of X from any superset of Z, that is, any set which contains all the
sentences of Z as well as one or more additional sentences. 1&3. SOUNDNESS AND COMPLETENESS Students often have
difficulty appreciating the difference between the question of whether an argument, Z\X, is valid (a semantic question) and the
question of whether there is a proof from Z to X (a syntactic question). And no wonder! The syntactic rules of proof have been
carefully crafted so that there is a proof from Z to X if and only if the argument, Z\X, is valid. Of course, we have done this so that
we can use proofs to ascertain validity. But this must not obscure the fact that Derivability-that is, the existence of a proof-is one
thing and validity is another. That these two very different concepts go together is something we must demonstrate. Indeed, this
fundamental result about logic is what the rest of this book is about. To help in talking about these ideas, we will use two new
abbreviations in the metalanguage. (The following definitions also use the abbreviation 'iff', which is just shorthand for the
metalanguage expression 'if and only if'.) Dl: ZkX iff X is Derivable from Z, that is, iff there is a formal proof of X using only
sentences in 2. D2: Z~X iff the argument Z\X is valid, that is, iff every interpretation which makes all of the sentences in Z true
also makes X true. The symbol 't' is called the Single Turnstyle. ZtX asserts that a syntactic relation holds between the sentences in
the set of sentences, Z, and the sentence, X, that the latter is derivable from the former. The symbol '1' is called the Double
Turnstyle. Z~X asserts that a semantic relation holds between the set of sentences, Z, and the sentence, X, that any interpretation
which makes all of the sentences in Z true will also make X true. Here's a mnemonic to help remember which turnstyle is which. 'k'
has more bars and so has to do with meaning. 't' has kss bars and so has to do with the form of language. Using the turnstyle
notation, we can express the close relation between derivability and validity in two convenient parts: D3: A system of formal proof
is Sound iff for all 2, X, if ZkX, then Z~X. To say that a system of formal proof is sound is to say that whenever you have a proof
from Z to X, then any interpretation which makes all of the sentences in Z true also makes X true. D4: A system of formal proof is
Complete iff for all Z, X, if Z~X, then ZkX To say that a system of formal proof is complete is to say that in every case of an
argument, Z\X, which is valid (that is, any interpretation which makes every sentence in Z true also makes X true), there exists a
pr~of from Z to X. Completeness means that there is a proof in every case in which there ought to be a proof. Once more,
derivability and validity are distinct concepts. But derivability has been set up so that it can be used as a surefire test of validity. To
give a crude analogy, derivability is like the litmus test for acids. If you put a piece of litmus paper in a liquid and the paper turns
red, you know that the liquid is an acid. If the litmus paper does not turn red, the liquid is not an acid. Derivability is a kind of
litmus test for validity. Proving that the test works, proving soundness and completeness, is the fundamental metatheoretical result
in logic. This litmus test analogy is a good way to emphasize the fact that derivability and validity are distinct but related ideas.
However, I must be sure that the analogy does not mislead you in the following respect. Derivability is a surefire test for validity in
the sense that if there is a proof, then 164 Metatheoty: The Basic Concepb 10-4. Some Further Notation and Definitions 165 the
correspondi.ng argument is valid, and if an argument is valid, then there exists a proof which establishes that validity. But there
may not be any surefire way to establish whether or not such a proof exists! We might look for a proof from Z to X until the cows
come home and still not know for sure whether or not a proof exists. In predicate logic there is no mechanical means to determine
whether or not a proof from Z to X exists, no means guaranteed to give a definite yes or no answer in some finite number of steps.
This fact about predicate logic is known as Undecidability, and constitutes a second fundamental metatheoretical result. (Sentence
logic is decidable.) If you learned the tree method, I can give you a hint of what is involved by reminding you of the problem of
infinite trees. The same fact will turn up for derivations when we get to chapter 15. However, further study of undecidability goes
beyond what you will study in this text. EXERCISES 10-3. Some one might propose a set of rules of inference different from our
natural deduction or truth tree rules. Explain what is involved in such a new set of rules being Unsound (not sound) or 1ncompbte
(not complete). In fact, logicians have proposed many, many sets of inferential rules. Some such sets are sound and complete, some
are not. Whenever someone proposes a new set of inference rules it is important to determine whether or not the rules are sound
and complete. Exercises 10-4 to 10-6 concern the idea of Rule Soundness. To say that an individual rule of inference is sound is to
say that if the rule is applied to a sentence or sentences which is (are) true in a case, then the sentence which the rule licenses you to
draw is also true in that case. We can state the rules of inference for derivations using the turnstyle notation, and we can also use
this notation to assert the soundness of these rules. For example, the rule &I is expressed by saying that if Zt-X and Zt-Y, then Zt-
X&Y. We can state, in one way, that the rule &I is sound by stating that if Z~X and Z~Y, then ZkX&Y. 10-4. Show that the rule &I
is sound. 10-5. State the other primitive rules for derivations using the turnstyle notation and show that they are sound. 10-6.
Consider the following new rules for derivations. Determine which are sound and which are not. In each case, give an informal
demonstration of your conclusion about the rules. a) If ZkX3Y and ZkY, then ZkX. b) If ZkX=Y and Zk-X, then k-Y. C) If
Zk[(Vu)P(u) v (Vu)Q(u)], then Zk(Vu)[P(u) v Q(u)]. d) If Zk(3u)P(u), then ZkP(s). e) If Zk[(3u)P(u) & (3u)Q(u)], then Zk(3u)[P(u)
& Q(u)]. 10-7. Refresh your memory of the truth table method of establishing validity in sentence logic (see exercise 4-2 in chapter
4 of volume I). Then show that this method provides a decision procedure for I sentence logic. That is, show that, given a sentence
logic argument, the truth table method is guaranteed in a finite number of steps to give you a yes or no answer to the question of
whether or not the argument is valid. 1W. SOME FURTHER NOTATION AND DEFINITIONS Some further notation and
definitions will prove very useful in the following chapters, and will also give you a chance to practice the concepts of the last three
sections. First, here's an obvious and trivial extension of the turnstyle notation, a fussy logician's point which you might not even
notice. For example, if I write 'Zt-X', I have used Z as a metavariable over sets of sentences. What if I want to look at the special
case in which Z contains just one sentence? Then I may just use 'Z', a metavariable over individual sentences, writing 'Zt-X'. Or, if I
want more explicitly to list the sentences that come before the turnstyle, I may do just that, explicitly giving the list, for example,
writing WJt-X. I may use the same latitude in notation with the double turnstyle. A little less trivially, I have glossed over an
important point about using the single turnstyle. 'Zt-X' means that there is a proof of X from the sentences in the set Z. But by
proof, do I mean a derivation or a closed tree? It is important to keep separate these very distinct kinds of formal proof. Strictly
speaking, I should use one kind of turnstyle, say, 'kd' to mean derivations. Thus 'ZFdX' means that there is a derivation which uses
premises in Z and has X as its last line. And I should use a second kind of turnstyle, say, 't-,', to mean trees. Thus 'Zt-,X' means that
there is a closed tree of which the initial sentences are -X and sentences taken from Z. Other systems of formal proof (and there are
many others) must be distinguished from these with further subscripts on the single turnstyle. When there is any danger of
confusion about what kind of proof is in question, we must use a disambiguating subscript on the turnstyle. Usu- 166 Metatheory:
The Baric Concepts 104. Some Further NokJion and De&irions 167 ally, context will make it quite plain which kind of proof is in
question, In which case we may omit the subscript. EXERCISE 10-8. Do we need corresponding subscripts on the double
turnstyle? Explain why or why not. Here is one more refinement. How should we understand the turnstyle notation when the set Z
has infinitely many sentences? In the case of 'ZkX' this should be clear enough. This asserts that every interpretation which makes
all of the infinitely many sentences in Z true also makes X true. But what do we mean by 'ZtX'? A formal proof can use only
finitely many sentences. So by 'ZtX' we mean that there is a proof of X each premise of which is a sentence in the set Z. This
formulation leaves it open whether all of the sentences in Z get used as premises. If Z is infinite, only finitely many sentences can
be used in a proof from Z. If Z is finite, all or only some of the sentences in Z may get used. Reread definition Dl and be sure that
you understand 'formal proof of X using only sentences in Z', as just explained, to mean a proof which uses any number of, but not
necessarily all, the sentences in Z as premises. We even allow the case of using no premises at all. Any proof of a sentence, X, from
no premises makes ZtX true for any set of sentences, Z. EXERCISES 10-9. 'ZCW means that every sentence in Z is a sentence in
W. We say that Z is a Subset of W. Show that if ZtX and ZCW, then WtX. 10-10. Show that if ZkX and ZCW, then W~X. 10-1 1. If
Z is the empty set, we write tX for ZtX and k~ for Z~X. Explain what tX and k~ mean. If you have studied truth trees, you have
already encountered (in section 9-2, volume I, and section 8-1, this volume) the idea of a Model of a set of sentences. It's not
complicated: An interpretation, I, is a model of a set of sentences, Z, iff every sentence in Z is true in I. That is, a model, I, of a set
of sentences, Z, makes all the sentences in Z true. For example, consider the truth value assignment, I which makes 'A' true, 'B'
false, and 'C' true. I is a model for the set of sentences {(AvB),(-B>C)), but is not a model for the set of sentences {(A&-B),C,
(B=C)). Be sure you understand why this is so. To check, work out the truth values of each sentence in the two sets in the truth
value assignment, I, and apply the definition of model just given. In the following chapter we will use the notion of a model so
often that it's worth introducing an abbreviation: D5: 'Mod' is a predicate in the metalanguage (an abbreviation in English), defined
as Mod(I,Z) iff all the sentences in the set Z are true in the interpretation, I. If Mod(I,Z), I is said to be a Model for the sentences in
Z. I is also said to Satufy the sentences in Z. As with the turnstyle notation, we can use metavariables for sentences, such as 'Z',
where the metavariable, 'Z', for sets of sentences occurs in the definition of 'Mod'. We will also lean heavily on the notations of
consistency and inconsistency, already introduced in exercise 7-8 and section 9-2 (in volume I) and in sections 6-3 and 8-1 (in this
volume). To get ready for this work, asd to practice this chapter's ideas, here is a pair of equivalent definitions for each of these
concepts. (The slash through the double turnstyle in D6' means just what a slash through an equal sign means-the double turnstyle
relation does not hold.) D6: The set Z of sentences is Consirtent iff (3I)Mod(I,Z). D6': The set Z of sentences is Consirlent iff
zkA&-A. D7: The set Z of sentences is Inconsistent iff (V1)-Mod(I,Z), that is, iff Z is not consistent, that is, iff there is no model
fdr all the sentences in Z. D7': The set Z of sentences is Inconsistent iff zkA&-A. EXERCISES 10-12. Show that D6 and D6' are
equivalent. 10-13. Show that D7 and D7' are equivalent. 10-14. Explain why the notions of consistency and inconsistency are
semantic and not syntactic notions. Modify definitions D6' and D7' to provide corresponding syntactic notions, and label your new
definitions D6" and D7". You will then have a pair of notions, Semanlic Consis- and Syntactic ConsistLncy, and a second pair,
Semantic Znconsistency and Syntactic Znconststmuy. You must always carefully distinguish between these semantic and syntactic
ideas. Whenever I speak about consistency and inconsistency without specifying whether it is the semantic or syntactic notion, I
will always mean the semantic notion. 10-15. What do you think the rehtion is hetween semantic and sptactic consLtcncy, and
between semanticand syntactic iaeoarttency? What would you is the connection between thia qdn and the idea of ~wndaeg and
cornpkfenam? Write a pagraph inforIdly Cxphining ttrere umnaaiom as best you can. CHAPTER CONCEPrS Here are the
iplportant'coraccpts which I have introduced and disaxwed in thir chapter. Rcview them carefully to be sure you undastand them. a)
04-- , - b) Mecshaeulsc c) Mctawwhbk d) Use e) Mention f) SpcPmcFact g) samdcFact h) -tY i) Ij) + w fhuhas I) -pktenar m) Set of
mtcmr n) Wmu 0) Model P) Conrbtency qI1-n~~
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SECTION OVERVIEW
3.11: Mathematical Induction

Mathematical induction is a form of direct proof, usually done in two steps. When trying to prove a given statement for a set of
natural numbers, the first step, known as the base case, is to prove the given statement for the first natural number.

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The point of metatheory is to establish facts about logic, as distinguished from using logic. Sentence and predicate logic themselves
become the object of investigation. Of course, in studying logic, we must use logic itself. We do this by expressing and using the
needed logical principles in our meta-language. It turns out, however, that to prove all the things we want to show about logic, we
need more than just the principles of logic. At least we need more if by 'logic' we mean the principles of sentence and predicate
logic which we have studied. We will need an additional principle of reasoning in mathematics called Mathematical Induction.
You can get the basic idea of mathematical induction by an analogy. Suppose we have an infinite number of dominos, a first, a
second, a third, and so on, all set up in a line. Furthermore, suppose that each domino has been set up close enough to the next so
that if the prior domino falls over, it will knock over its successor. In other words, we know that, for all n, if the nth domino falls
then the n + 1 domino will fall also. Now you know what will happen if you push over the first domino: They will all fall. To put
the idea more generally, suppose that we have an unlimited or infinite number of cases, a first case, a second, a third, and so on.
Suppose that we can show that the first case has a certain property. Furthermore, suppose that we can show, for all n, that if the nth
case has the property, then the n + 1 case has the property also. Mathematical induction then licenses us to conclude that all cases
have the property. If you now have the intuitive idea of induction, you are well enough prepared to read the informal sections in
chapters 12 and 13. But to master the details of the proofs in what follows you will need to understand induction in more detail.
3.11.1: Informal Introduction is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
Let's look at a more specific example. You may have wondered how many lines there are in a truth table with n atomic sentence
letters. The answer is 2". But how do we prove that this answer is correct, that for all n, an n-letter truth table has 2" lines? If n = 1,
that is, if there is just one sentence letter in a truth table, then the number of lines is 2 = 2'. So the generalization holds for the first
case. This is called the Basis Step of the induction. We then need to do what is called the Inductive Step. We assume that the
generalization holds for n. This assumption is called the Inductive Hy$othesis. Then, using the inductive hypothesis, we show that
the generalization holds for n + 1. So let's assume (inductive hypothesis) that in an n-letter truth table there are 2" lines. How many
lines are there in a truth table obtained by adding one more letter? Suppose our new letter is 'A'. 'A' can be either true or false. The
first two lines of the n + 1 letter truth table will be the first line of the n-letter table plus the specification that 'A' is true, followed
by the first line of the n-letter table plus the specification that 'A' is false. The next two lines of the new table will be the second line
of the old table, similarly extended with the two possible truth values of 'A'. In general, each line of the old table will give rise to
two lines of the new table. So the new table has twice the lines of the old table, or 2" x 2 = 2"+ l. This is what we needed to show in
the inductive step of the argument. We have shown that there are 2" lines of an n-letter truth table when n = 1 (basis step). We have
shown that if an n-letter table has 2" lines, then an n + 1 letter table has 2"+' lines. Our generalization is true for n = 1, and if it is
true for any arbitrarily chosen n, then it is true for n + 1. The princple of mathematical induction then tells us we may conclude that
it is true for all n. We will express this principle generally with the idea of an Inductive Property. An inductive property is, strictly
speaking, a property of integers. In an inductive argument we show that the integer 1 has the inductive property, and that for each
integer n, if n has the inductive property, then the integer n + 1 has the inductive property. Induction then licenses us to conclude
that all integers, n, have the inductive property. In the last example, All n Ltter truth tubb have exact4 2" lines, a proposition about
the integer n, was our inductive property. To speak generally, I will use 'P(n)' to talk about whatever inductive property might be in
question:
 Principle of Weak Induction

a. Let P (n) be some property which can be claimed to hold for (is defined for) the integers, n = 1, 2, 3, . . . (the Inductive
Property).
b. Suppose we have proved P (l) (Basis Step).
c. Suppose we have proved, for any n, that if P(n), then P(n + 1) (IndzLchue Skp, with the assumption of P(n), the Inductive
Hypothesis).
d. Then you may conclude that P (n) holds for all n from 1 on.
e. If in the basis step we have proved P (i), we may conclude that P (n) holds for n = i, i + 1, i + 2, … .
(e) simply says that our induction can really start from any integer, as long as the inductive property is defined from that integer
onward. Often it is convenient to start from 0 instead of from 1, showing that P(n) holds for n = 0,1,2,. . . .
Most of the inductions we will do involve facts about sentences. To get you started, here is a simple example. The conclusion is so
obvious that, ordinarily, we would not stop explicitly to prove it. But it provides a nice illustration and, incidentally, illustrates the
fact that many of the generalizations which seem obvious to us really depend on mathematical induction. Let's prove that if the
only kind of connective which occurs in a sentence logic sentence is '-', then there is a truth value assignment under which the
sentence is true and another in which it is false. (For short, we'll say that the sentence "can be either true or false.") Our inductive
property will be: All sentences with n occurrences of '-' and no other connectives can be either true or false. A standard way of
expressing an important element here is to say that we will be doing the induction on th number of connectives, a strategy for
which you will have frequent use. We restrict attention to sentences, X, in which no connectives other than '-' occur. Suppose (basis
case, with n = 0) that X has no occurrences of '-'. Then X is an atomic sentence letter which can be assigned either t or f. Suppose
(inductive hypothesis for the inductive step) that all sentences with exactly n occurrences of '-' can be either true or false. Let Y be
an arbitrary sentences with n + 1 occurrences of '-'. Then Y has the form -X, where X has exactly n occurrences of '-'. By the
inductive hypothesis, X can be either true or false. In these two cases, -X, that is, Y, is, respectively, false and true. Since Y can be
any sentence with n + 1 occurrences of '-', we have shown that the inductive property holds for n + 1, completing the inductive
argument. 172 Mathematied Induction 11 -3. Strong Induction 173 EXERCISES 11-1. By a Restricted Conjunctive Sentence, I
mean one which is either an atomic sentence or is a conjunction of an atomic sentence with another restricted conjunctive sentence.
Thus the sentences 'A' and '[C&(A&B)]&D' are restricted conjunctive sentences. The sentence 'A &[(C&D)&(H&G)]' is not,
because the component, '(C&D)&(H&G)', fails to be a conjunction one of the components of which is an atomic sentence letter.
Here is a rigorous definition of this kind of sentence:
a. Any atomic sentence letter is a restricted conjunctive sentence.
b. Any atomic sentence letter conjoined with another restricted conjunctive sentence is again a restricted conjunctive sentence.
c. Only such sentences are restricted conjunctive sentences. Such a definition is called an Inductive Definition. Use weak
induction to prove that a restricted conjunctive sentence is true iff all the atomic sentence letters appearing in it are true. 11-2.
Prove that the formula is correct for all n.
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 Learning Objectives
In reviewing this chapter, be sure you understand clearly the following ideas:
1. Weak Induction
2. Inductive Property
3. Basis Step
4. Inductive Hypothesis
5. Inductive Step
6. Induction on the Number of Connectives
7. Strong Formulation of Weak Induction
8. Strong Induction
9. Least Number Principle
Let's drop the restriction in exercise 11-1 and try to use induction to show that any sentence in which '&' is the only connective is
true iff all its atomic sentence letters are true. We restrict attention to any sentence logic sentence, X, in which '&' is the only
connective, and we do an induction on the number, n, of occurrences of '&'. If n = 0, X is atomic, and is true iff all its atomic
sentence letters (namely, itself) are true. Next, let's assume, as inductive hypothesis, that any sentence, X, in which there are exactly
n occurrences of '&' is true iff all its atomic sentence letters are true. You should try to use the inductive hypothesis to prove that the
same is true of an arbitrary sentence, Y, with n + 1 occurrences of '&'. If you think you succeeded, you must have made a mistake!
There is a problem here. Consider, for example, the sentence '(A&B)&(C&D)'. It has three occurrences of '82. We would like to
prove that it has the inductive property, relying on the inductive hypothesis that all sentences with two occurrences of '&' have the
inductive property. But we can't do that by appealing to the fact that the components, '(A&B)' and '(C&D)', have the inductive
property. The inductive hypothesis allows us to appeal only to components which have two occurrences of '&' in them, but the
components '(A&B)' and '(C&D)' have only one occurrence of '&' in them. The problem is frustrating, because in doing an
induction, by the time we get to case n, we have proved that the inductive property also holds for all previous cases. So we should
be able to appeal to the fact that the inductive property holds, not just for n, but for all previous cases as well. In fact, with a little
cleverness one can apply weak induction to get around this problem. But, more simply, we can appeal to another formulation of
mathematical induction: Wed Induction, Strong Formulation: Exactly like weak induction, except in the inductive step assume as
inductive hypothesis that P(i) holds for all i 5 n, and prove that P(n + 1). 11-3. Using the strong formulation of weak induction,
prove that any sentence logic sentence in which '&' is the only connective is true iff all its atomic sentence letters are true. You
could have done the last problem with yet another form of induction:
Strong Induction
Suppose that an inductive property, P(n), is defined for n = 1, 2, 3, . . . . Suppose that for arbitrary n we use, as our inductive
hypothesis, that P(n) holds for all i < n; and from that hypothesis we prove that P(n). Then we may conclude that P(n) holds for all
n from n = 1 on. If P(n) is defined from n = 0 on, or if we start from some other value of n, the conclusion holds for that value of n
onward.
Strong induction looks like the strong formulation of weak induction, except that we do the inductive step for all i < n instead of all
i 5 n. You are probably surprised to see no explicit statement of a basis step in the statement of strong induction. This is because the
basis step is actually covered by the inductive step. If we are doing the induction from n = 1 onward, how do we establish P(i) for
all i < l? There aren't any cases of i < l! When n = 1, the inductive hypothesis holds vacuously. In other words, when n = 1, the
inductive hypothesis gives us no facts to which to appeal. So the only way in which to establish the inductive step when n = 1 is
just to prove that P(1). Consequently, the inductive step really covers the case of the basis step. Similar comments apply if we do
the induction from n = 0 onward, or if we start from some other integer. You may be wondering about the connections among the
three forms of induction. Weak induction and weak induction in its strong formulation are equivalent. The latter is simply much
easier to use in problems such as the last one. Many textbooks use the name 'strong induction' for what I have called 'weak
induction, strong formulation'. This is a mistake. Strong induction is the principle I have called by that name. It is truly a stronger
principle than weak induction, though we will not use its greater strength in any of our work. As long as we restrict attention to
induction on the finite integers, strong and weak induction are equivalent. Strong induction shows its greater strength only in
applications to something called "transfinite set theory," which studies the properties of mathematical objects which are (in some
sense) greater than all the finite integers. Since, for our work, all three principles are equivalent, the only difference comes in ease
of use. For most applications, the second or third formulation will apply most easily, with no real difference between them. So I
will refer to both of them, loosely, as "strong induction." You simply need to specify, when doing the inductive step, whether your
inductive hypothesis assumes P(i) for all i < n, on the basis of which you prove P(n), or whether you assume P(i) for all i 5 n, on the
basis of which you prove P(n + 1). In either case, you will, in practice, have to give a separate proof for the basis step. I should
mention one more pattern of argument, one that is equivalent to strong induction:
 LEAST Number Principle
To prove that P (n), for all integers n , assume that there is some least value of n, say m, for which P(m) fails and derive a
contradiction. The least number principle applies the reductio argument strategy. We want to show that, for all n, P(n). Suppose
that this is not so. Then there is some collection of values of n for which P(n) fails. Let m be the least such value. Then we
know that for all i < m, P(i) holds. We then proceed to use this fact to show that, after all, P(m) must hold, providing the
contradiction. You can see that this form of argument really does the same work as strong induction: We produce a general
argument, which works for any value of m, which shows that if for all i < m P(i) holds, then P(m) must hold also.
You will notice in exercises 11-7 to 11-9 that you are proving things which in the beginning of Volume I we simply took for
granted. Again, this illustrates how some things we take for granted really turn on mathematical induction.
EXERCISES 114. Prove that any sentence logic sentence in which 'v' is the only connective is true iff at least one of its atomic
sentence letters is true.
11-5. Consider any sentence logic sentence, X, in which '&' and 'v' are the only connectives. Prove that for any such sentence, there
is an interpretation which makes it true and an interpretation which makes it false. Explain how this shows that '&' and 'v', singly
and together, are not expressively complete for truth functions, as this idea is explained in section 34, (volume I).
11-6. Consider any sentence logic sentence, X, in which '-' does not appear (so that '&', 'v', 'S, and '=' are the only connectives).
Prove that for any such sentence there is an interpretation which makes X true. Explain how this shows that '&', 'v', 'S, and '=' are,
singly and together, not expressively complete for truth functions. 11-7. Prove for all sentence logic sentences, X, and all
interpretations, I, that either I makes X true or I makes X false, but not both. 11-8. Prove for all sentence logic sentences, X, that if
two truth value assignments, I and If, agree on all the atomic sentence letters in X, then I and I' assign X the same truth value. 11-9.
Prove the law of substitution of logical equivalents for sentence logic.
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SECTION OVERVIEW
3.12: Soundness and Completeness for Sentence Logic Tree

3.12.1: Preliminaries
3.12.2: Soundness and Completeness of the Tree Method- Informal Statement

3.12: Soundness and Completeness for Sentence Logic Tree is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
12.1: Preliminaries
12-2. Soundness and Completeness of the Tree Method: lnfotnral Statement 177 Soundness and Completeness for Sentence Logic
Trees 12-1. PRELIMINARIES This chapter will explain the soundness and completeness of sentence logic for the tree method.
Section 12-2 gives an informal statement which you will be able to follow without having studied more than the first short section
of chapter 11, on mathematical induction. Section 12-3 gives full details. Before getting started, I want to make a general point
which will be useful in discussing both trees and derivations. I am going to make a statement which uses a new bit of notation. 'U'
indicates set union. That is, ZUW is the set consisting of all the members of the set Z together with the members of the set W. Also,
if X is a sentence, {X} is the set which has X as its only member. Now, to say that the X is valid, that is, that Z~X, is to say that
every interpretation which makes all the sentences in Z true also makes X true. Keep in mind that X is true in an interpretation iff -
X is false in that interpretation. Consequently L1: z~X iff ZU{-X} is inconsistent. (The 'L' in 'Ll' stands for 'lemma'. A lemma is a
statement which may not be of great interest in itself but which we prove because it will be useful in proving our main results.) L1
shows that validity of an argument comes to the same thing as the inconsistency of a certain set of sentences, namely, the premises
and negation of the conclusion of the argument. You will soon see that Ll's equivalent formulation of validity provides a
particularly convenient way to study soundness and completeness. EXERCISE 12-1. Prove L1 12-2. SOUNDNESS AND
COMPLETENESS OF THE TREE METHOD: INFORMAL STATEMENT Soundness and completeness tell us that there is an
exact correspondence between a semantic concept-validity-and a corresponding syntactic concept-proofs. Let's be explicit about
what counts as a proof in the tree method: Given some premises and a conclusion, a tree method proof is a closed tree (a tree with
all its branches closed) which has the premises and negation of the conclusion as its initial sentences. Closed trees are the syntactic
objects which need to correspond to the semantic concept of validity. So proving soundness and completeness for the tree method
means proving that we have the right sort of correspondence between validity and closed trees. To become clear on what
correspondence we need, let's go back to the way I introduced the tree method. I said that, given an argument, X, the argument is
valid just in case it has no counterexamples, that is, no interpretations in which the premises, Z, are all true and the conclusion, X,
is false. I then went on to develop truth trees as a method of looking for counterexamples, a way which is guaranteed to find a
counterexample if there is one. If the whole tree closes, there are no counterexamples, and we know the argument is valid. But a
closed tree is what counts as a proof. So if there is a proof, the argument is valid. If you look back at definition D3 in chapter 10,
you will see that this is what we call soundness. On the other hand, if there is an open branch (and so no proof), there is a
counterexample, and thus the argument is invalid. A little thinking indicates that the last statement is just completeness: "If no
proof, then invalid" comes to the same as "If valid, then there is a proof," which is completeness, as defined by D4 in chapter 10. I
have used the law of contraposition: X3Y is logically equivalent to -Y>-X. The first time through, this argument is bound to seem
very slick. It is 178 Soundness and Completeness foz Sentence Logic Trees 12-2. Soundness and Completeness of the Tree
Method: Infnmal Stat-t 179 also incomplete: I have yet to prove that the truth tree method is guaranteed to find a counterexample if
there is one. To sort all of this out more carefully, we need to examine the connection between a counterexample and lemma L1. A
counterexample to the argument Z'X is just an interpretation which shows the set of sentences, ZU{-X} to be consistent.
(Remember that such an interpretation is called a model of the set of sentences.) Now, look at lemma L1, and you will start to see
that all this talk about counterexamples is just another way of saying what lemma L1 says. EXERCISE 12-2. Show that lemma L1
is equivalent to the statement that an argument is valid iff it has no counterexamples. Lemma L1 tells us that we can forget about
validity and talk about consistency and inconsistency instead. Indeed, conceptually, the tree method is really a method for
determining whether the initial sentences on a tree form a consistent set. It is a method which is guaranteed to find a model for a
tree's initial sentences if there is one, thereby showing the set of sentences to be consistent. Conversely, if a set is inconsistent, it
has no model, and a tree starting with the sentences in the set is bound to close. The real work we have to do is to show that the tree
method is guaranteed to find a model for a set of sentences if the set has a model. We'll worry later about connecting this up with
validity-lemma L1 assures us that we will be able to do so. For now, we will connect the semantic concept of a model with the
syntactic concept of an open branch. Remember that an open branch always represents an interpretation in which all sentences on
the branch are true. Hence, if there is an open branch, there is an interpretation in which all the sentences on the branch, including
the tree's initial sentences, are true. Here is how we proceed: We will show that a finite set of sentences is consistent if and only if
we always get an open branch on a finished tree which starts from the sentences in the set. Equivalently, a set is inconsistent if and
only if we always get a closed tree if we start from the sentences in the set. This gives us the connection between a syntactic
concept--open and closed trees-and a semantic concept+onsistency and inconsistency. Lemma L1 tells us we will be able to
connect the latter with validity and invalidity. To keep track of how we will carry out this program, let's talk about it in terms of an
example, say, the tree which results from using as initial sentences the sentences in the set {-(DvB), (A&B)v(C>D)}: We must first
show that tree method is what I will call Downwardly Adequute. This means that the tree method applied to a consistent set of
sentences always results in at least one open branch. Why is the tree method downwardly adequate? Remember that the rules are
written so that when a rule is applied to a sentence, the rule instructs you to write, on separate branches, all the minimally sufficient
ways of making the original sentence true. In effect, this means that, for any assignment of truth values to sentence letters which
makes the original sentence true, all the sentences of at least one of the resulting stacksbf sentences will be true also for the same
assignment of truth values. This fact is easiest to grasp by example. In applying the rule v to line 2, we produce two branches, on
line 5. Suppose that we have an assignment that makes line 2 true. This can only be because it makes at least one of the two
disjuncts true. But then, on this assignment, at least one of the sentences on the two branches of line 5 will be true for the same
assignment. Or consider application of the rule -V to line 1, and suppose that we have a truth value assignment which makes line 1
true. A negated disjunction can be true only if both disjuncts are false. Consequently, an assignment which makes line 1 true will
make lines 3 and 4 true. As I introduced the rules, I made sure that they are all, as in these two examples, what I will call
Downwardly Correct. In outline, the downward correctness of the rules works to show that the tree method is downwardly
adequate as follows: Suppose that the initial sentences on a tree are consistent, so that there is an interpretation, I, which assigns
truth values to sentence letters making all the initial sentences true. N~W apply a rule to one of the sentences. The downward
correctness of the rules means that -- - applying the rule will result in at least one branch on which all the new sentences are true in
I. Of course, all the former sentences along this " branch were true in I. So at this stage we have at least one branch along which all
sentences are true in I. Now we just do the same thing over again: Apply one of the rules to a sentence on this branch. We get at
least one extension of the branch on which all sentences are true in I. 180 Soundness and Completeness for Sentence Logic Treer
12-3. Soundness and Completeness for Sentence La@ Trees: Formal Details 181 This process will eventually come to an end
because each application of a rule produces shorter sentences. At the end we have at least one branch on which all the sentences are
true in I. But this branch must be open. Since all the sentences along the branch are true in I, no sentence and its negation can both
appear on the branch! In sum, if the original sentences are consistent, there will be an open branch. We are half done. We must still
show that the tree method is Upwardly Adequate, that is, that if there is an open branch, then the set of initial sentences is
consistent. So now let us suppose that we have a tree with an open branch. Since an open branch never has both a sentence and its
negation, I can consistently assign the truth value t to all atomic sentences on the branch and the truth value f to all those atomic
sentences whose negations occur on the branch. Call this assignment I. I will also make the longer sentences on the branch true.
Look, for instance, at the open branch in the last example. Reading up from the bottom, this branch specifies the assignment 'C', 'B',
and 'D' all false. Call this assignment I. If 'C' is false, that is, if '-C' is true in I, then 'C>D' is true in I. In turn, 'C>D' being true in I
will make line 2, '(A&B)v(C>D)' true in I. Likewise, lines 3 and 4, '-D' and '-B', true in I will make line 1, '-(DvB)', true in I. All the
rules have the property just used, called Upward Correctness: If I makes true the sentence or sentences which a rule produced from
a previous sentence, I makes that previous sentence true also. Upward correctness will apply to any open branch in any tree just as
it did in the example. Choose an interpretation, I, as the one which makes all the atomic sentences on the open branch true and all
the negated atomic sentences false. Apply upward correctness again and again. You can see that, finally, all the sentences along the
open branch are true in I. So the open branch provides an interpretation, I, which makes all the sentences along the branch true,
including the initial sentences. So if there is an open branch there is a model for the initial sentences, which is to say that the initial
sentences form a consistent set, which is just what we mean by upward adequacy. Let's pull the threads together. The tree method is
downwardly adequate. That is, if the initial sentences are consistent, then there is an open branch. By contraposition, if there is no
open branch, that is, if there is a proof, then the initial sentences form an inconsistent set. Lemma 1 tells us that then the
corresponding argument is valid. This is soundness. The tree method is also upwardly adequate. If there is an open branch, and so
no proof, then the initial set of sentences is consistent. By contraposition, if the set of initial set of sentences is inconsistent, then
there is a proof. Lemma 1 then connects the inconsistency with validity: If the corresponding argument is valid, there is a proof.
This is completeness. If you are starting to see how soundness and completeness work for trees, this section is doing its job. Doing
the job fully requires further precision and details, presented in the next section. If in the next sectionyou start to feel lost in a forest
of definitions (as I often do) reread this section, which contains all the concepts. Reviewing the intuitively presented concepts will
help you to see how the following formal details fit together. 12-3. SOUNDNESS AND COMPLETENESS FOR SENTENCE
LOGIC TREES: FORMAL DETAILS In this section I am going to make a very natural simplifying assumption: I will restrict
discussion to Ute sets of sentences Z. This restriction is natural because intuitively we think of arguments as only having finitely
many premises anyway. Generalization to the case of infinite sets of sentences involves a complication which would only distract
us from the main line of argument. Chapter 14 will take care of the generalization. For precision and efficiency of statement, we
need the following definitions: D8: A Minimal Sentence is a sentence which is either atomic or the negation of an atomic sentence.
D9: A truth tree is Finished iff each branch is either closed or has all applicable rules applied to all nonminimal sentences. D10:
'Tr', 'Op', and 'Cl' are predicates of the metalanguage (abbreviations in English) which are defined as a) Tr(T,Z) iff T is a finished
tree with all the sentences in Z its initial sentences. b) Op(T) iff the tree T is open, that is, if T has an open branch. c) Cl(T) iff -
Op(T), that is, if T is closed, that is, if all of T's branches are closed. A proof of Z\X is just a closed tree which starts with sentences
in Z and the sentence -X. Expressed with our abbreviations, this is Dl 1: A tree, T, is a proofofx from Z iff Tr(T,ZU{-X}) and
CI(T). Next, recall that ZkX just means that there exists a proof of X using premises in Z, where here a proof just means a tree as
described in Dl 1. So applying Dl 1 to the definition of l-, we have L2: For finite sets, Z, Zl-X iff (3T)[Tr(T,ZU{-X}) & CI(T)]. 182
Soundness and Completeness for Sentence Logic Trees Of course, throughout this section 'k' means kt, that is, derivability
according to the tree method. EXERCISE 12-3. In chapter 10 I specified that ZkX means that there is a proof of X using any
number of sentences in Z, but not necessarily all of them. (I did this to accommodate the eventual generalization to infinite sets.)
But Dl 1 defines T as being a proof of X from Z in terms of Tr(T,ZU{-XI), which specifies a tree, T, which has dl the sentences of
ZU{-X} as its initial sentences. Prove L2, taking care to deal with this apparent difficulty. Use the fact that L2 is stated with the
existential quantifier, '(3T)'. Now remember how we used L1 to show that we could exchange the study of validity and invalidity
for the study of the consistency and inconsistency of a certain set of sentences, namely, the premises together with negation of the
conclusion. Our next step is to connect the consistency of this set with the syntactic notion of an open branch. We do this with the
idea of upward and downward adequacy of the tree method. Downward adequacy says that if the set Z is consistent, that is, if there
is a model for Z, then the tree starting from Z has an open branch. Using definitions D5 and D6, this becomes D12: The tree
method is Downwardly Adequate iff for all finite, nonempty sets of sentences Z, if (3I)Mod(I,Z), then (VT)[Tr(T,Z) 3 Op(T)].
Upward adequacy is the converse: If there is an open branch, the initial set is consistent: D13: The tree method is Upwardly
Adequute iff for all finite, nonempty sets of sentences Z, if (VT)[Tr(T,Z) 3 Op(T)], then (3I)Mod(I,Z)]. A detail in Dl2 and Dl3
requires comment. If we start a tree with the sentences in Z, we can come up with more than one tree because we can apply the
rules in different orders. So when I give a formal definition of upward and downward adequacy, I must make a choice whether to
define these in terms of dl open trees starting from Z or some open tree starting from Z. In terms of the proof of upward and
downward adequacy, I could do either because, in essence, the proof will show that, for a given set of initial sentences, one tree is
open iff all are. I choose to define upward and downward adequacy in terms of all open trees for the following rea- 12-3.
Soundness and Completeness for Sentence Logic Trees: Formal Details 183 son: When we connect adequacy with soundness and
completeness, I will be taking a converse. This will introduce a negation sign, and when the negation sign gets pushed through the
quantifier, 'all' turns into 'some'. At that point I will be talking about "some closed tree". That is just what we will need to get a
smooth fit with derivability, which is defined in terms of "there is some proof', where a proof is just a closed tree. If I had defined
upward and downward adequacy in terms of some instead of all open trees, it would be a mess to make the connection with
soundness and completeness. 124. Assume (as we will prove shortly) that if a tree has at least one open branch, then the initial
sentences of the tree form a consistent set. Also assume downward adequacy. Prove that for all the finished trees starting from the
same set of initial sentences, one is open iff all are. The next step is to show that upward adequacy is equivalent to soundness and
downward adequacy is equivalent to completeness. The connection will not sink in unless you do the work! But I will break the job
down into several steps. First we define a version of soundness and completeness for the tree method: D3': The tree method is
Sound iff for all finite, nonempty sets of sentences Z, if (3T)[Tr(T,Z) & CI(T), then (V1)-Mod(1,Z). D4': The tree method is
Complete iff for all finite, nonempty sets of sentences Z, if (V1)-Mod(I,Z), then (3T)[Tr(T,Z) & Cl(T)]. Now it is not hard to prove
that downward adequacy is soundness and upward adequacy is completeness in the form of four new lemmas: L3: The tree method
is sound according to D3 iff it is sound according to D3'. L4: The tree method is complete according to D4 iff it is complete
according to D4'. L5: The tree method is sound according to D3' iff it is downwardly adequate. L6: The tree method is complete
according to D4' iff it is upwardly adequate. 184 Soundness and Completeness for Sentence Logic Trees 12-3. Soundness and
Completeness for Sentence Logic Trees: Farma1 DetaiLc 185 1 EXERCISES 12-5. Prove lemmas L3 and L4. You will need to use
lemmas L1 and L2. 124. Prove lemmas L5 and L6. You will need to use contraposition and the laws of logical equivalence for
negated quantifiers as laws applied to statements in the metalanguage. We have reduced the problem of proving soundness and
completeness to that of proving that the tree method is downwardly and upwardly adequate, which the last section indicated we
would do by appealing to the downward and upward correctness of the rules. Some further terminology will enable us to state rule
correctness more clearly. When we apply a truth tree rule to a sentence, the rule instructs us to write one or two branches and on
each branch one or two new sentences. For example, the = rule is We will call the sentence to which the rule is applied, X=Y in the
example, the Input Sentence. The rule instructs you to write one or two lists of sentences (each "list" containing one or two
sentences). We will call each such list an Output List. In the example, X,Y is one output list and -X,-Y is the second output list. The
rule has only one output list, namely, X,Y. In summary D14: The sentence to which a tree method rule is applied is called the Input
Sentence. The sentence or (sentences) along one branch which the rule directs you to write is (are) called an Outjmt List of
Sententes. Here is what we must require of a correct truth tree rule. Suppose that I give you an interpretation (an assignment of
truth values to sentence letters) which makes true the input sentence of a rule. Then that same interpretation must make true all the
sentences on at least one (but perhaps not all) output lists. This is downward correctness. And suppose f give you an interpretation
which makes all the sentences on one output list true. Then that same interpretation must make the input sentence true. This is
upward correctness. D15: A tree method rule is Downwardly Correct iff any interpretation which makes the input sentence true
also makes true all the sentences of at least one output list. D16: A tree method rule is Upwardly Correct iff any interpretation
which makes all the sentences of one output list true also makes the input sentence true. EXERCISES 12-7. Show that all of the
truth tree rules for sentence logic are downwardly and upwardly correct. 12-8. Consider the following two proposed truth tree rules:
Determine which of these is downwardly correct and which is upwardly correct. In each case show correctness or failure of
correctness. We are now ready to prove T1: The truth tree method for sentence logic is downwardly adequate. (The 'T' stands for
'theorem'.) Suppose we are given a finite nonempty set of sentences, Z, and a tree, T, which has the sentences of Z as its initial
sentences. Now suppose that there is a model, I, of the sentences in Z. What we will do is to look at successively larger initial
segments of one branch of T and show that all these initial segments of the branch are open. Start with just the sentences in Z, that
is, the initial sentences of T. This initial segment of a branch must so far be open. Why? Well, a branch 186 Soundness and
Completeness for Sentence Logic Trees 12-3. Soundness and Completeness fw Sentence Logic Trees: Fonnal Details 187 closes
only if it contains both a sentence and the negation of that same sentence. But Z can't contain a sentence and its negation. This is
because there is a model, I, of all the sentences in Z. That is, I makes all the sentences in Z true. But no sentence and its negation
can both be true in the same interpretation! If I makes one sentence true, it makes its negation false. So far we have an initial
segment-let's call it the first segment--of a branch, all the sentences of which are true in I, and which consequently is (so far) open.
Next, in constructing the tree T, we apply a rule to one of the sentences in this first initial segment of our so far open branch. The
input sentence for this rule is true in I. By the downward correctness of the rules, there will be at least one output list all the
sentences of which are true in I. Pick one such output list (say, the left one if there are more than one). Look at the extension of the
first segment of our branch extended by this output list. Call this extension the second initial segment. This second segment now
has all its sentences true in I. You can see the handwriting on the wall. We just do the same thing over and over again. At the nth
stage we start with a branch all the sentences of which are true in I. The tree grows by application of some rule to some sentence on
the nth initial segment. Downward correctness guarantees that at least one output list will have all its sentences true in I also. We
pick the leftmost such output list as the extension of the nth initial segment to the n + 1st initial segment. Then the n + 1st initial
segment has all its sentences true in I, and we can start all over again. In a sentence logic tree, the sentences get shorter with each
application of a rule, so this process must eventually end. When it does, we have a branch all the sentences of which are true in I.
For exactly the same reason that the first initial segment must be open, this final branch must be open also: All its sentences are
true in I, and no sentences and its negation can both be true in the same interpretation. EXERCISE 12-9. Formulate the foregoing
argument sketch into an explicit use of mathematical induction to prove T1. There are many correct ways to apply induction. For
example, begin by supposing that you are given a finite, nonempty set of sentences, Z, a model I of Z, and a finished tree, T, with
initial sentences Z. Break the tree up into stages: The nth stage of the tree includes all lines written down in the first through nth
application of a rule. Your inductive property will be: There is a branch through the nth stage of the tree all the sentences of which
are true in I. Or you can similarly organize the inductive property around the number of lines to be checked: The first line to be
checked, the first and second lines to be checked, and so on. Be sure to show explicitly how the results from the induction establish
downward adequacy. I have suggested a formulation for this proof which I hope you will find to be relatively intuitive, but the
logical form of the suggested proof is actually a bit convoluted. In this formulation you use both universal introduction and
induction. That is, for an arbitrary, finite, nonempty set Z, model I of Z, and tree T with initial sentences in Z, you show how
induction gives the desired result in that case. Then, since you have assumed nothing else about the Z, I, and T, what you have
shown is true for all such Z, 1, and T. In addition, the induction is a finite induction. In a specific case it runs only from the first
through the number of stages in the tree in question. Logicians prefer a more abstract but "pure" way of doing this kind of problem.
In the inductive step you assume that in any tree with n stages (or n checked sentences) and interpretation I which makes all initial
sentences true, there is a path all the sentences of which are true in I. You then use downward rule correctness to show that the
same is true in any n + 1-stage tree. To do this you consider an arbitrary n + 1-stage tree and the n-stage tree (or trees) which result
by deleting the first sentence to which a rule was applied in the original n + 1-stage tree. The downward rule correctness of the
applied rule shows that if the inductive hypothesis holds of the subtree, it holds of the full n + 1-stage tree. But I will leave the
details to you and your instructor! Let's turn to T2: The truth tree method for sentence logic is upwardly adequate. The proof works
similarly to that for downwardly adequate, differing in that we use upward correctness of the rules and we look at successively
longer and longer sentences on a branch instead of successively longer and longer initial segments of a branch. Suppose we are
given a tree with an open branch. Take one open branch (say, the leftmost). Because this branch is open, and so has no sentence and
its negation, we can consistently assign the truth value t to all the atomic sentence letters which appear on the branch and the truth
value f to all atomic sentence letters the negation of which appear on the branch. This constitutes an interpretation I-an assignment
of truth values to sentence letters. We are going to show that all the sentences along this branch are true in I. By the Length of a
sentence let us understand the total number of con- 188 Soundness and Compkteness for Sentence Logic Trees 12-3. Soundness
and Compkteness for Sentence Logic Trees: Ford Details 189 nectives and sentence letters that appear in the sentence. So far, all
minimal sentences along the branch are true in I-that is, all sentences of length 1 or 2. Now, consider any sentence along the branch
(if there are any) of length 3. When a rule was applied to such a sentence, the rule produced an output list the sentences of which
are each shorter than the input sentence; that is, each has fewer total connectives plus sentence letters. (You should check this.) But
all such shorter sentences of the branch, that is, sentences of length 1 or 2, are already known to be true in I. Upward rule
correctness then tells us that the sentence of length 3 is true in I. The same goes for any length 3 sentence on the branch. So now we
know that all sentences of length 1, 2, and 3 on the branch are true in I. Again, you can see how this will go: We do the same thing
over and over again. At stage n we know that all sentences of the branch of length n or less are true in I. Consider any sentence of
length n + 1. The rule applied to it produced shorter sentences, already known to be true in I. By upward correctness of the applied
rule, the sentence of length n + 1 is then also true in I. The same goes for any sentence of length n + 1 on the branch, so that we
now have shown that all of the branch's sentences of length n + 1 are true in I. Ultimately, the process shows that all the sentences
in the branch are true in I. This includes the initial sentences, which are the initial sentences of the tree. EXERCISE 12-1 0.
Formulate the foregoing argument sketch into an explicit inductive argument. That is, given a tree and an open branch on the tree,
show that there is an interpretation which can be shown by induction to make all sentences (and hence the initial sentences) along
the branch true. Comments exactly parallel to those on your proof of T1, about the logical "purity" of the proof, also apply here.
Just as for T1, one can also do the induction on the "size" of the tree. In the inductive step, you assume that all open trees with no
more than n checked sentences have the desired characteristic-that open paths represent interpretations which make all the
sentences on the path true-and you then use upward rule correctness to show that all trees with n + 1 checked sentences also have
this characteristic. In outline, the idea is that any tree with n + 1 checked sentences has one or more subtrees with no more than n
checked sentences-namely, the tree or trees obtained by deleting the first checked sentence in the original tree. You then apply the
inductive hypothesis assumed to hold for the shorter trees. We have shown that, given some tree with an open branch, there is an
interpretation, I, in which all of the tree's initial sentences are true. How does this show upward adequacy? Suppose we are given a
finite, nom empty set of sentences, Z. Assume the antecedent in the statement of upward adequacy. That is, assume that any tree
starting from Z is open. There is always at least one tree, T, starting from Z. Since all such trees are being assumed to be open, T is
open, that is, T has an open branch. But in the previous paragraphs we have shown that this open branch provides an interpretation
in which all initial sentences of T, that is, all the sentences in Z, are true. We have now completed the proof of T2. T1 and T2, with
the help of lemmas L3, L4, L5, and L6, complete our proof of soundness and completeness for the tree method. As you can check
in a moment, T1, L3, and L5 immediately give T3: The tree method for sentence logic is sound. T2, L4, and L6 immediately give
T4: The tree method for sentence logic is complete. EXERCISES 12-1 1. This exercise makes precise the work you did informally
in exercises 10-14 and 10-15. Recall that when I refer to consistency and inconsistency without qualification, I always mean
semantic consistency and inconsistency. We want a notion of Syntactic Consistency and Inconsistency, that is, a syntactic notion
which will work as a test for semantic consistency and inconsistency. These are D17: Z is Syntactically Consistent iff (VT)[Tr(T,Z)
3 Op(T)]. D18: Z is Syntactically Inconsistent iff (3T)[Tr(T,Z)&Cl(T)]. (Throughout this problem, be sure to assume that Z is finite
and nonempty.) a) Show that a set of sentences is syntactically consistent according to Dl7 iff it is not syntactically inconsistent
according to D18. b) Show that Z is syntactically consistent iff ZYA&-A. c) Show that Z is syntactically inconsistent iff ZkA&-A.
d) Show that Z is syntactically inconsistent iff for any X, ZkX. e) Reexpress lemma L2 and definitions D12, D13, D3', and D4' in
terms of semantic and syntactic consistency. CHAPTER CONCEPTS To check your understanding of this chapter, make sure that
you understand all of the following: Input Sentence of a Rule Output Sentence of a Rule Downward Rule Correctness Upward Rule
Correctness Downward Adequacy Upward Adequacy Minimal Sentence Finished Tree Tr(TZ) oPo cw7 Tree T is a proof of X from
Z Syntactic Consistency Semantic Consistency
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12.2: Soundness and Completeness of the Tree Method: Informal Statement

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SECTION OVERVIEW
3.13: Soundness and Completeness for Sentence Logic Derivation

3.13.1: Soundness for Derivations- Informal Introduction
3.13.2: Soundness for Derivations- Formal Details
3.13.3: Completeness for Derivations- Informal Introduction
3.13.4: Completeness for Derivations- Formal Details

3.13: Soundness and Completeness for Sentence Logic Derivation is shared under a not declared license and was authored, remixed, and/or
13.1: Soundness for Derivations: Informal Introduction
Let's review what soundness comes to. Suppose I hand you a correct derivation. You want to be assured that the corresponding
argument is valid. In other words, you want to be sure that an interpretation which makes all the premises true also makes the final
conclusion true. Soundness guarantees that this will always be so. With symbols, what we want to prove is
T5 (Soundness for sentence logic derivations): For any set of sentences, Z, and any sentence, X, if ZkX, then Z~X.
with 'k' meaning derivability in the system of sentence logic derivations.
The recipe is simple, and you have already mastered the ingredients: We take the fact that the rules for derivations are truth
preserving. That is, if a rule is applied to a sentence or sentences (input sentences) which are true in I, then the sentence or
sentences which the rule licenses you to draw (output sentences) are likewise true in I. We can get soundness for derivations by
applying mathematical induction to this truth preserving character of the rules.
Consider an arbitrary derivation and any interpretation, I, which makes all of the derivation's premises true. We get the derivation's
first conclusion by applying a truth preserving rule to premises true in I. So this first 192 Soundness and Completeness for
Sentence Logic Derivations 13-2. Soundness for Derivations: Fonnol Detaik 193 conclusion will be true in I. Now we have all the
premises and the first conclusion true in I. Next we apply a truth preserving rule to sentences taken from the premises andor this
first conclusion, all true in I. So the second conclusion will also be true in I. This continues, showing each conclusion down the
derivation to be true in I, including the last.
Mathematical induction makes this pattern of argument precise, telling us that if all the initial premises are true in I (as we assume
because we are interested only in such I), then all the conclusions of the derivation will likewise be true in I.
This sketch correctly gives you the idea of the soundness proof, but it does not yet deal with the complication arising from rules
which appeal to subderivations. Let's call a rule the inputs to which are all sentences a Sentence Rub and a rule the inputs to which
include a subderivation a Subderivation Rub. My foregoing sketch would be almost all we need to say if all rules were sentence
rules. However, we still need to consider how subderivation rules figure in the argument.
What does it mean to say that the subderivation rule, >I, is truth preserving? Suppose we are working in the outermost derivation,
and have, as part of this derivation, a subderivation which starts with assumption X and concludes with Y. To say that >I is truth
preserving is to say that if all the premises of the outer derivation are true in I, then X>Y is also true in I. Let's show that >I is truth
preserving in this sense.
We have two cases to consider. First, suppose that X is false in I. Then X>Y is true in I simply because the antecedent of X>Y is
false in I. Second, suppose that X is true in I. But now we can argue as we did generally for outer derivations. We have an
interpretation I in which X is true. All prior conclusions of the outer derivation have already been shown to be true in I, so that any
sentence reiterated into the subderivation will also be true in I. So by repeatedly applying the truth preserving character of the rules,
we see that Y, the final conclusion of the subderivation, must be true in I also. Altogether, we have shown that, in this case, Y as
well as X are true in I. But then X>Y is true in I, which is what we want to show.
This is roughly the way things go, but I hope you haven't bought this little argument without some suspicion. It appeals to the truth
preserving character of the rules as applied in the subderivation. But these rules include 31, the truth preserving character of which
we were in the middle of proving! So isn't the argument circular?
The problem is that the subderivation might have a sub-subderivation to which >I will be applied within the subderivation. We can't
run this argument for the subderivation until we have run it for the sub-subderivation. This suggests how we might deal with our
problem. We hope we can descend to the deepest level of subderivation, run the argument without appealing to >I, and then work
our way back out.
Things are sufficiently entangled to make it hard to see for sure if this strategy is going to work. Here is where mathematical
induction becomes indispensable. In chapter 11 all my applications of induction were trivial. You may have been wondering why
we bother to raise induction to the status of a principle and make such a fuss about it. You will see in the next section that, applied
with a little ingenuity, induction will work to straighten out this otherwise very obscure part of the soundness argument.
EXERCISES 13-1. Using my discussion of the >I rule as a model, explain what is meant by the rule -I being truth preserving and
argue informally that -I is truth preserving in the sense you explain. 13-2. Explain why, in proving soundness, we only have to deal
with the primitive rules. That is, show that if we have demonstrated that all derivations which use only primitive rules are sound,
then any derivation which uses any derived rules will be sound also.
13.1: Soundness for Derivations: Informal Introduction is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
13.2: Soundness for Derivations: Formal Details
The straightforward but messy procedure in our present case is to do a double induction. One defines the complexity of a derivation
as the number of levels of subderivations which occur. The inductive property is that all derivations of complexity n are sound. One
then assumes the inductive hypothesis, that all derivations with complexity less than n are sound, and proves that all derivations of
complexity n are sound. In this last step one does another induction on the number of lines of the derivation. This carries out the
informal thinking developed in the last section. It works, but it's a mess. A different approach takes a little work to set up but then
proceeds very easily. Moreover, this second approach is particularly easy to extend to predicate logic.
This approach turns on a somewhat different way of characterizing the truth preserving character of the rules, which I call Rule
Soundness, and which I asked you to explore in exercises 10-4, 10-5, and 10-6. One might argue about the extent to which this
characterization corresponds intuitively to the idea of the rules being truth preserving. I will discuss this a little, but ultimately it
doesn't matter. It is easy to show that the rules are truth preserving in the sense in question. And using the truth preserving character
thus expressed, proof of soundness is almost trivial.
194 Sowrdness and Compkteness for Sentence Logic Derivations 13-2. Soundness for Derivations: Fd Details 195 Here is the
relevant sense of rule soundness, illustrated for the case of &I. Suppose we are working within a derivation with premises Z.
Suppose we have already derived X and Y. Then we have ZtX and ZtY. &I then licenses us to conclude X&Y. In other words, we
can state the &I rule by saying
&I Rule: If ZtX and ZtY, then ZtX&Y.
There is a fine point here, about whether this really expresses the &I rule. The worry is that 'ZtX' means there exists a derivation
from Z to X, and 'ZtY' means that there exists a derivation from Z to Y. But the two derivations may well not be the same, and they
could both differ extensively from some of the derivations in virtue of which 'ZtX&Y' is true.
For sentence rules, this worry can be resolved. But it's really not important because, as with rule soundness, this way of stating the
rules will provide us with all we need for the soundness proof. We proceed by introducing the sense in which the &I rule is sound.
We do this by taking the statement of the rule and substituting 'k' for 't':
L7 (Soundness of &I): If Z~X and Z~Y, then Z~X&Y.
Why should we call this soundness of the &I rule? First, it has the same form as the rule &I. It is the semantic statement which
exactly parallels the syntactic statement of the &I rule. And it tells us that if we start with any interpretation I which makes the
premises Z true, and if we get as far as showing that X and Y are also true in I, then the conjunction X&Y is likewise true in I.
In particular, you can show that L7 directly implies that &I is truth preserving in the original sense by looking at the special case in
which Z = {X,Y}. {x,Y}~x and {X,Y}~Y are trivially true. So L7 says that {x,Y}~x&Y, which just says that any interpretation
which makes X true and also makes Y true makes the conjunction X&Y true.
We treat the other sentence rules in exactly the same way. This gives
L8 (Soundness of &E: If z~x&Y, then Z~X; and if Z~X&Y, then Z~Y.
L9 (Soundness of vI): If ZkX, then Z~XVY; and if Z~Y, then Z~XVY.
L10 (Soundness of vE): If Z~XVY and Zk-X, then Z~Y; and if Z~XVY and Zk - Y, then ZkX.
L11 (Soundness of -E): 1f Zk--x, then Z~X.
L12 (Soundness of 3E: If Z~X~Y and Z~X, then Z~Y.
L13 (Soundness of =I): If Z~X~Y and Z~Y~X, thenZ~XEY.
L14 (Soundness of =E): If Z~XEY, then Z~X~Y; and if Z~X~Y, then zky3x.
EXERCISES 13-3. Prove lemmas L7 to L14. Note that in proving these you do not need to deal with t at all. For example, to prove
L7, you need to show, using the antecedent, that Z~X&Y. So you assume you are given an I for which all sentences in Z are true.
You then use the antecedent of L7 to show that, for this I, X&Y is also true. 13-4. In this problem you will prove that for sentence
rules, such as the rules described in L7 to L14, what I have called rule soundness and the statement that a rule is truth preserving
really do come to the same thing. You do this by giving a general expression to the correspondence between a syntactic and a
semantic statement of a rule:
Suppose that X, Y, and W have forms such that
(i) (VI){[Mad(I,X) & Mod(I,Y)] 3 Mod(I,W)}.
That is, for all I, if I makes X true and makes Y true, then I makes W true. Of course, this won't be the case for just any X, Y, and
W. But in special cases, X, Y, and W have special forms which make (i) true. For example, this is so if X = U, Y = UIV, and W = V.
In such cases, thinking of X and Y as input sentences of a rule and W as the output sentence, (i) just says that the rule that allows
you to derive W from X and Y is truth preserving in our original sense.
Now consider
(ii) If Z~X and ZkY, then Z~W.
This is what I have been calling soundness of the rule stated by saying that if ZtX and ZtY, then ZtW. (ii) gives turnstyle expression
to the statement that the rule which licenses concluding W from X and Y is truth preserving.
Here is your task. Show that, for all X, Y, and W, (i) holds iff and (ii) holds. This shows that for sentence rules (rules which have
only sentences as inputs) the two ways of saying that a rule is truth preserving are equivalent. Although for generality, I have
expressed (i) and (ii) with two input sentences, your proof will work for rules with one input sentence. You can show this trivially
by letting Y = Av-A for rules with one input sentence.
I have not yet discussed the two subderivation rules, 31 and -I. Soundness of these rules comes to
L15 (Soundness of >I): 1f ZU{X}~Y, then ZkX>Y.
L16 (Soundness of -I): 1f ZU{X)~Y and ZU{X}~-Y, then Zk-X.
In the case of >I and -I there is a more substantial question of whether, and in what sense, L15 and L16 also express the intuitive
idea that these rules are truth preserving. The problem is that the turnstyle notion makes no direct connection with the idea of
subderivations. Thus, if the syntactic counterpart of L15 is assumed (if ZU{X}kY, then ZkX>Y), it is not clear whether, or in what
sense, one can take this to be a statement of the >I rule. (The converse is clear, as you will show in exercise 13-6.) However, this
issue need not sidetrack us, since L15 and L16 will apply directly in the inductive proof, however one resolves this issue.
EXERCISES 13-5. Prove L15 and L16. 13-6. Prove that if the system of derivations includes the rule >I, then if ZU{X}kY, then
ZkX>Y. Also prove that if the system of derivations includes the rule -I, then if both ZU{X}kY and ZU{X}k-Y, then Zk-X.
We are now ready to prove T5, soundness for derivations. Here is an outline of the proof: W; will start with an arbitrary derivation
and look at an arbitrary line, n. We will suppose that any interpretation which makes governing premises and assumptions true
makes all prior lines true. Rule soundness will then apply to show that the sentence on line n must be true too. Strong induction will
finally tell us that all lines are true when their governing premises and assumptions are true. The special case of the derivation's last
line will constitute the conclusion we need for soundness.
To help make this sketch precise, we will use the following notation:
X,, is the sentence on line n of a derivation. Z, is the set of premises and assumptions which govern line n of a derivation.
Now for the details. Suppose that for some Z and X, ZkX. We must show that Z~X. The assumption ZkX means that there is some
derivation with premises a subset of Z, final conclusion X, and some final line number which we will call n*. The initial premises
are the sentences, Z,., governing the last line, n*; and the final conclusion, X, is the sentence on the last line, which we are calling
X,.. We will show that z,.~x,.. This will establish ZkX because X,,. = X and Z,. is a subset of Z. (Remember exercise 10-10.)
We will establish Z,.kX,. by showing that z,kX, for all n, 1 5 n 5 n*. And in turn we will establish this by applying strong induction.
We will use the
Inductive property: &k&.
and the
Inductive hypothesis: ZikX, holds for ail i < n.
So let's consider an arbitrary line, n, and assume the inductive hypothesis. What we have to do is to consider each of the ways in
which line n might be justified and, applying the inductive hypothesis, show that the inductive property holds for line n. First, X,,
might be a premise or an assumption. Notice, by the way, that this covers the special case of the first line (n = l), since the first line
of a derivation is either a premise or, in the case of a derivation with no premises, the assumption of a subderivation. But if X,, is a
premise or assumption, X,, is a member of Z,. Therefore, z,~x,.
Next we consider all the sentence rules. I'll do one for you and let you do the rest. Suppose that X,, arises by application of &I to
two previous lines, Xi and Xj, so that X,, = X,&X,. By the inductive hypothesis
4k~i and Zjk% (Inductive hypothesis)
Since we are dealing with a sentence rule, Xi, Xj, and X,, all occur in the same derivation. Consequently, Zi = Zj = Z,. So
Z,kxi and Z,kxj.
This is just the antecedent of lemma 7, which thus applies to the last line to give z,kX,,.
EXERCISE 13-7. Apply lemmas L8 to L14 to carry out the inductive step for the remaining sentence rules. Your arguments will
follow exactly the same pattern just illustrated for &I.
Turning to the other rules, suppose that X, arises by reiteration from line i. That is just to say that X,, = Xi. We have as inductive
hypothesis that zikxi. If lines i and n are in the same derivation, Z, = Zi, so that z,kX,,, as we require. If we have reiterated Xi into a
subderivation, Z, 198 Soundness and Completeness for Sentence Logic Derivations 13-3. Completeness fm Dm'votionr: Znfd
Zntruduetion 199 differs from Zi by adding the assumption of the subderivation (or the assumptions of several subderivations if we
have reiterated several levels down). That is, Zi is a subset of Z,. But as you have shown in exercise 10- 10, if ~~1% and Zi is a
subset of Z,, then z,kX,.
Now suppose that X,, arises by >I. Then on previous lines there is a subderivation, beginning with assumption Xi and concluding
with Xj, so that X,, = Xi>Xj. By inductive hypothesis,
zjkxj (Inductive hypothesis for line j)
The trick here is to notice that the subderivation has one more assumption than Z,. Though not perfectly general, the following
diagram will give you the idea:
Set of Premises and Assumptions
When we start the subderivation with the assumption of Xi, we add the assumption Xi to Z, to get Zi = Z,U{XJ as the total set of
premises and assumptions on line i. When we get to line n and discharge the assumption of Xi, moving back out to the outer
derivation, we revert to Z, as the set of governing premises and assumptions.
Since Zj = Z,u{Xi), we can rewrite what the inductive hypothesis tells us about line j as
But this is just the antecedent of lemma L15! Thus lemma L15 immediately applies to give z,kXi>Xj, or z,~x,, since X, = Xi>Xj.
13-8. Carry out the inductive step for the case in which X, arises by application of -I. Your argument will appeal to lemma L16 and
proceed analogously to the case for >I.
We have covered all the ways in which X,, can arise on a derivation. Strong inducton tells us that z,kX,, for all n, including n*, the
last line of the derivation. Since Z,* is a subset of Z and X,. = X, this establishes ZkX, as was to be shown.

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13.3: Completeness for Derivations: Informal Introduction
We still need to prove
T6 (Completeness for sentence logic derivations): For any finite set of sentences, Z, and any sentence, X, if Z~X, then ZtX.
where 'I-' is understood to mean kd, derivability in our natural deduction system. The proof in this section assumes that Z is finite.
Chapter 14 will generalize to the case of infinite Z. -
The proof of completeness for derivations is really an adaptation of the completeness proof for trees. If you have studied the tree
completeness proof, you will find this and the next section relatively easy. The connection between trees and derivations on this
matter is no accident. Historically, the tree method was invented in the course of developing the sort of completeness proof that I
will present to you here.
Begin by reading section 12-1, if you have not already done so, since we will need lemma L1 and the notation from that section.
Also, do exercises 12-1 and 12-2. (If you have not studied trees, you will need to refresh your memory on the idea of a
counterexample; see section 4-1, volume I.) For quick reference, I restate L1:
L1: Z~X iff ZU{-X} is inconsistent.
The basis of our proof will be to replace completeness with another connection between semantic and syntactic notions. Let us say
that
D19: Z is Syntactically Inconsirtent iff ZtA&-A.
Semantic inconsistency is just what I have been calling 'inconsistency', defined in chapter 10, D7, as (V1)-Mod(1,Z). L1 says that
an argument is valid iff the premises together with the negation of the conclusion form a semantically inconsistent set. Analogously
L17: ZU{-X}tA&-A iff ZtX.
says that -X together with the sentences in Z form a syntactically inconsistent set iff there is a proof using sentences in Z as
premises to the conclusion X. Together, L1 and L17 show that T6 is equivalent to
T7: For any finite set of sentences, Z, if Z is semantically inconsistent, then Z is syntactically inconsistent; that is, if (W)-
Mod(I,Z), then ZkA&-A.
EXERCISES 13-9. Prove L17. 13-10. Using L1 and L17, prove that T6 is equivalent to T7.
We have boiled our problem down to proving T7. We do this by developing a specialized, mechanical kind of derivation called a
Semantic Tableau Derivation. Such a derivation provides a systematic way of deriving a contradiction if the original premises form
an inconsistent set.
If you haven't done trees, it is going to take you a little time and patience to see how this method works. On a first reading you may
find the next few paragraphs very hard to understand. Read them through even if you feel quite lost. The trick is to study the two
examples. If you go back and forth several times between the examples and the text you will find that the ideas will gradually come
into focus. The next section will add further details and precision.
A semantic tableau derivation is a correct derivation, formed with a special recipe for applying derivation rules. Such a derivation
is broken into segments, each called a Semantic Tableau, marked off with double horizontal lines. We will say that one tableau
Generates the next tableau. Generating and generated tableaux bear a special relation. If all of a generated tableau's sentences are
true, then all the sentences of previous generating tableaux are true also. In writing a derivation, each tableau we produce has
shorter sentences than the earlier tableaux. Thus, as the derivation develops, it provides us with a sequence of tableaux, each a list
of sentences such that the sentences in the later tableaux are shorter. The longer sentences in the earlier tableaux are guaranteed to
be true if all of the shorter sentences in the later tableaux are true.
A tableau derivation works to show that if a set, Z, of sentences is semantically inconsistent, then it is syntactically inconsistent.
Such derivations accomplish this aim by starting with the sentences in Z as its premises. The derivation is then guaranteed to have
'A&-A' as its final conclusion if Z is (semantically) inconsistent.
To see in outline how we get this guarantee, suppose that Z is an arbitrary finite set of sentences, which may or may not be
inconsistent. (From now on, by 'consistent' and 'inconsistent' I will always mean semantic consistency and inconsistency, unless I
specifically say 'syntactic consistency' or 'syntactic inconsistency'.) A tableau derivation, starting from Z as premises, will continue
until it terminates in one of two ways. In the first way, some final tableau will have on it only atomic and/or negated atomic
sentences, none of which is the negation of any other. You will see that such a list of sentences will describe an interpretation which
will make true all the sentences in that and all previous tableaux. This will include the original premises, Z, showing this set of
sentences to be consistent. Furthermore, we will prove that if the initial sentences form a consistent set, the procedure must end in
this manner.
Consequently, if the original set of sentence forms an inconsistent set, the tableau procedure cannot end in the first way. It then
ends in the second way. In this alternative, all subderivations end with a contradiction, 'A&-A'. As you will see, argument by cases
will then apply repeatedly to make 'A&-A' the final conclusion of the outermost derivation.
Altogether we will have shown that if Z is (semantically) inconsistent, then Z!-A&-A, that is, Z is syntactically inconsistent.
To see how all this works you need to study the next two examples. First, here is a tableau derivation which ends in the first way (in
writing lines 3 and 4, I have omitted a step, '-B&-C', which gives 3 and 4 by &E):
You can see that this is a correct derivation in all but two respects: I have abbreviated by omitting the step '-B&-C', which comes
from 1 by DM and gives 3 and 4 by &E; and I have not discharged the assumptions of the subderivations to draw a final conclusion
in the outer derivation.
Each tableau is numbered at the end of the double lines that mark its -(BvC) P BvD P 1 -B 1, DM, &E -C 1, DM, &E BvD 2, R 6 7
8 9 A&-A 6, 7, CD X 3 11 12 -C 4, R 4 + PO2 Soundness and Completeness for Sentence Lo& Mvations 134. Completeness for
Denenvathis: Ponnal Details PO3 end. A tableau may generate one new tableau (Sequential Generation): In this example tableau 1
generated tableau 2 by applying the rules DM, &E, and R. Or a tableau may generate two new tableaus (Branching Generatian): In
the example tableau 2 generated tableaux 3 and 4 by starting two new subderivations, each using for its assumption one of the
disjuncts, 'B' and 'D' of 'BvD' on line 5, and each reiterating the rest of tableau 2.
Tableau 3 ends in a contradiction. It can't describe an interpretation. We mark it with an 'x' and say that it is Closed. Tableau 4,
however is Open. It does not contain any sentence and the negation of the same sentence; and all its sentences are Minimal, that is,
either atomic or negated atomic sentences. Tableau 4 describes an interpretation by assigning f to all sentence letters which appear
negated on the tableau and t to all the unnegated sentence letters. In other words, the interpretation is the truth value assignment
which makes true all the sentences on this terminal tableau.
Note how the interpretation described by tableau 4 makes true all the sentences on its generator, tableau 2. The truth of '-B' and '4'
carries upward simply because they are reiterated, and the truth of 'D' guarantees the truth of 'BvD' by being a disjunct of the
disjunction. You should check for yourself that the truth of the sentences in tableau 2 guarantees the truth of the sentences in
tableau 1.
Examine this example of a tableau derivation which ends in the second way:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 -(BvC) BvC 1 - B -C BvC -C A&-A P P 1, DM, &E 1, DM, &E 2, R A 3, R 4, R 6, 7, CD
In this example, all terminal tableaux (3 and 4) close, that is, they have both a sentence and the negation of the same sentence, to
which we apply the rule CD. We can then apply AC to get the final desired conclusion, 'A&-A'.
Again, here is the key point: I am going to fill in the details of the method to guarantee that a consistent initial set of sentences will
produce a derivation like the first example and that an inconsistent set will give a result like the second example. More specifically,
we will be able to prove that if there is an open terminal tableau, like tableau 4 in the first example, then that tableau describes an
interpretation which makes true all its sentences and all the sentences on all prior tableaux. Thus, if there is an open terminal
tableau, there is an interpretation which constitutes a model of all the initial sentences, showing them to form a consistent set.
Conversely, if the original set is inconsistent, all terminal tableaux must close. We will than always be able to apply argument by
cases, as in the second example, to yield 'A&-A' as a final conclusion. But the last two sentences just state T7, which is what we
want to prove.
To help you get the pattern of the argument, here is a grand summary which shows how all our lemmas and theorems connect with
each other. We want to show T6, that if ZkX, then ZkX. We will assume ZkX, and to take advantage of lemmas L1 and L17, we
then consider a semantic tableau derivation with the sentences in ZU{-X} as the initial tableau. Then we argue
(1) Z~X. (Assumption)
(2) If Z~X, then ZU{-X} is inconsistent. (By L1)
(3) If some terminal tableau is open, then ZU{-X} is consistent. (By L18, to be proved in the next section)
(4) If ZU{-X} is inconsistent, then all terminal tableaux close. (Contrapositive of (3))
(5) If all terminal tableaux close, then ZU{-X}l-A&-A. (L20, to be proved in the next section)
(6) If ZU{-X}l-A&-A, then Zl-X. (By L17)
Now all we have to do is to discharge the assumption, (I), applying it to (2), (4), (5), and (6), giving T6: If Z~X, then Zl-X.
In the next section we carry out this strategy more compactly by proving T7 (corresponding to (4) and (5) above), which you have
already proved to be equivalent to T6.

13.3: Completeness for Derivations: Informal Introduction is shared under a not declared license and was authored, remixed, and/or curated by
LibreTexts.
13.4: Completeness for Derivations: Formal Details
To keep attention focused on the main ideas, I'm going to restrict consideration to sentences in which '-' and 'v' are the only
connectives used. 404 Soundness and Completenesr for Sentence Logic DctiYotions 13-4. Completeness for Derivations: Fond
Details 205 Once you understand this special case, extension to the other connectives will be very easy. As I mentioned, I will also
carry out the proof only under the restriction that the initial set of sentences, Z, is finite. Chapter 14 will generalize the result to
infinite sets, Z.
To help fix ideas, I'll start with a slightly more extended example. Skip over it now and refer back to it as an illustration as you read
the details.
The method of semantic tableau derivations constitutes a way of testing a finite initial set of sentences for consistency. Here are the
rules for generating such a derivation:
R1 Znicial Tabbau: The method begins by listing the sentences in the set to be tested as the premises of the derivation. This
initial list constitutes the initial tableau.
Lines 1 and 2 in the example are an initial tableau.
Each further tableau (the Generabd Tableau) is generated from some prior tableau (the Generating Tableau) by one of two methods:
R2 Sequatial generation
a) Each line of the generated tableau is a new line of the same derivation as the generating tableau.
b) If a sentence of the form --X occurs on the generating tableau, enter X on the generated tableau.
c) If a sentence of the form -(XVY) occurs on the generating tableau, enter -X and -Y as separate lines on the generated
tableau. -
d) Reiterate all remaining sentences of the generating tableau as new lines of the generated tableau.
Tableaux 2 and 3 in the example illustrate sequentially generated tableaux. c) is illustrated in the example by lines 3, 4, 6, 7, 8, and
9. d) is illustrated by lines 5 and 10. Note that the rule I apply for c), which I have called '-v', is a new derived rule, constituted by
simply applying DM followed by &E.
R3 Branching generation:
a) If a sentence of the form XVY occurs on the generating tableau, start two new subderivations, one with assumption X and
the other with assumption Y.
b) Reiterate all the remaining sentences of the generating tableau on each of the subderivations.
c) Each of the (initial parts of) the subderivations started by steps a) and b) constitutes a generated tableau.
Branching generation is illustrated in the example by tableaux 4, 5, 6, 7.
Tableaux 4, 6, and 7 illustrate what happens when both a sentence and the negation of a sentence appear on a tableau. No
interpretation will make all the sentences on such a tableau true. So such a tableau will never provide an interpretation which will
prove the original sentences consistent. We record this fact by extending the tableau by applying CD to derive 'A&-A'. We say that
such a tableau is Closed and mark it with an ' x '.
We have applied CD to draw the explicit contradiction, 'A&-A', on closed tableaux because this contradiction will be helpful in
deriving 2 06 Soundness and Completeness for Sentence Logic Derivations 134. Completeness for Derivations: Formal Details 207
'A&-A' in the outermost derivation. We will see that, if the original set of sentences is inconsistent, then all chains of tableaux will
terminate with a closed tableau. Argument by cases will then allow us to export 'A&-A' from subderivations to outer derivations,
step by step, until we finally get 'A&-A' as the final conclusion of the outermost derivation.
We make these ideas more precise with two further instructions:
R4: If both a sentence and the negation of the same sentence appear on a tableau, apply CD to derive 'A&-A' as the last line
of the tableau, and mark the end of the tableau with an 'X ' to indicate that it is Closed. Do not generate any new tableaux
from a closed tableau.
R5: If 'A&-A' appears on two subderivations, both generated by the same disjunction in the outer derivation, apply AC to
write 'A&-A' as the final conclusion on the outer derivation.
Look again at tableaux 4, 6, and 7, as illustrations of R4. Lines 30 and 31 illustrate R5.
We now need to prove that semantic tableau derivations do what they are supposed to do. Here is the intuitive idea. We start with a
set of sentences. The tableau procedure constitutes a way of determining whether or not this set is consistent. This works by
systematically looking for all possible ways of making the original sentences true. If the systematic search turns up a way of
making all the original sentences true (a model), then we know that the original set is consistent. Indeed, we will prove that if the
original set is consistent, the procedure will turn up such an interpretation. Thus we know that if the procedure fails to turn up such
an interpretation, the original set must be inconsistent. This is signaled by all chains of tableaux terminating with a closed tableau.
The procedure accomplishes these aims by resolving the original sentences into simpler and simpler sentences which enable us to
see what must be true for the original set to be true. Each new tableau consists of a set of sentences, at least some of which are
shorter than previous sentences. If all of the generated tableau's sentences are true, then all of the sentences on the generating
tableau will be true. For a sequentially generated tableau, the new sentences give us what has to be true for the sentences on the
generating tableau to be true. When we have branching generation, each of the two new tableaux gives one of the only two possible
ways of making all sentences of the generating tableau true. In this way the procedure systematically investigates all ways in which
one might try to make the original sentences true. Attempts that don't work end in closed tableaux.
We need to work these ideas out in more detail. We will say that A tableau is a
Tenninul Tableau if it has not generated any other tableau, and no rule for tableau generation applies to it.
It can happen that no rule applies to a tableau for one of two reasons: The tableau can be closed. Or it might be open but have only
minimal sentences (atomic or negated atomic sentences). We will discuss these two - cases separately.
First we will prove
L18: An open terminal tableau describes an interpretation in which all sentences of the initial tableau are true.
An open terminal tableau has only minimal sentences, none of which is the negation of any other. The interpretation such a tableau
specifies is the one which makes all its sentences true, that is, the assignment of t to all the tableau's unnegated atomic sentences
and f to the atomic sentences which appear negated on the tableau. Let's call such an interpretation a Tminul Interpretation, for
short.
Our strategy will be to do an induction. Suppose we are given an open terminal tableau, and so the terminal interpretation, I, which
it specifies. The fact that all the sentences of the terminal tableau are true in 1 provides our basis step. For the inductive step you
will show that instructions for constructing a tableau derivation guarantee that if all the sentences of a generated tableau are true in
an interpretation, then all the sentences of the generating tableau are true in the same interpretation. Thus all the sentences of the
tableau which generated the terminal tableau will be true in I. In turn, that tableau's generator will have all its sentences true in I.
And so on up. In short, induction shows that all the Ancestors of the open terminal tableau are true.
To fill in the details of this sketch, you will first prove the inductive step:
L19: If tableau Tp is generated from tableau TI and all sentences of T, are true in interpretation I, then all the sentences of TI are
also true in I.
EXERCISE 13-11. Prove L19.
Since the proof of L18 will be inductive, we need to specify more clearly the sequence of cases on which to do the induction:
A terminal tableau's generator will be called the tableau's first Ancestor. In general, the i + 1st ancestor of a terminal tableau
is the generator of the ith ancestor.
We will do the induction starting from a 0th case, namely, the terminal tableau. The ith case will be the terminal tableau's ith
ancestor.
We are now ready to prove L18. Suppose we are given a semantic tableau derivation, with an open terminal tableau. This tableau
specifies an interpretation, I, in which all the terminal tableau's sentences are true. The inductive property is: The nth ancestor of
the terminal tableau has all its sentences true in I. The terminal tableau provides the basis case. By L19, if the nth ancestor of the
terminal tableau has all its sentences true in I, then so does the n + 1st ancestor. Then, by induction, all the terminal tableau's
ancestors have all their sentences true in I, which includes the derivation's initial tableau, as required to prove L18.
I have now said all I need about tableau derivations which terminate with one or more open tableaux. What happens if all the
terminal tableaux are closed? In a word, rule R5 applies repeatedly until, finally, 'A&-A' appears as the final conclusion of the
outermost derivation:
L20: If in a semantic tableau derivation all the terminal tableaux are closed, then 'A&-A' appears as the derivation's final
conclusion.
We will prove this with another induction.
We need a sequence of cases on which to do the induction. The natural choice is the level or depth of subderivations, as measured
by the number of nested scope lines. But we want to start with the deepest level of subderivation and work our way back out. So we
need to reverse the ordering: The first level of subderivations will be the deepest, the second will be the subderivations one level
less deep, and so on. More exactly defined
Given a tableau derivation, let k be the largest number of nested scope lines on the derivation (including the outermost scope
line). The Inverted Level of each subderivation is k less the number of scope lines to the left of the subderivation.
(I will henceforth omit the word 'inverted' in 'inverted level'.) The key to the proof will be the inductive step:
L21: Let D be a semantic tableau derivation in which all terminal tableaus are closed. Then, if all of D's subderivations of
level n have 'A&-A' as their final conclusion, so do all the subderivations of level n + 1.
(I construe 'subderivation' broadly to include the outermost derivation, a sort of null case of a subderivation.)
EXERCISE 13-12. Prove L2 1. I
We are now ready to prove L20. Let D be a semantic tableau derivation in which all terminal tableaux are closed. Our inductive
property will be: All the subderivations of level n have 'A&-A' as their final conclusion. Ai level 1 all subderivations have no sub-
subderivations. So all of the subderivations must end in terminal tableaux. By assumption, all of these are closed. So the inductive
property holds for level 1. L21 gives the inductive step. By induction, the derivations at all levels conclude with 'A&-A', which
includes the outermost derivation.
We are at long last ready to prove T7. Suppose that Z, a finite set of sentences, is inconsistent. (Note that, if inconsistent, Z must
have at least one sentence.) Make the sentences of this set the first tableau of a semantic tableau derivation. Suppose that the
derivation has an open terminal tableau. Then, by L18, there is an interpretation which makes true all the sentences in Z. But this is
impossible since Z is supposed to be inconsistent. Therefore all terminal tableaux are closed. Then L20 tells us that the derivation
terminates with 'A&-A', so that ZkA&-A, as was to be shown.
We have one more detail to complete. My proof of T7 is subject to the restriction that 'v' and '-' are the only connectives which
appear in any of the sentences. We easily eliminate this restriction by exchanging sentences with other connectives for logical
equivalents which use 'v' and '-' instead. At each stage we deal only with the main connective or, for negated sentences, with the
negation sign and the main connective of the negated sentence. We rewrite rule R2 for sequential generation to read:
R2 Sequential generation:
a) Each line of the generated tableau is a new line of the same derivation as the generating tableau.
b) If a sentence of the form --X occurs on the generating tableau, enter X on the generated tableau.
c) If a sentence of the form -(XvY) occurs on the generating tableau, enter both -X and -Y as separate lines on the generated
tableau.
d) If a sentence of the form X&Y occurs on the generating tableau, enter both X and Y as separate lines on the generated tableau.
e) If a sentence of the form X>Y occurs on the generating tableau, enter -XvY on the generated tableau.
f) If a sentence of the form XEY occurs on the generating tableau, enter (X&Y)v(-X&-Y) on the generated tableau.
g) If a sentence of the form -(X&Y) occurs on the generating tableau, enter -XV-Y on the generated tableau.
h) If a sentence of the form -(X>Y) occurs on the generating tableau, enter both X and -Y as separate lines on the generated
tableau.
i) If a sentence of the form -(X=Y) occurs on the generating tableau, enter (X&-Y)v(-X&Y) on the generated tableau.
j) Reiterate all remaining sentences of the generating tableau as new lines of the generated tableau.
We could provide a more complicated version of R2 which would produce more efficient tableau derivations, but it's not worth the
effort since true efficiency is only obtained with the truth tree method. In the next exercises you will show that the proof for the
special case, using only the connectives 'v' and '-', extends to the general case covered by our reformulated R2.
EXERCISES Generalizing the proof of T7 only requires checking three points. 13-13. I argued that a tableau derivation always
comes to an end because each new tableau shortens at least one sentence of the previous tableau. This argument no longer works, at
least not as just stated. Show that tableau derivations, with sentences using any sentence logic connectives and the new rule R2,
always come to an end. 13-14. Check that when all terminal tableaux close, a tableau derivation created using the new rule R2 is a
correct derivation. You will have to prove two new derived rules, one for biconditionals and one for negated biconditionals.
13-15. Reprove lemma L19 for our fully general tableau derivations.
13-16. Explain why the proof of completeness in this section shows that the primitive sentence logic derivation rules of chapter 5
(volume I) are complete for sentence logic.
CHAPTER CONCEPTS
As a check on your mastery of this material, review the following ideas to make sure you understand them clearly:
a) Rule Soundness
b) Sentence Rule
c) Subderivation Rule
d) Semantic and Syntactic Inconsistency
e) Semantic Tableau Derivation (or Tableau Derivation)
f) Tableau
g) Initial Tableau
h) Generating Tableau
i) Generated Tableau
j) Sequential Generation
k) Branching Generation
I) Derived Rule -v
m) Closed Tableau
n) Minimal Sentence
o) Terminal Tableau
p) Terminal Interpretation
q) Ancestors of a Tableau

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SECTION OVERVIEW
3.14: Compactness, and Generalization to Infinite Sets of Premises

3.14.1: Koenig's Lemma

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14.1: Koenig's Lemma
My proofs of completeness, both for trees and for derivations, assumed finiteness of the set Z in the statement ~k-X. Eliminating
this restriction involves something called 'compactness', which in turn is a special case of a general mathematical fact known as
'Koenig's lemma'. Since we will need Koenig's lemma again in the next chapter, we will state and prove it in a form general enough
for our purposes.
Suppose we have a branching system of points, or Nodes, such as the following:
The nodes are connected by branching lines running downward; these 212 are called Paths, or Branches. I have numbered the
horizontal lines to help in referring to parts of the tree. We will consider only tree structures which have Finite Branching-that is,
from any one node, only finitely many branches can emerge. To keep things simple, I will always illustrate with double branching,
that is, with at most two branches emerging from a node. The restriction to two branches won't make an important difference.
Truth trees are one example of such a tree structure. Semantic tableau derivations are another, with each branch representing the
formation of a new subderivation and each node representing all the tableaux on a subderivation before starting new
subderivations. Some of the paths end with a 'x ', as when we close a path in a truth tree or close a tableau in a tableau derivation.
We say that such a path is Closed. A tree might have only finitely many horizontal lines, That is, there might be a line number, n,
by which all paths have ended, or closed. Or such a tree might have infinitely many lines. What we want to prove is that if such a
tree is infinite (has infinitely many horizontal lines with at least one open path extending to each line), then there is an infinite path
through the tree.
Perhaps this claim will seem obvious to you (and perhaps when all is said and done it is obvious). But you should appreciate that
the claim is not just a trivial logical truth, so it really does call for demonstration. The claim is a conditional: If for every line there
is an open path extending to that line, th there is an open path which extends to every line. The antecedent of the conditional is a
doubly quantified sentence of the form (Vu)(3v)R(u,v). The consequent is the same, except that the order of the quantifiers has
been reversed: (3v)(Vu)R(u,v). Conditionals of this form are not always true. From the assumption that everyone is loved by
someone, it does not follow that there is someone who loves everyone. The correctness of such conditionals or their corresponding
arguments requires special facts about the relation R.
The tree structure provides the special facts we need in this case. Let's assume that we have an infinite tree, that is, a tree with
infinitely many horizontal lines and at least one open path extending to each line. The key is to look at infinite subtrees. For
example, look at line 3. The first, third, and fourth nodes can each be viewed as the first node in its own subtree, that is, the system
of paths which starts with the node in question. The first node of line 3 heads a subtree which does not end, at least not as far as we
can tell by as much of the tree as I have drawn. The same is true for the third node of line 3. But the fourth node heads a subtree
that we can see is finite: All paths starting from that node close.
Now consider all of the nodes of line 3 again. Suppose that all of the subtrees headed by these nodes are finite. Then the whole tree
would be finite. Line 3 has only four nodes, and if each has below it only finitely many nodes, then there are only finitely many
nodes in the whole tree. 414 Koenig'r Lemma, Compactness, and Generalization to Infinite Sets of Premises 14-2. Compactness
and Infinite Sets of Premises 415 In such cases there are no more than four times the maximum number of nodes in the subtrees
headed by line 3 nodes, plus the three nodes in lines 1 and 2. Conversely, if the whole tree is infinite, at least one node of line 3
must head an infinite subtree.
We can use induction to prove that the same will be true of any line of an infinite tree:
L22: In any infinite tree, every line has at least one node which heads an infinite subtree.
Suppose we have an infinite tree. Our inductive property will be: The nth line has at least one node which heads an infinite tree.
Line 1 has this property, by assumption of the argument. This gives the basis step of the induction. For the inductive step, assume
the inductive hypothesis that line n has the inductive property. That is, line n has at least one node which heads an infinite tree. Let
N be the leftmost such node. Consider the nodes on line n + 1 below node N. If both of these nodes were to head only finite
subtrees, then N would also head only a finite subtree, contrary to the inductive hypothesis. So at least one of these nodes of line n
+ 1 must also head an infinite subtree. In sum, if line n has the inductive property, so does line n + 1, completing the inductive
proof of L22. It is now easy to establish
L23 (Koenig's lemma): In any infinite tree there is an infinite path.
Proof: Given an infinite tree, start with the top node and extend a path from each line to the next by choosing the leftmost node in
the next line which heads an infinite tree. L22 guarantees that there will always be such a node. Since at each stage we again pick a
node which heads an infinite tree, the process can never end. (See Exercise 14-1.)

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In my proofs of completeness, the statement that if ZkX, then ZtX, I assumed that Z is finite. But in my original definition of z~X
and ZtX, I allowed Z to be infinite. Can we lift the restriction to finite Z in the proofs of completeness?
There is no problem with t. By ZtX, for infinite Z, we just mean that there is a proof which uses some finite subset of Z as
premises. Counting Z as a subset of itself, this means that (whether Z is finite or infinite) X can be derived from Z iff X can be
derived from some finite subset of Z. That is (using 'Z'CZ' to mean that Z' is a subset of Z)
(1) ZkX iff (3Z1)(Z'CZ and Z' is finite and Z'kX).
EXERCISE 14-1. Consider a tree that looks like this:
This tree differs from the ones we have been considering because it allows Infinite Branching-that is, from one node (here, the first
node) infinitely many new branches emerge. These branches also extend farther and farther down as you move from left to right, so
that the tree extends infinitely downward as well as to the right. For each integer, n, there is an open path that extends to the nth
line. But there is no infinite path through the tree!
This example helps to show that Koenig's lemma is not just a trivial truth. Thinking about this example will also help to make sure
you understand the proof of Koenig's lemma.
Explain why the proof of Koenig's lemma breaks down for trees with infinite branching. My proof actually assumed at most double
branching. Rewrite the proof to show that Koenig's lemma works when the tree structure allows any amount of finite branching.
What we need is a similar statement for 1:
(2) Z~X iff (3Z1)(Z'C Z and Z' is finite and ~'kx).
(1) and (2) will enable us quickly to connect completeness for finite Z' with completeness for infinite Z. Using L1 we see that (2) is
equivalent to
(3) ZU{-X) is inconsistent iff (3Zr)(Z'CZ and Z' is finite and Z'U{-X) is inconsistent).
Compactness is just (3), but stated slightly more generally, without the supposition that the inconsistent set has to include the
negation of some sentence:
T8 (Compactness): Z is inconsistent iff Z has an inconsistent finite subset. Equivalently, Z is consistent iff all its finite
subsets are consistent.
Compactness with the help of L1 will immediately give us
T9 (Completeness): If Z~X, then ZtX, where Z now may be infinite.
t may be derivability by trees or derivations (or, indeed many other systems of proof). All that we require here is (l), compactness,
and completeness for finite sets Z in the system of proof at hand.
EXERCISES
14-2. Prove the equivalence of the two statements of compactness in T8.
14-3. Prove completeness for arbitrary sets of sentences. That is, prove that if Z~X, then ZtX, where Z may be infinite. Do this by
using compactness and L1 to prove (2). Then use (2) and (I), together with the restricted form of completeness we have already
proved (with Z restricted to being a finite set) to lift the restriction to finite Z.
The key here is compactness, and the key to compactness is Koenig's lemma. In outline, we will create a tree the paths of which
will represent lines of a truth table. Finite subsets of an infinite set of sentences, Z, will be made true by paths (truth table lines)
reaching down some finite number of lines in our tree. Koenig's lemma will then tell us that there is an 14-2. Compactness and
Infinite Sets ofPremires 2 17 infinite path, which will provide the interpretation making everything in Z true, showing Z to be
consistent.
Here goes. Since our language has infinitely many sentence letters, let's call the sentence letters 'A,', 'A2', . . . , 'A,'. . . . Consider the
tree which starts like this:
Each branch through the third line represents one of the eight possible truth value assignments to 'A,', 'A2', and 'As'. Branch (1)
represents 'A,', 'A;, and 'A3' all true. Branch (2) represents 'A,' and 'A2' true and 'A3' false. Branch (3) represents 'Al' true, 'A2'
false, and 'A3' true. And so on. Line 4 will extend all branches with the two possible truth value assignments to 'A;, with 'A; true on
one extension and 'A; false on the other. Continuing in this way, each initial segment of a branch reaching to line n represents one
of the truth value assignments to 'A,' through 'A,', and every possible truth value assignment is represented by one of the branches.
Now let us suppose that the set, Z, is composed of the sentence logic sentences X1, Xp, . . . , X, . . . , all written with the sentence
letters 'A,', 'A,', . . . , 'A,'. . . . Let Z, = {XI, Xp, . . . XJ. That is, for each n, Z, is the finite set composed of the first n sentences in
the list XI, XI. . . . Finally, let us suppose that each Z, is consistent, that is, that Z, has a model, an interpretation, I, which assigns
truth values to all sentence letters appearing in the sentences in Z, and which makes all the sentences in Z, true.
Our tree of truth value assignments will have initial path segments which represent the models which make the 2,'s consistent.
Koenig's lemma will then tell us that there will be an infinite path which makes all the XI, Xp, . . . . true. To show this carefully, let
us prune the truth value tree. For each Z,, starting with ZI, let in be the first integer such that all the sentence letters in the sentence
in Z, occur in the list 'A,', 'A2', . . . , 'Ai,'. Then the initial paths through line in will give all the possible interpretations to the
sentences in 2,. Mark as closed any path which does not 418 Koenig'r Lemma, Compactness, and Generalwtion to Injnite Sets of
Premises 14-2. Compactness and lnjnite Sets of Premises 2 19 represent a model of Z,, that is, which makes any sentence in Z,
false. Since each Z, is consistent, there will be at least one open path reaching to line in.
I have provided an outline of a proof of lemma 24:
L24: Let XI, X, . . . 2C,, be an infinite sequence of sentences, each initial subsequence of which is consistent. Let T be a tree
the paths which represent all the truth value assignments to the sentence letters occurring in XI, X,. . . . Let each path be
closed at line in if the path's initial segment to line in makes any sentence XI through X. false, where line in is the first line
paths to which assign truth values to all sentence letters in XI through %. Then, for every line in T, there is an open path that
reaches to that line.
EXERCISE 14-4. Prove lemma L24. Wait a minute! What remains to be done to prove L24? That depends on how thorough you
want to be. There are details I didn't discuss. What if the vocabulary used is finite? What if the vocabulary of some Z, already
includes the vocabulary of Z,+,? More interestingly, perhaps you can find a simpler proof of L24 than the one I suggested. Or better
still, you may be able to reformulate L24 so that your L24 is less complicated to prove but still functions to make the proof of
compactness easy, in something like the way I will describe in the following paragraphs.
Proving compactness is now easy. Suppose that all of 2's finite subsets are consistent. If Z itself is finite, then, because any set
counts as one of its own subsets, Z is consistent. If Z is infinite, we can order its sentences in some definite order. For example,
write out each connective and parenthesis with its English name ('disjunction', 'negation', 'right parenthesis', etc.) and think of each
sentence logic sentence thus written out as a very long word. Then order the sentences (as words) as one does in a dictionary. (This
is called a Lexicographical Ordering.) Since all finite subsets of Z are consistent, each initial segment of the ordered list of
sentences is a consistent set. L24 applies to tell us that there is a tree, the initial finite open paths of which represent models of the
initial segments of the list of sentences. L24 further tells us that for each line of the tree, there will be at least one open path that
reaches that line. Koenig's lemma then tells us that there will be at least one path through the whole tree (an infinite path if the tree
is infinite). This path will represent a model for all the sentences in the set, establishing the consistency of Z.
EXERCISES 14-5. Complete the proof of compactness by showing that if Z is consistent, then so are all of its finite subsets. 14-6.
In my proof of soundness for trees I also limited Z in the statement ZkX to be a finite set. There was no reason for doing so other
than the fact that for trees it was convenient to treat soundness and completeness together, and I needed the restriction to finite Z in
the proof of completeness.
Assume soundness for finite Z, that is, assume that for all finite Z, if ZkX, then ZkX. Prove the same statement for infinite Z. Your
proof will be perfectly general; it will not depend on which system of proof is in question. You will not need to use compactness,
but you will need to use the result of exercise 10-9. I
CHAPTER CONCEPTS
Here are this chapter's principal concepts. In reviewing the chapter, be sure you understand them.
a. Tree Structure
b. Node of a Tree
c. Path (or Branch) in a Tree
d. Koenig's Lemma
e. Compactness
f. Finite Branching
g. Infinite Branching
h. Tree of Truth Value Assignments
i. Lexicographical Ordering

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SECTION OVERVIEW
3.15: Interpretations, Soundness, and Completeness for Predicate Logic

3.15.2: Soundness and Completeness for Tree
3.15.3: Soundness for Predicate Logic Derivations
3.15.6: Conclusion

3.15: Interpretations, Soundness, and Completeness for Predicate Logic is shared under a not declared license and was authored, remixed, and/or
15.1: Interpretations
In chapter 2 I introduced the idea of an interpretation for a predicate logic sentence, that is, of a case which determines the truth
value for closed sentences of predicate logic. In the definition of chapter 2 I required that every object in the domain of an
interpretation have at least one name, I included this requirement because with it I could give a sim- , ple and intuitive truth
definition for existentially and universally quantified sentences: I said that an existentially quantified sentence is true in any
interpretation just in case at least one of its substitution instances is true in the interpretation. And I said that a universally
quantified sentence is true in an interpretation just in case all of its substitution instances are true in the interpretation.
Requiring every object to have a name may have been expedient for teaching fundamentals, but ultimately the requirement is
unsatisfactory. Our system of logic should be able to deal with situations in which some objects go unnamed. So henceforth, by an
interpretation for predicate logic, I will mean exactly what I meant in chapter 2, except that I will no longer require every object to
have a name. I also will streamline the definition somewhat by counting atomic sentence letters as Zero Place Predicates:
D20: An In&rprehtion consists of a nonempty domain of objects, a list of names, and a list of (zero place, one place, two
place, and in general many place) predicates. The list of names may be empty, but there must be at least one predicate. For
each name, the interpretation specifies the object in the domain which is named by that name; and for each predicate the
interpretation specdies its truth value if it is a zero place predicate (an atomic sentence letter), or the objects in the domain of
which the predicate is true if it is a one place predicate, or the ordered lists of objects of which the predicate is true if it is a
two, three, or many place predicate. If a predicate is not true of an object or ordered lit of objects, it is false of that object or
list of objects.
This definition allows us to consider situations in which there are objects without names in the object language. But it makes hash
of my definition of truth in an interpretation for quantified sentences.
Before we begin, precision requires a comment on notation. Remember that '(3u)P(u)' is an expression of the metalanguage ranging
over closed existentially quantified sentences, with u the existentially quantified variable. Ordinarily, P(u) will be an open sentence
with u the only free variable, which is the way you should think of 'P(u)' while getting an intuitive grasp of the material. But strictly
speaking, '(3u)P(u)' ranges over closed existentially quantified sentences, the s-substitution instances of which are P(s), the
expressions formed by substituting s for all free occurrences of u in P(u)-if there are any free occurrences of u. This detail
accommodates vacuously quantified sentences, such as '(3x)A1, as discussed in exercise 3-3.
To work toward new truth definitions for the quantifiers, let's think through what we want these definitions to do. Intuilively,
(3u)P(u) should be true in an interpretation iff there is some object in the domain of the interpretation of which the open sentence,
P(u), is true. When all objects in the domain had names, we could express this condition simply by saying that there is at least one
name, s, in the interpretation for which the substitution instance, P(s), is true in the interpretation. But now the object or objects in
virtue of which (3u)P(u) is true might have no names, so this strategy won't work.
We can get around this problem by appealing to the fact that, even if the interpretation we are considering does not include a name
for the object we need, there will always be another interpretation which does have a name for this object and which is otherwise
exactly the same.
In more detail, here is how the idea works. Suppose we have an interpretation, I, and a sentence (3u)P(u). Intuitively speaking,
(3u)P(u) is true in I when I has an object, o, of which, intuitively speaking, the open sentence P(u) is true. We cannot say that
(3u)P(u) is true of o by saying that o has a name, s, in I such that P(s) is true in I. We are considering an example in which o has no
name in I. But we get the same effect in this way: We consider a second interpretation, If, which is exactly like I. except that in I'
we assign o a name. We can always do this, because if I 422 Interpretations, Soundness, and Completeness for Predicate Logic 15-
1. Interpretations 223 is one interpretation, we get a second interpretation, 1', which has exactly the same domain of objects, the
same list of predicates, the same specification of what is true of what, but which differs from I only by assigning the name s to
object o.
We do also have to require that s not be a name which occurs in (3u)P(u). If, in going from I to If, we move a name from one object
to another, and this name occurs in (3u)P(u), we may disturb some other aspect of the truth conditions for (3u)P(u). Some new
terminology will help in transforming this intuitive idea into a precise definition:
D21: I, is an s-Variant of I iff I, assigns the name s to some object in its domain and I. differs from I at most by having name
s or by assigning s to a different object.
With the help of the idea of an s-variant, we can say
D22: (3u)P(U) is true in interpretation I iff, for some name, s, which does not appear in (3u)P(u), there is an s-variant, I., of I
in which P(s) is true. 1
EXERCISE 15-1. Give an example of a sentence and an interpretation which shows that D22 would not work as intended if it did
not include the requirement that s not appear in (3u)P(u).
The truth definition for the universal quantifier works in exactly the same way, except that we use 'all s-variants' instead of 'some s-
variant'. We want to specify the conditions under which (Vu)P(u) is true in I. Intuitively, the condition is that P(u) be true of all
objects in I. We capture this idea with the requirement that P(s) be true in all s-variants of I:
D23: (Vu)P(u) is true in interpretation I iff, for some name, s, which does not appear in (Vu)P(u), P(s) is true in all s-variants
of I.
EXERCISE 15-2. Give an example of a sentence and an interpretation which shows that D23 would not work as intended if it did
not include the requirement that s not appear in (Vu)P(u).
I hope you will find these new truth definitions for quantifiers to have some plausibility. But they are a bit abstract and take some
getting used to. The only way to become comfortable with them is to work with them. We can get the needed practice, and at the
same time lay the groundwork . for the next sections, by proving some basic lemmas.
Consider a predicate logic sentence, X, and an interpretation, I. Now consider some name which does not occur in X. If we reassign
the name to some new object in the interpretation, this should make no difference to the truth value of X in I. X does not constrain
the referent of the name in any way. The same thing goes for a predicate symbol not occurring in X. Intuitively, X and the unused
predicate have no bearing on each other. So what the predicate is true of (or the truth value of a zero place predicate) should make
no difference to the truth or falsity of X:
L25: Let X be a sentence and I and I' two interpretations which have the same domain and which agree on all names and
predicates which occur in X. Then X is true in I iff X is true in 1'.
By 'agreeing on all names and predicates which occur in X', I mean that, for each name which appears in X, I and I' assign the same
object to that name, and for each predicate appearing in X, I and I' specify the same truth value or the same collection of objects of
which the predicate is true. For names and predicates not appearing in X, I and I' may make different assignments.
We prove L25 by induction on the number of connectives in X. For the basis case, consider an atomic X and an I and I' with the
same domain which agree on all names and predicates in X. An interpretation explicitly provides the truth values in terms of the
extensions of the used predicates and names (e.g., 'Pa' is true in I just in case the thing named 'a' is in the extension which I assigns
to 'P'). Since I and I' agree on the predicates and names in X, they assign X the same truth value.
For the inductive case, assume, as inductive hypothesis, that L25 holds for all X with n or fewer connectives and all I and I'
agreeing on X, as before. We must separately consider each of the connectives. For example, suppose that X has the form Y&W.
Then X is true in I iff both Y and W are true in I. But since Y and W both have fewer connectives than X, we can apply the
inductive hypothesis to conclude that Y is true in I iff Y is true in 1'; and W is true in I iff W is true in 1'. Finally, Y and W are both
true in I' iff X (=Y&W) is true in It, which is what we need to show in this part of the argument.
EXERCISE 15-3. Carry out the inductive step of the proof of L25 for the other sentence logic connectives, modeling your proof on
the example just given for '82.
Now assume that X has the form (3u)P(u). The ideas are not hard, but keeping everything straight can be confusing. So let's
introduce some further terminology: For I' I will write I(X) to remind us that I(X) is an interpretation with the same domain as I and
just like I so far as names and predicates in X are concerned, but differing arbitrarily from I on other predicates and names. In
considering the case of X = (3u)P(u), instead of writing out I((3u)P(u)), I will write just I(P). Finally, I will write I(P,s) for an
otherwise arbitrary interpretation agreeing with I on domain, on P, and on s.
So suppose that (3u)P(u), I, and I(P) have been given. Suppose that I makes (3u)P(u) true. Definition D22 then tells us that there is
a name, s, not appearing in (3u)P(u), and an s-variant of I, I,, where P(s) is true in I,, Now we change I,. We keep 1,'s assignment of
s and of all the names and predicates in (3u)P(u), and we change everything else to look just like I(P). The resulting interpretation,
I(P,s), is an s-variant of I(P). Furthermore, the inductive hypothesis applies to tell us that, since P(s) is true in I,, P(s) is true in
I(P,s). D22 applies to these facts to yield the conclusion that (3u)P(u) is true in I(P).
I have shown that if (3u)P(u) is true in I, it is true in I(P). But exactly the same argument works in the reverse-direction-if I(P)
agrees with I on all vocabulary in (3u)P(u), then I agrees with I(P) on this vocabulary. So we may conclude that (3u)P(u) is true in I
iff it is true in I(P), as was to be shown. (I did not use an iff in the chain of inferences in the previous paragraph because doing so
makes it harder to keep clear about the existential ~uantifiers, 'there is an s' and 'there is an I,'. I will avoid certain 'iffs' in the proof
of the next lemma for the same reason.)
EXERCISE 15-4. Carry out the inductive step of the proof of L25 for the universal quantifier.
Let's move on to another very intuitive fact, but one which is a bit tricky to prove. Consider a sentence of the form R(s,t), a perhaps
very complex sentence in which the names s and t may (but do not have to) occur. Let I be an interpretation in which s and t refer to
the same object. Then it should not make any difference to the truth of R(s,t) in I if we replace any number of occurrences of s with
occurrences oft or occurrences of t with occurrences of s. In I, s and t are just two different ways of referring to the same thing.
R(s,t) says something about this thing, and how one refers to this thing should not make any difference to the truth of R(s,t) in I.
(At this point it would be a good idea to review the discussion of extensional semantics in section 9-2.)
L26: Let R(s,t) be a closed sentence in which the names s and t may occur. Let I be an interpretation in which the names s
and t refer to the sameobject. Let R'(s,t) arise by replacing any number of instances of s by t or instances oft by s. then R(s,t)
is true in I iff R1(s,t) is true in I.
I have stipulated that s and t do not have to occur in R(s,t) to cover the important case in which all occurrences of s in a sentence
P(s) get replaced by occurrences of t.
EXERCISE 15-5. Begin the proof of L26 by carrying out the basis step and the inductive step for the sentence logic connectives.
The complications in the inductive step for L26 call for writing it out in some detail. In what follows, take care to understand what I
mean by 'r = s'. 'r' and 's' are metavariables over names. So 'r = s' means that the name picked out by 'r' is identical to the name
picked out by 's', that is, that r and s are the same name. 'r = s' does not mean the object referred to by the name picked out by 'r' is
the same as the object referred to by a different name picked out by 's'.
Now let's assume (inductive hypothesis) that L26 holds for all R(s,t) with n or fewer connectives. And let's consider the case of
R(s,t) with the form (3u)Q(u,s,t). R1(s,t) is then the sentence (3u)Q1(u,s,t). Let interpretation I be given with names s and t having
the same referent. In outline, the argument runs as follows:
(1) Suppose that I makes (3u)Q(u,s,t) true. (Assumption)
(2) Then there is a name, r, and an r-variant, I, of I, such that I, makes Q(r,s,t) true. (By (1) and D22)
(3) Suppose that r f s and r f t. (Assumption, to be discharged)
(4) Then I, makes Qf(r,s,t) true. (By the inductive hypothesis applied to (2) and (3))
(5) Then I makes (3u)Qf(u,s,t). (By D22 applied to (4))
I want to be sure you understand step (4) and the role of step (3). First, you might have thought that D22 guarantees (3). But that
happens only if both s and t actually occur in (3u)Q(u,s,t). Since we want out proof to cover, for example, a sentence in which just t
occurs and in which we replace all occurrences oft with occurrences of s, we have allowed that s and t don't have to occur. Next,
remember that to apply the inductive hypothesis to switch around the names s and t, we need to be considering an interpretation in
which s and t both refer to the same object. By assumption, I is such an interpretation. But in step (4) we need this to be 426
Inkrjnutatiar, Soundness, and Cmnpkteness for Predicate Logic 15-1. Interpretations 227 true of I,. If r # s and r # t, we're OK.
According to D22, I, arises from I by at most reassigning r to a new referent. When r # s and r # t, s and t still have their mutual
referent, so the inductive hypothesis can be applied.
To get ready to discharge the assumption (3), let's see what can go wrong if (3) fails. Let's suppose that r = s. In this case, when we
apply D22 to make I, out of I, we might have the situation pictured for I, in figure 15-1.
In I, s and t both refer to the object 0,. We apply D22, which says that there is an r-variant, I,, of I, differing at most from I by
assigning a new referent, which I'm calling 'oi, to r (or is an object which makes the existential quantification true). But if r = s, this
means assigning r, that is, s, to the object or, which in general will be distinct from q. So in I, we may not have available the
condition that s and t have the same referent, the condition needed to apply the inductive hypothesis.
To get around this difficulty I will argue by cases. Case 1: Neither s nor t actually occurs in (3u)Q(u,s,t). Then there is nothing to
prove, since there are no occurrences of s and t to switch around. Case 2: s and t both occur in (3u)Q(u,s,t). D22 requires that r not
occur in (3u)Q(u,s,t). So in this case r # s and r # t, we have assumption (3) available, and the proof (1)-(5) can proceed.
Case 3: t but not s actually occurs in (3u)Q(u,s,t). (The case in which s but not t occurs is the same.) To remind us that s occurs
vacuously, I will put parentheses around s, like this: (3u)Q(u,(s),t). If, in this case, r happens by luck to be distinct from s, the proof
(1)-(5) applies. So I will also assume that r = s. In this case we have the situation for I, pictured in figure 15-1, and the inductive
hypothesis will not apply because s and t no longer have the same referent. In addition, we won't be able to apply
Figure 15-1
D22 in step (5). When r = s, r, that is, s, will get put in for occurrences oft when we exchange Q(r,(s),t) for Q1(r,(s),t). Then when
we try to apply D22 to reform the existential quantification, the u will get put into the wrong places.
To resolve these difficulties, I must accomplish two things. I must show that I can pick another name, r', with r' # s and f t, and
assign r' to 0,. Then I must reassign s as a name of q. If I do these two things, then s and t will again have the same referent, so that
I can apply the inductive hypothesis in step (4); and I will again be using a name, r' # s and # t, so that D22 will unproblematically
apply in step (5).
Once this problem is clearly explained it is easy to solve, with the help of lemma L25. I pick a new name, r', not occurring in Q(r,
(s),t), # s and # t. L25 tells us that Q(r,(s),t) has the same truth value in a new interpretation, I,., that it had in I,, where I,, is just like
I, except that r' has been assigned as an additional name of or. Next I apply the inductive hypothesis to Q(r,(s),t) and the
interpretation I,,. In I,,, r' and r (that is, s) both name or. So the inductive hypothesis allows me to replace all occurrences of r with
r'. I now have Q(rl,(s),t) true in I,,, with s not actually occurring in Q(rl,(s),t). Consequently, I can again apply L25 to tell me that
Q(rl,(s),t) is also true in 13, an interpretation just like I,, except that s has been reassigned to 0,. At this point 13 is an rt-variant of I,
r' # s and C t, and s and t are both referents of ot so that I can carry out steps (4) and (5) of the foregoing proof using 13, instead of
I,,.
We are almost done. I have shown that if I makes (3u)Q(u,s,t) true, then I makes (3u)Q1(u,s,t) true. But the argument works in
exactly the same way in the opposite direction. So we have shown that I makes (3u)Q(u,s,t) true iff it makes (3u)Q1(u,s,t) true,
completine: this Dart of - " the proof of L26.
EXERCISES 15-6. In a more advanced logic text, this sort of informal proof would be written up much more briefly. I have spelled
it out in some detail to help you learn how to read and study such a proof. To further practice study of an informal proof and to
appreciate better how complicated it really is, formalize the proof as a natural deduction. Use 'Mod(1,X)' for 'X is true in 1', '(3r)'
for 'There is a name, r', '(31,)' for 'There is an r-variant, I,, of 1', and so on. I suggest that you formalize the initial proof (1)-(5), with
the undischarged assumption of step (3), being sure to make explicit the tacit appeal to 3E. Then fill in the full argument explained
in the discussion which follows (1)-(5). The most efficient natural deduction may have a significantly different organization than
the informal presentation, 228 Int+ekrrionr, Soundness, and Completeness for Predicate Logic which was designed to help you see
what is going on as opposed to presenting the argument in as few steps as possible. Students of truth trees may also have fun doing
this argument as a truth tree proof, although this is less helpful in exposing the structure of the informal argument in English.
15-7. Carry out the inductive step of the proof of L26 for universally quantified sentences. You may do this most efficiently by
commenting on how to modify the proof for the case of existentially quantified sentences.
Now that I have shown you how to proceed with this kind of argument, I am going to ask you to prove the rest of the lemmas we
will need. When not otherwise specified, P(u) can be any open sentence with u the only free variable, I any interpretation, and so
on.
Lemmas L27 and L28 show that the truth definitions for the quantifiers are equivalent to conditions which, superficially, look
stronger than the definitions:
L27: Let s be any name not appearing in (3u)P(u). Then Mod[I,(3u)P(u)] iff there is an s-variant, I., of I such that Mod[I,,
P(s)J.
L28: Let s be any name not appearing in (Vu)P(u). Then Mod[I,(Vu)P(u)] iff Mod[I,,P(s)] for all s-variants, I., of I.
EXERCISE 15-8. Prove L27 and L28. Apply L25 and L26 to D22 and D23. You will not need to do an induction.
L29: -(Vu)P(u) is logically equivalent to (3u)-P(u) and -(3u)P(u) is logically equivalent to (Vu)-P(u). 15-8. Prove L27 and
L28. Apply L25 and L26 to D22 and D23. You will not need to do an induction.
EXERCISE 15-9. Prove L29. Remember that logical equivalence is the semantic notion of having the same truth value in all
interpretations. You will not need to use induction. Instead, simply apply L27 and L28.
When you have finished your proof of L29, look it over and find the places at which you used, as informal logical principles
applied in the metalanguage, just the negated quantifier rules which you were proving as generalizations about the object language!
It is a noteworthy, and perhaps disturbing, fact that we cannot prove anything about the object language formulation of logic
without assuming logical principles at least as strong in the metalanguage. What, then, do we gain in the process? Pre- - cision and
clarity.
L30: Suppose that Mod[I,P(s)]. Then Mod[I,(3u)P(u,s)], where P(u,s) arises from P(s) by substituting u for any number of
occurrences of s in P(s).
L31: Suppose that Mod[I,(Vu)P(u)]. Let I' differ from I only in assignment of names not occurring in (Vu)Pu, and let s be any
name in 1'. Then Mod[I1,P(s)].
Note that in L31, s may be a name appearing in (Vu)P(u). L31 is a generalization of the principle that all substitution instances of a
universally quantified sentence are true, a generalization we will need in the following sections.
EXERCISES 15-10. Prove L30. You will use L26 and D22 and no induction. 15-1 1. Prove L31, using L25, L26, and L28. The fact
that s may appear in (Vu)P(u) may give you trouble in this problem. The trick is not to use the name s for the s-variant in D23. Use
some other name, t, which does not appear in (Vu)P(u) and then apply L25 and L26 to s and t.
L32: Let I be an interpretation in which every object in its domain has a name. Then
a) Mod[I,(3u)P(u)J iff Mod[I,P(s)J for some name, s, that appears in I.
b) Mod[I,(Vu)P(u)] iff Mod[I,P(s)] for all names, s, that appear in I.
L32 simply says that the truth definitions for quantifiers given in chapter 2 work in the special case in which all objects in an
interpretation's domain have names.
EXERCISE 15-12. Using any prior definitions and lemmas from this section that you need, prove L32.
We are now ready to extend our previous proofs of soundness and completeness for sentence logic to predicate logic. Most of the
real work has been done in the lemmas of this section and in Koenig's lemma from 230 Znterpntationr, Soundness, and
Compkteness for Predicate Logic 15-2. Soundness and Completeness for Trees 231 chapter 14. I am only going to outline the
proofs and ask you to fill in the details. In the next three sections I will only treat predicate logic without identity or function
symbols, and 1 will treat only finite sets of sentences in the completeness proofs

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15.2: Soundness and Completeness for Tree
When we extend trees for sentence logic to predicate logic, we add four new rules: 3, -3, V, and -V. Roughly speaking, what we
need to do is to check that these rules are downwardly and upwardly correct. There is, however, a complication: infinite trees.
Before going further, please review section 8-4. In sentence logic every tree ends, including all open trees. That is because we need
to work on each sentence only once, and when working on a sentence the sentences on the output list are all shorter than the input
sentence. But in predicate logic we may have to work on sentences of the form (Vu)(3v)R(u,v) more than once. When we
instantiate (Vu)(3v)R(u,v) with a name, s, we get an existentially quantified sentence, (3v)R(s,v). When we apply 3 to this sentence,
we must use a new name, t, which we must then substitute back into (Vu)(3v)R(u,v), producing another existentially quantified
sentence, which will produce another new name, and so on.
The overall tree strategy still works as before: We make longer sentences true by making shorter sentences true, until we get to
minimal sentences which describe an interpretation. The process may now go on forever, but we can still think of infinite open
paths as describing interpretations in which all sentences on the paths are true.
Because trees can be infinite, we need to reconsider what is involved in a finished tree. We do not need to revise our definition, D9,
but we do need to make sure that if a tree does not close in a finite number of steps that it can be finished in the sense given by D9.
That is, we must make sure that there is some systematic way of applying the rules which guarantees that, for each sentence to
which a rule can be applied, eventually the rule is applied.
Here's a system which supplies the guarantee. We segment our work on a tree into stages. At stage n we work only on sentences
that appear on lines 1 through n. Stage n continues until all sentences on lines 1 through n which can be checked have been checked
and until all names occurring in lines 1 through n have been substituted into all universally quantified sentences occurring in lines 1
through n. Of course, at the end of stage n, the tree may have grown to many more than n lines. But that does not matter. Every
checkable sentence occurs by some line, n, and so will eventually get checked by this process, and every name and every
universally quantified sentence occurs by some line n, so every universally quantified sentence will eventually be instantiated by
every name. Of course, this system is not efficient. But efficiency is not now the point. We want to show that there is a system
which is guaranteed not to leave anything out.
The next point to establish is that if a tree is infinite it has an infinite open branch. Koenig's lemma tells us that if a tree is infinite, it
has an infinite branch, and since closed branches are finite, this infinite branch must be open.
Open branches, infinite or finite, describe interpretations in pretty much the way they do for sentence logic. Given an open branch,
collect all the names that occur on the branch and set up a domain of objects, I each one named by one of the names on the branch,
with no two names assigned to the same object. Then let the minimal sentences on the branch specify what is true of what. Atomic
sentence letters are treated as in sentence logic. If an atomic sentence of the form P(s) appears on the branch, in the branch's
interpretation P is true of s. If an atomic sentence of the form -R(s,t) appears, then in the branch's interpretation R is false of the
pair of objects named by s and t (in that order). And so on. The minimal sentences will generally fail to specify all atomic facts.
The unspecified facts may be filled in arbitrarily.
For sentence logic we formulated rule correctness in terms of any interpretation: Any interpretation which makes an input sentence
true makes at least one output list true. And any interpretation which makes an output list true makes the input sentence true. This
won't work for quantified sentences.
For upward correctness of the V rule, consider some sentence, (Vu)P(u), and some interpretation, I, in which there are more objects
than are named on an open branch. Even if all of the output sentences of the V rule-that is, even if all of the substitution instances
of (Vu)P(u) which appear on this branch-are true in I, (Vu)P(u) might not be true in I. To be true in I, (Vu)P(u) must be true for all
objects in I, whether the object concerned has a name or not.
For downward correctness we need the following: Given an interpretation in which the first n lines of a branch are true, there is an
interpretation which makes true all of these sentences as well as the sentences in an output list resulting from applying a rule. But
for the 3 rule, not just any interpretation in which the first n lines, including (3u)P(u), are true will serve. Such an interpretation
might not have a name for an object which makes (3u)P(u) true. Worse, the interpretation might have such a name but the resulting
substitution instance might conflict with another sentence already on the branch.
This last problem is what necessitated the new name rule, and it is essential that you understand how that requirement fits in here.
Suppose that our branch already has '(3x)Bx' and '-Ba' and that the interpreta- 232 Interpretdionr, Soundness, and Compkteness for
Predicate Logic 15-2. Soundness and Compkteness for Trees 233 tion, I, which makes these two sentences true has just one name,
'a', and two objects, the first, named by 'a', which is not B and the second, which has no name in I and is B. This I is a consistent
interpretation for '(3x)Bx' and '-Ba', but we cannot use it in forming a substitution instance which shows '(3x)Bx' to be true. We
must extend or change our interpretation by assigning a new name, 'b', to the unnamed object. Then the truth of '(3x)Bx' is made
explicit by including 'Bb' on the branch.
The new name feature of the 3 rule ensures that we always proceed in the way just described. When it comes time to describe
downward correctness of the 3 rule, the downward correctness must be given a corresponding description. As in the last example,
the I which makes the initial sentences on the branch true may not have the required new name. Or I may have the name but, since
the name does not occur in any of the sentences so far considered on the branch, the name could refer to the wrong object. (Think
of lemma 25 in making sure you understand this last point.) For lack of the right name referring to the right object, the I which
makes true the first n sentences on a branch may not also make true the substitution instance which comes by applying the 3 rule
with its new name requirement. But there will always be an s-variant of I, I,, resulting by assigning the new name s to the right
object, which will make true (3u)P(u)'s substitution instance, P(s). Since s is new to the branch, lemma 25 guarantees that all the
prior sentences in the branch will still be true in I,.
The foregoing remarks should motivate the following revisions of Dl5 and Dl 6:
D15': A tree method rule is Downwardly Correct iff it meets the following condition for all interpretations, I, and all line
numbers, n: Suppose that I is an interpretation which makes true all sentences along a branch from lines 1 through n.
Suppose that the input sentence for the rule lies on this branch, on one of the lines 1 through n, and the sentences on the
output lists lie on the lines immediately succeeding n. Then there is an s-variant of I which makes true all of the sentences on
the original branch, lines 1 through n, and also all of the sentences on at least one of the output lists.
D 16': A tree method rule is Upwardly Cmect iff in any interpretation, I, which is described by an open branch, if all the
sentences on an output list on that branch are true in I, then the input sentence is true in I.
Note that upward correctness concerns only interpretations which are described by the open branch in question.
Before checking rule correctness, we need to clarify what is to count as the output list for an application of the V rule. For upward
correctness, the output list resulting when V is applied to (Vu)P(u) includes all the substitution instances of (Vu)P(u) on the
finished branch. For downward correctness the output list includes only those substitution instances on the branch as it exists just
after the V rule is applied to (Vu)P(u) but before any further rules are applied.
You can now proceed to check downward and upward correctness of the quantifier rules.
EXERCISES 15-13. Using lemma L29, show that the rules -3 and -V are downwardly and upwardly correct according to D15' and
D16' (though, for these two rules, the difference with Dl5 and Dl6 is inessential). 15-14. Prove that the 3 rule is upwardly correct.
You only need apply definition D22. 15-15. Prove that the V rule is upwardly correct. You need to apply lemma L32. 15-16. Prove
that the 3 rule is downwardly correct. You need lemmas L25 and L27. Note carefully the role of the new name requirement in your
proof. 15-17. Prove that the V rule is downwardly correct. You need lemma L31. Don't forget to treat the case in which V applies to
a sentence on a branch with no names. This case will require L25.
We have now done all the real work in proving downward and upward adequacy:
T10: The truth tree method for predicate logic is downwardly adequate.
T11: The truth tree method for predicate logic is upwardly adequate.
Given the revised definitions of upward rule correctness, the proof of upward adequacy works pretty much as it does for sentence
logic. Downward adequacy requires some change, in ways which I have already indicated. Suppose that an initial set of sentences
has a model. For sentence logic we showed that each time we applied a rule there is at least one extension of the initial segment of
a branch all the sentences of which are true in the original model. Now we show instead that each time we apply a rule there is at
least one extension of the initial segment of a branch all the sentences of which are true in an s-variant of the model for the prior
branch segment. D15' has been designed to make the inductive proof of this statement straightforward.
EXERCISES 15-18. Prove downward adequacy for predicate logic trees. 15-19. Prove upward adequacy for predicate logic trees.
To extend the proof of section 12-2, you will need to revise the definition of 254 Interpretations, Soundness, and Cornpkteness for
Predicate Logic 15-3. Soundess for Predicate Logic Derivations 235 'length of a sentence'. The natural alternative is to let the
length of a predicate logic sentence be the number of predicates and connectives. But on this definition the input and output
sentences of the -3 and -V rules have the same length. With a little care you can still do the induction with this definition. Or you
can define length by letting an initial negation followed by a quantifier count as three units of length and letting each occurrence of
'=' count as two units.
T10 and T11, downward and upward adequacy, immediately give
T12: The truth tree method for predicate logic is sound.
and
T13: The truth tree method for predicate logic is complete.
in exactly the way they do for sentence logic.

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15.3: Soundness for Predicate Logic Derivations
To extend the proof for sentence logic, we need to prove rule soundness for the four new predicate logic rules. Two are easy
applications of definitions and lemmas given in section 15-1:
L33 (Soundness for 31): If Zk~(s), then Zk(3u)~(u,s),where (3u)P(u,s) is an existential generalization of P(s), that is, P(u,s)
results from P(s) by replacing any number of occurrences of s with u.
L34 (Soundness for VE): If z~(vu)P(u), then ZkP(s), where P(s) is a substitution instance of (Vu)P(u), that is, s is substituted
for all free occurrences (
EXERCISES 15-20. Apply lemma L30 to prove lemma L33. 15-2 1. Apply lemma L3 1 to prove lemma L34.
Let's look at VI in a bit more detail. We want to prove
L35 (Soundness for VI): Assume that the name s does not occur in Z or in (Vu)P(u). On this assumption, if ~kP(s), then
Zk(Vu)P(u), where (Vu)P(u) is the universal generalization of P(s), that is, P(u) results by replacing all occurrences of s in
P(s) with u.
Let's consider an arbitrary interpretation, I, in which all the sentences in Z are true. What will it take for (Vu)P(u) to be true also in
I? Lemma L28 tells us that given any name, s, not appearing in (Vu)P(u), we need only show that P(s) is true in all s-variants of I.
What we need to do is squeeze the conclusion that P(s) is true in all the s-variants of I out of the assumption that Z~P(S) and the
hypothesis that Mod(1.Z).
But this is easy. The assumption that s does not occur in Z allows us to apply lemma L25 as follows: I is a model for Z. Since s
does not occur in Z, L25 tells us that any s-variant of I is also a model of Z. Then the assumption that ~kP(s) tells us that any s-
variant of I makes P(s) true.
You should carefully note the two restrictions which play crucial roles in this demonstration. In order to apply lemma L25, s must
not appear in Z. Also, in order to apply lemma L28, s must not appear in (Vu)P(u). The latter restriction is encoded in the VI rule
by requiring that (Vu)P(u) be the universal generalization of P(s).
In a similar way, the restrictions built in the 3E rule play a pivotal role in proving
L36 (Soundness for 3E): Assume that s does not appear in Z, in (3u)P(u), or in X. Then if ZU{(~U)P(U),P(~))~X, then
ZU{(~U)P(U)}~X.
You will immediately want to know why the restrictions stated in L36 are not the same as the restriction I required of the 3E rule,
that s be an isolated name. If you look back at section 5-6, you will remember my commenting that requiring s to be an isolated
name involves three more specific requirements, and that other texts state the 3E rule with these three alternative requirements.
These three requirements are the ones which appear in the assumption of L36. Requiring that s be an isolated name is a
(superficially) stronger requirement from which the other three follow. Since we are proving soundness, if we carry out the proof
for a weaker requirement on a rule, we will have proved it for any stronger requirement. You can see this immediately by noting
that if we succeed in proving L36, we will have proved any reformulation of L36 in which the assumption (which states the
requirement) is stronger.
Of course, by making the requirement for applying a rule stronger (by making the rule harder to apply), we might spoil
completeness-we might make it too hard to carry out proofs so that some valid arguments would have no corresponding proofs. But
when we get to completeness, we will check that we do not get into that problem.
Let's turn to proving L36. The strategy is like the one we used in proving L35, but a bit more involved. Assume that I is a model for
Z and (3u)P(u). Since s does not appear in (3u)P(u), there is an s-variant, I, of I, such that P(s) is true in I,. Since s does not appear
in (3u)P(u) or in Z, and since I and I, differ only as to s, lemma L25 tells us that (3u)P(u) and 236 Zntcrpretationr, Soundness, and
Completeness for Predicate Logic 154. Completeness for Predicate Logic Derivations 237 Z are also true in I,. The hypothesis, that
ZU{(~U)P(U),P(S)}~X, then tells us that X is true in I,. Finally, since s is assumed not to appear in X and I and I, differ only as to
s, lemma L25 again applies to tell us that X is true in I.
The soundness of the quantifier rules immediately gives us
T14 (Soundness for predicate logic derivations): For any set of sentences, Z, and sentence, X, if ZkX, then Z~X.
The proof is a trivial extension of the proof for sentence logic, but to fix the ideas you should carry out this extension.
EXERCISE 15-22. Prove T14. You only need to extend the inductive step in the proof of T5 to cover the cases of the four
quantifier rules.

15.3: Soundness for Predicate Logic Derivations is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.
For completeness, we also follow the same overall strategy as we did for sentence logic. Starting with an initial tableau of
sentences, we generate a new tableau the sentences of which make the sentences on the original tableau true. The sentences on the
generated tableau are, on the whole, shorter than on the generating tableau. Roughly speaking, we eventually get down to minimal
sentences which characterize an interpretation on which all the sentences of ancestor tableaux are true. But there will be some new
wrinkles.
We have to say how quantified and negated quantified sentences will be treated on a tableau. For negated quantified sentences, we
apply the rules of logical equivalence for negated quantifiers, pushing the negation sign through the quantifier and switching the
quantifier. That will leave us with only quantified sentences, with no negation signs in front, with which we have to deal.
We will make a universally quantified sentence true by making all its substitution instances true. We will make an existentially
quantified sentence true by making one substitution instance true. But we will have to make this substitution instance the
assumption of a new subderivation so that we will be able to apply the 3E rule to contradictions to get 'A&-A' as the final
conclusion of the outermost derivation.
These ideas get incorporated by extending the rules for sequential and branching generation:
R2' Sequential generation: Extend the statement of the rule with the following steps (to be applied before the instruction to
reiterate remaining sentences).
a) If a sentence of the form -(3u)P(u) appears on the generating tableau, enter (Vu)-P(u) on the generated tableau.
b) If a sentence of the form -(Vu)P(u) appears on the generating tableau, enter (3u)-P(u) on the generated tableau.
c) If a sentence of the form (Vu)P(u) occurs on the generating tableau, enter on the generated tableau all the
substitution instances formed with names which appear on the generating tableau. If no names appear on the
generating tableau, pick one name arbitrarily and use it to form a substitution instance entered on the generated
tableau. Also reiterate (Vu)P(u) on the generated tableau.
We must reiterate (Vu)P(u) on the generated tableau because, as you will soon see, new names can arise on later tableaux. These
new names must be substituted into (Vu)P(u) to make (Vu)P(u) true for all its substitution instances. So we must carry (Vu)P(u)
along on each tableau to have it for forming substitution instances any time that a new name arises.
R3' Branching generation: If any sentence of the form XVY occurs on the generating tableau, apply R3 exactly as stated. If
no XVY occurs but there is a sentence of the form (3u)P(u) on the generating tableau, pick a New Name, that is, a name
which does not appear anywhere on the generating tableau. Use the new name to form a substitution instance of (3u)P(u),
and use this substitution instance as the assumption starting a new subderivation. Reiterate all other sentences on the
generating tableau in the subderivation to complete the generated tableau, just as in R3.
As students of the tree method already know, these rules create a problem. Suppose that a sentence of the form (Vu)(3v)R(u,v)
appears on a tableau. R2' tells us to enter at least one substitution instance, (3v)R(s,v), on the next tableau and to reiterate (Vu)
(3v)R(u,v) itself. R3' will then tell us to start a new subderivation with R(s,t), t a new name. Of course, (Vu)(3v)R(u,v) also gets
reiterated onto the subderivation. But now we will have to do the same thing all over again. The new name, t, will have to go into
(Vu)(3v)R(u,v), giving a new existentially quantified sentence, . (3v)R(t,v), which will call for a new subderivation with yet
another new name, which will have to go back into the reiterated (Vu)(3v)R(u,v). We are off and running in a chain of
subderivations that will never end.
A first impulse is to wonder if the generation rules couldn't be written better, so as to avoid this problem. They can be written so as
to avoid exactly this form of the problem, but it turns out that no matter how the rules are written, some problem with essentially
the same import will arise. Indeed, proving this is a further important fact about logic.
Here is an overview of what the problem involves. The semantic tableau procedure provides a mechanical method for searching for
a derivation 238 Interpretatwns, Soundness, and Completeness for Predicate Logic 154. Completeness for Predicate Logic
Derivations 439 which establishes the validity of a given argument, or equivalently, a mechanical method for searching for an
interpretation of a given finite set of sentences. In sentence logic, the method is guaranteed to terminate. A method which thus
terminates is guaranteed to give a definite yes or no answer to the original question ('Is the argument valid?' or 'Is the initial set of
sentences consistent?'). Such a method, guaranteed eventually to turn,up a yes or no answer, is called a Decision Procedure.
Now, here is the general form of our current problem. Given an exceedingly plausible assumption about what will count as a
mechanical decision procedure, one can prove that there is no decision procedure for predicate logic. In our formulation we fail to
get a decision procedure because we may get an infinite sequence of sub-sub . . . -sub-derivations. If our tableau procedure has
failed to close at some stage, we may not be able to tell for sure whether that is because we just haven't pursued it far enough, or
because it will go on forever. This is not just a weakness of our rules. One can prove that any sound and complete system of
predicate logic will suffer in the same way. Roughly speaking, the problem arises from the requirement on the 3E rule, which we
must have in order to ensure soundness.
Since there is no point in searching for better rules, we will have to see what we can make of our R2' and R3' in fashioning a
completeness proof.
Consider a set of sentences, for example, just the sentence (Vu)(3v)R(u,v) for which our tableau procedure generates an infinite
sequence of tableaux. We will need the fact that we can then, so to speak, draw an unending path through the nested sequence of
subderivations. Koenig's lemma assures us that we can always do so. Refer back to the tree structure at the beginning of chapter 14
and imagine that each node represents a subderivation, beginning with the outermost derivation at the top node. Moving from one
node to two nodes beneath represents the process of starting two new subderivations by working on a sentence of the form XVY.
When we start one new subderivation by working on a sentence of the form (3u)P(u), we start one new node, that is, a "branch"
with one rather than two new forks. When a subderivation closes, the corresponding path on the tree structure closes. Koenig's
lemma tells us that if such a tree structure is infinite, then there is an infinite open path through the tree.
We now know that if a tableau derivation does not close (is infinite or does not have all its terminal tableaux closed) then there is an
open path of subderivations through the derivation. The path might be finite or it might be infinite. Each such path provides an
interpretation, which we will again call a Terminal Interpretation. But we want to characterize the idea of a terminal interpretation
so that it will work for infinite as well as finite cases. Since an infinite path through a derivation has no terminal tableau, we cannot
let the terminal interpretation simply be one provided by the terminal tableau.
Here's the recipe for the terminal interpretation represented by an infinite path. Collect all the names that occur on the path, and set
up _a domain of objects, each one named by one of the names on the path, with no two names assigned to the same object. Then
look at all the minimal sentences which appear on the path. If an atomic sentence letter appears, the interpretation will make it true.
If an atomic sentence letter appears negated, the interpretation will make the atomic sentence letter false. If an atomic sentence of
the form P(s) appears, the interpretation will make the predicate P true of the object named by s. Similarly, if -P(s) appears, the
interpretation will make P false of the object named by s. Two and more place predicates are treated similarly. If this recipe fails to
specify all the atomic facts of the interpretation, fill in the missing facts arbitrarily. In sum
D24: A Terminal Interpretation represented by an open path has as its names all the names which occur on the path and as its
domain a set of objects, each named by exactly one of the names. The interpretation assigns truth values to atomic sentence
letters and determines which predicates are true of which objects (pairs of objects, and so on) as described by the minimal
sentences on the path. Any facts not so specified by the minimal sentences may be filled in arbitrarily.
Note, incidentally, that this recipe gives a consistent interpretation. Since the path is open, it cannot contain both an atomic sentence
and its negation. So this recipe will not make an atomic sentence both true and false. That is, it will not both say and deny that a
predicate is true of an object. The main work we need to do is to prove the analogy of lemma 18, namely
L37: The sentences of the initial tableau are all true in a terminal interpretation represented by an open path.
We prove this by proving that a terminal interpretation makes true all the sentences in all the tableaux along its path, arguing by
induction on the length of the sentences.
We need to take a little care in saying what the length of a sentence is. To keep things initially simple, let us first consider a special
case-analogous to our procedure in section 13-4: Suppose that 'v', '-', and the quantifiers are the only connectives occurring in any
of the initial sentences. Then we can take the length of a sentence simply to be the number of connectives occurring in the sentence.
To carry out the inductive argument, suppose that we have an open path and a terminal interpretation, I, represented by that path.
By the definition of a terminal interpretation, all atomic and negated atomic sentences, and so all sentences of length 0 or 1, along
this path are true in I. 240 Zntcrpntafimu, Soundness, and Cmnpleteness for Predicate Logic 154. COrnpIctmess for Predicate
Logic Derivorionr 241 For the inductive hypothesis, suppose that all sentences of length no greater than n along the path are true in
I. Let X be a sentence of length n + 1. Suppose that X has the form -(YvZ). Then rule R2 for sequential generation tells us that -Y
and -Z will both be on the path, since they will be on the tableau generated by the tableau on which -(YvZ) occurs. -Y and -Z are
both shorter than -(YvZ), and the inductive hypothesis tells us that -Y and -Z are both true in I. Hence -(YvZ), that is, X, is true in I.
When X has the form --Y the argument goes quite like the case of -(YvZ).
Next, we must consider X of the form YvW. Such X gives rise to two generated tableaux, one including Y and one including W.
One of these generated tableaux must be on the open path. Suppose it is the one with Y. Since (by the inductive hypothesis) all
sentences along this path with n or fewer connectives are true, Y, and so YvW, are true. If W rather than Y is on the path, the same
argument applies. Suppose that X has the form (3u)P(u). Then rule R3' specifies that there is a subderivation along the path that
includes a substitution instance, P(s), which the inductive hypothesis tells us is true in I. Definition D22 applies to tell us that then
(3u)P(u), that is, X, is true in I.
Now suppose that X has the form (Vu)P(u). Rule R2' specifies that, for each tableau in which (Vu)P(u) appears, all its substitution
instances formed with names in that tableau appear in the next sequentially generated tableau. (Vu)P(u) is also reiterated, so that
any name which comes up will eventually get instantiated along the path. By the inductive hypothesis, all these substitution
instances are true in I. Remember that in a terminal interpretation there is exactly one object named by each name, and we have just
seen that all of these names eventually get used to form true substitution instances of (Vu)P(u). So lemma L32 applies to tell us that
(Vu)P(u), that is, X, is also true in I.
Make sure that you understand how this last step in the inductive proof makes essential use of the fact that (Vu)P(u) is always
reiterated, to ensure that when new names come up in later tableaux, they will always be used to instantiate (Vu)P(u).
We still need to consider sentences of the form -(3u)P(u) and -(Vu)P(u). Rule R2' applies to such sentences to produce sentences,
respectively, of the form (Vu)-P(u) and (3u)-P(u). There might seem to be a problem here because -(-Ju)P(u) and (Vu)-P(u) have
the same number of connectives, as do -(Vu)P(u) and (3u)-P(u). But we can still complete the inductive step. Suppose that -(-
Ju)P(u) has n + 1 connectives apd appears on the path. R2' tells us that (Vu)-P(u), also having n + 1 connectives, also appears on
the path. But we have already seen that the inductive hypothesis ensures us of the truth of (Vu)-P(u) in the terminal interpretation, I.
Lemma L29 then tells us that -(3u)P(u) is also true in 1. Of course, the case for -(Vu)P(u) works the same way.
To complete the proof of L37 we must lift the restriction and allow sentences to include all the sentence logic connectives. This
creates a new difficulty. For example, R2 instructs us to generate (X&Y)V(-X&-Y) from X=Y. But (X&Y)v(-X&-Y) has four more
connectives than X=Y rather than fewer.
We can resolve this impasse by assigning weights to the connectives. I-', 'v', and the quantifiers are each worth one "point," '3' and
'&' each get three "points," and '=' gets six "points." The length of a sentence is now just the number of these "points" added up for
all the connectives in the sentence. (This technique can also be applied to arrange for -(3u)P(u) and -(Vu)P(u) to be longer than
(Vu)-P(u) and (3u)-P(u).)
EXERCISE 15-23. Complete the inductive step of the argument for lemma L37 with all of the sentence logic connectives.
We have proved L37, the analogue of L18 needed for proving completeness for sentence logic derivations. The proof for sentence
logic derivations also used L20, which says that if all terminal tableaux close, then 'A&-A' appears as the derivation's final
conclusion. We must reformulate the statement ever so slightly because, with the possibility of infinite derivations, some paths
might not have terminal tableaux. So we will say
D25: a semantic tableau derivation is Closed if all sequences of subderivations terminate in a closed tableau.
You will then prove the analogy of L20:
L38: If a semantic tableau derivation is closed, then 'A&-A' appears as the derivation's final conclusion.
The key to L20 is the inductive step, L21. Again, we only need to reformulate to accommodate our more specific definition of a
closed tableau derivation:
L39: Let D be a closed semantic tableau derivation. Then, if all of D's subderivations of level i have 'A&-A' as their final
conclusion, so do all the subderivations of level i + 1.
EXERCISE 15-24. Prove L39. You only need to check the inductive step for rule R3', involving subderivations started with a
substitution instance of a sentence of the form (3u)P(u). Be sure you see how the new 444 Intcrprctation, Soundness, and
Cinnpktcncss fm Predicate Logic 15-5. Cinn~~s, Identity, and Functions 443 name requirement in the statement of R3'
functions~crucially in your proof.
If you now go back and read the short paragraph proving T7 and change just the words 'L18' and 'L20' to 'L37' and 'L38', you will
see that we have a proof of T7, where the set of sentences Z may now include predicate logic sentences. T7 applies exactly as it did
in section 12-3 to establish
T15 (Completeness for predicate logic derivations): For any finite set of sentences, Z, and any sentence X, if Z~X, then ZkX.

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LibreTexts.
In this section I am going to get started in cleaning up some details. But I am going to let you do most of the work. Students of
truth trees and of derivations will be able to apply the material of this section appropriately to what they have learned.
My completeness proofs for predicate logic assumed a finite set of sentences, Z. To get a full statement of completeness, where Z
can be infinite, we need to show that the compactness result, T8, which we proved in chapter 14, also holds for predicate logic. To
accomplish this we need to modify the idea of a tree of truth value assignments.
Here's what we do. We can consider all possible closed atomic sentences written out in some definite order: the first atomic
sentence letter, the second, the first one place predicate with the first name, the second . . . : 'A', 'B', 'Pa', 'Pb', 'Raa'. . . . To make
sure that this is possible, again consider that we could write each such description of the atomic sentences in English and order
them as in a dictionary.
Say the closed atomic sentences are XI, X2, XS, . . . Then we can diagram all possible truth value assignments to these atomic
sentences in the form of a tree:
The third line will catalogue the alternative truth values for X; underneath all the possibilities covered in lines 1 and 2, and so on.
Note that each path through this tree represents an interpretation, indeed, just the sort of interpretations represented by open paths
on a truth tree or semantic tableau derivation. We have seen. in the com~leteness proofs, how there must be at least one such
interpretation for each consistent finite set of sentences. We now proceed very much as we did in predicate logic. Let Z be an
infinite set of sentences all of the finite subsets of which are consistent. We list the sentences in some definite order, and consider
the initial finite segments of this ordering: Z,, Zp, 5, . . . As we work down the lines of the tree, we close branches which conflict
with some sentences in one of the 2;. Since all of the Z; are consistent. for each line of the tree, there will be an open branch
reaching down to that line. Koenig's lemma tells us that there is then an infinite path through the tree. But (if you make the right
sort of arrangement of when paths get closed) you will see that this infinite path represents an interpretation which makes true all
the sentences in all the Zi. That is, this interpretation is a model of Z.
EXERCISE 15-25. Following the suggestions. of the argument sketch given in the last paragraph, give a detailed proof of
compactness for predicate logic.
Actually, we have done the work to prove the Lowenheim Skolem Theorem, a much stronger result, of fundamental importance in
logic and set theory. In all my discussion of infinite interpretations, I have not mentioned the fact that there are different kinds of
infinities. The infinity of the integers is the smallest, called, for obvious reasons, a Countable Infinity. However, other infinities are,
in a certain sense, "larger." Consider, for example, the infinity of the real numbers (numbers representable by a finite or an infinite
decimal fraction, such as 27.75283 . . .). The infinity of the real numbers is larger, or Uncountable, in the sense that there is no one-
to-one correspondence between the integers and the real numbers. We cannot list the real numbers with the integers the way we can
an infinite set of sentences.
The Lowenheim Skolem theorem says that if a set of sentences has a model with a finite, countable, or uncountable domain, then it
has a finite or a countable model. For finite sets of sentences, these models are generated by open paths on a truth tree or semantic
tableau derivation. If a finite set has a model (finite, countable, or uncountable) then there is an open path. But then the open path
represents a finite or countably infinite model. The compactness theorem then shows how the same is true of infinite consistent sets
of sentences. (If our object language does not include identity, then there is always a countable model. But '=' allows us to write a
sentence which, for example, is only true in an interpretation with exactly one object. Can you produce such a sentence?)
My soundness and consistency proofs assumed that our object language contained neither identity nor function symbols. For the
moment, let's consider just identity. To begin with, we must refine the characterization of an interpretation with requirements which
should seem natural if '=' really means 'identity':
D20' (Interpretations for languages with identity): An interpretation is as described in D20 with the following two additional
requirements:
a) A sentence of the form s=t is true in an interpretation iff s and t name the same object.
b) For all atomic sentences of the form R(s,t), if s=t is true in an interpretation, then R(s,t) and R1(s,t) have the same
truth value in the interpretation, where R1(s,t) arises from R(s,t) by replacing any number of occurrences of s with t or
oft with s.
Clause b) covers sentences such as 'Qab': If 'a=c' is true in an interpretation, then 'Qab' and 'Qcb' have the same truth value in the
interpretation.
A good many of the semantical facts surrounding identity turn on the following lemma, which simply generalizes clause b) to the
case of any closed sentence:
L40: Let I be an interpretation for predicate logic with identity. Then, for all sentences of the form R(s,t), if s=t is true in I,
R(s,t) and R1(s,t) have the same truth value in I, where R1(s,t) arises from R(s,t) by replacing any number of occurrences of s
with t or oft with s. / I 1 1
15-26. Prove LAO.
You are now in a position to examine how our soundness proofs need to be modified if our language includes identity. Identity
involves new rules, the roles of which need to be checked in the proofs.
15-27. (Trees) Show that the truth tree = rule is downwardly correct. To treat the f rule, note that we can reconstrue it in the
following way: Whenever a sentence of the form sf s appears on a branch, also write the sentence s= s on that branch. Explain why
this rule comes to the same as the # rule as stated in chapter 9. Prove that the rule in this form is downwardly correct.
15-28. (Derivations) State and prove rule soundness for the two derivation rules for identity. Comment on whether and, if so, how
these rules require any changes in the inductive proof of soundness for derivations.
We can turn now to completeness. For semantic tableau derivations we must add two new parts to the rules for sequential
generation, corresponding exactly to the =I and = E rules: Whenever a name s occurs on a tableau, include the sentence s = s on the
sequentially generated tableau. And if two sentences of the form s=t and R(s,t) appear on a tableau, include the sentences Rt(s,t) on
the sequentially generated tableau. Then, for both trees and semantic tableau derivations, we change how we read an interpretation
off an open branch. Before, every name was assigned a distinct object. Now each name will be assigned a distinct object unless a
sentence of the form s=t appears on the branch. Then s and t are assigned the same object. This corresponds to clause a) in D20'.
Clause b) in D20' is already ensured by the identity rules for trees and for tableau generation.
 Exercise 3.15.5.1: Trees
EXERCISES 15-29. Show that clause b) of D20' will be satisfied in the interpretation represented by an open branch.
Comment on the status of lemma L40 in describing an open branch. That is, note the way in which, in effect, proof of upward
adequacy automatically covers the work done by lemma L40. Then check that the tree method with identity is upwardly
adequate. Though intuitively quite clear, a formal proof requires care, since the input and output sentences for the = rule all
have the same predicates and connectives, so that none of our prior methods of attributing lengths to sentences will apply here.
 Exercise 3.15.5.1: Derivations
15-30. Show that clause b) of D20' will be satisfied in the interpretation represented by an open branch. Comment on the status
of lemma L40 in describing an open branch. That is, note the way in which, in effect, proof of lemma L37 automatically covers
the work done by lemma L40. Then check that lemma L37 is still correct. Just as with the case for trees, proof requires care,
since none of our prior means of assigning lengths to sentences will work here.
Finally, let's take a brief look at function symbols. Again, we must extend the definition of an interpretation:
D20" (Interpretations for languages with function symbols): An interpretation is as described in D20 or D201, with the
following addition: For each function symbol, f, and each object, o, in the domain of the interpretation, the interpretation
assigns a unique object o' = f(o), as the value off applied to o. If s is a closed term referring to object o*, then f(s) is a term
referring to f(o*).
The last sentence in D20" constitutes a recursive definition. If s is a name, referring to o, then f(s) refers to f(o), ff(s) refers to ff(o),
and so on.
As with identity, once we have made this extension of the notion of an interpretation, mbst of the work is done.
EXERCISES 15-31. (Trees) Check the downward correctness of the quantifier rules when the language includes function symbols.
15-32. (Derivations) Check the proof of rule soundness for the quantifier rules when the language includes function symbols.
15-33. (Trees) Check that the proof of upward adequacy works when interpretations are read off open branches in accord with
definition D20".
15-34. (Derivations) Check lemma L37 when interpretations are read off open branches in accord with definition D20".

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15.6: Conclusion
You have worked hard trying to understand these proofs of soundness and completeness. I too have worked hard, first in
understanding them and then in my efforts to write them up in a clear and accessible form. Working on the strength of the
presentations of others, I will be very happy if I have made some small contribution to improving the accessibility of soundness and
completeness and if I have avoided both horns of the dilemma of too much complication versus inaccuracies in the proofs.
Whatever I have accomplished, I am sure that my presentation can be improved. I welcome your comments and suggestions. In the
meantime, you should not be discouraged if you have found part I1 of this text to be very difficult. Soundness and completeness are
substantial mathematical results. If you understand them only in a fragmentary way, you can greatly improve your grasp by
patiently going over these chapters again.
CHAPTER CONCEPTS
In reviewing this chapter, be sure you have a firm grasp on the following:
a. Interpretation
b. Unnamed Object
c. s-Variant Truth of an Existentially Quantified Sentenced)
d. Truth of a Universally Quantified Sentence
e. Infinite Tree
f. Infinite Semantic Tableau Derivation
g. Downward correctness of a Truth Tree Rule
h. Upward correctness of a Truth Tree Rule
i. Interpretation Represented by an Infinite Truth Tree Branch
j. Restrictions on the Derivation Rule for VI
k. Restrictions on the Derivation Rule for 3E
l. Terminal Interpretation in a Semantic Tableau Derivation
m. Closed Semantic Tableau Derivation
n. Compactness for Predicate Logic
o. Interpretation for a Language with Identity
p. Interpretation for a Language with Function Symbols

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Index
C L Rules of Inference
contradictions LEAST NUMBER PRINCIPLE 2.5.3: The Complete Rules of Inference
2.3.3: Logical Truths and Contradictions 3.11.3: Strong Induction
logical truths S
D 2.3.3: Logical Truths and Contradictions Scope Line
derivation Logically Equivalent 2.6.3: Derivations- Overview, Definitions, and Points
to Watch out for
2.6.3: Derivations- Overview, Definitions, and Points 3.3.4: Some Logical Equivalences
to Watch out for 3.6.3: Logical Truth, Contradictions, Inconsistency, Scope of a quantifier
and Logical Equivalence 3.3.2: Quantifier Scope, Bound Variables, and Free
Disjunctive Normal Form
Variables
2.3.4: Disjunctive Normal Form and the Sheffer
Stroke M Sheffer Stroke
Domain of the interpretation metatheory 2.3.4: Disjunctive Normal Form and the Sheffer
Stroke
3.2.1: Interpretations 3.11.1: Informal Introduction
Strong Induction
G O Subderivation
Good Deductive Argument Outer Derivation 2.6.3: Derivations- Overview, Definitions, and Points
2.1.1: Logic As the Science of Argument 2.6.3: Derivations- Overview, Definitions, and Points to Watch out for
to Watch out for
SUBSTITUTION INSTANCE
I 3.2.2: Truth in an Interpretation
induction P
3.11: Mathematical Induction Predicate Logic V
Interpretation 3: Volume II- Predicate Logic
Valid Deductive Argument
3.2.1: Interpretations 2.1.1: Logic As the Science of Argument
3.2.2: Truth in an Interpretation R
Interpretation of the sentence Rule of Inference W
3.2.1: Interpretations 2.6.3: Derivations- Overview, Definitions, and Points
to Watch out for
Weak Induction
Glossary
Sample Word 1 | Sample Definition 1
Detailed Licensing
Overview
Title: A Modern Formal Logic Primer (Teller)
Webpages: 129
All licenses found:
Undeclared: 100% (129 pages)
By Page
A Modern Formal Logic Primer (Teller) - Undeclared 2.4.2: Invalidity and Counter Examples -
Front Matter - Undeclared Undeclared
TitlePage - Undeclared 2.4.3: Soundness - Undeclared
InfoPage - Undeclared 2.4.4: The Conditional - Undeclared
Table of Contents - Undeclared 2.4.5: The Biconditional - Undeclared
Licensing - Undeclared 2.5: Natural Deduction for Sentence Logic -
1: Preface- A Guide to the Primer - Undeclared Fundamentals - Undeclared
2: Volume I- Sentence Logic - Undeclared 2.5.1: The Idea of Natural Deduction -
2.1: Basic Ideas and Tools - Undeclared Undeclared
2.5.2: Subderivations - Undeclared
2.1.1: Logic As the Science of Argument -
2.5.3: The Complete Rules of Inference -
Undeclared
Undeclared
2.1.2: Sentences and Connectives - Undeclared
2.1.3: Truth Tables and the Meaning of '~', 'and', 2.6: Natural Deduction for Sentence Logic -
and 'v' - Undeclared Strategies - Undeclared
2.1.4: Truth Functions - Undeclared 2.6.1: Constructing Correct Derivations -
2.1.5: Compounding Compound Sentences - Undeclared
Undeclared 2.6.2: Recognizing the Main Connective -
2.1.6: Rules of Formation and Rules of Valuation - Undeclared
Undeclared 2.6.3: Derivations- Overview, Definitions, and
2.1.S: Basic Ideas and Tools (Summary) - Points to Watch out for - Undeclared
Undeclared 2.7: Natural Deduction for Sentence Logic - Derived
2.2: Transciption Between English and Sentence Rules and Derivations without Premises - Undeclared
Logic - Undeclared 2.7.1: Derived Rules - Undeclared
2.2.1: Grouping in Transcription - Undeclared 2.7.2: Argument by Cases - Undeclared
2.2.2: GROUPING IN TRANSCRIPTION - 2.7.3: Further Derived Rules - Undeclared
Undeclared 2.7.4: Derivations without Premises - Undeclared
2.2.3: Adequacy of Transcriptions - Undeclared 2.8: Truth Trees for Sentence Logic - Fundamentals -
2.3: Logical Equivalence, Logical Truths, and Undeclared
Contradictions - Undeclared 2.8.1: Proving Validity with Truth Trees -
2.3.1: Logical Equivalence - Undeclared Undeclared
2.3.2: Substitution of Logical Equivalents and 2.8.2: The General Strategy for the Rules -
Some More Laws - Undeclared Undeclared
2.3.3: Logical Truths and Contradictions - 2.8.3: Proving Invalidity with Truth Trees -
Undeclared Undeclared
2.3.4: Disjunctive Normal Form and the Sheffer 2.8.4: The Complete Rules for the Connectives -
Stroke - Undeclared Undeclared
2.4: Validity and Conditionals - Undeclared 2.8.5: In Which Order Should You Work on the
Sentences in a Tree? - Undeclared
2.4.1: Validity - Undeclared
2.9: Truth Trees for Sentence Logic - Applications - Undeclared
Undeclared 3.7: Truth Trees for Predicate Logic - Fundamentals -
2.9.1: Application of the Rules to Complex Undeclared
Sentences - Undeclared 3.7.1: The Rule for Universal Quantification -
2.9.2: Other Uses for Truth Trees - Undeclared Undeclared
3: Volume II- Predicate Logic - Undeclared 3.7.2: The Rule for Existential Quantification -
3.1: Predicate Logic - Syntax - Undeclared Undeclared
3.7.3: Applying the Rules - Undeclared
3.1.1: We Need More Logical Form - Undeclared
3.7.4: Negated Quantified Sentences - Undeclared
3.1.2: Quantifiers and Variables - Undeclared
3.1.3: The Sentences of Predicate Logic - 3.8: More on Truth Trees for Predicate Logic -
Undeclared Undeclared
3.2: Predicate Logic - Semantics and Validity - 3.8.1: Contradictions, Logical Truth, Logical
Undeclared Equivalence, and Consistency - Undeclared
3.2.1: Interpretations - Undeclared 3.8.2: Truth Trees with Multiple Quantifiers -
3.2.2: Truth in an Interpretation - Undeclared Undeclared
3.2.3: Validity in Predicate Logic - Undeclared 3.8.3: Three Shortcuts - Undeclared
3.8.4: Infinite Trees - Undeclared
3.3: More about Quantifiers - Undeclared
3.9: Identity, Functions, and Definite Descriptions -
Undeclared
- Undeclared
3.9.1: Identity - Undeclared
3.3.2: Quantifier Scope, Bound Variables, and
3.9.2: Inference Rules for Identity - Undeclared
Free Variables - Undeclared
3.9.3: Functions - Undeclared
3.3.3: Correct Definitions of Substitution Instance
3.9.4: Definite Descriptions - Undeclared
and Truth in an Interpretation - Undeclared
3.3.4: Some Logical Equivalences - Undeclared 3.10: Metatheory - The Basic Concepts - Undeclared
3.4: Transcription - Undeclared 3.10.1: Object Language and Metalanguage -
3.4.1: Restricted Quantifiers - Undeclared Undeclared
3.4.2: Transcribing from English into Logic - 3.10.2: Syntax and Semantics - Undeclared
Undeclared 3.10.3: Soundness and Completeness -
3.4.3: Transcription Strategies - Undeclared Undeclared
3.10.4: Some Further Notation and Definitions -
3.5: Natural Deduction for Predicate Logic -
Undeclared
Fundamentals - Undeclared
3.11: Mathematical Induction - Undeclared
3.5.1: Review and Overview - Undeclared
3.5.2: The Universal Elimination Rule - 3.11.1: Informal Introduction - Undeclared
Undeclared 3.11.2: The Principle of Weak Induction -
3.5.3: The Existential Introduction Rule - Undeclared
Undeclared 3.11.3: Strong Induction - Undeclared
3.5.4: The Existential Elimination and Universal 3.12: Soundness and Completeness for Sentence
Introduction Rules- Background in lnformal Logic Tree - Undeclared
Argument - Undeclared 3.12.1: Preliminaries - Undeclared
3.5.5: The Universal Introduction Rule - 3.12.2: Soundness and Completeness of the Tree
Undeclared Method- Informal Statement - Undeclared
3.5.6: The Existential Elimination Rule - 3.12.3: Soundness and Completeness for Sentence
Undeclared Logic Trees- Formal Details - Undeclared
3.6: More on Natural Deduction for Predicate Logic - 3.13: Soundness and Completeness for Sentence
Undeclared Logic Derivation - Undeclared
3.6.1: Multiple Quantification and Harder 3.13.1: Soundness for Derivations- Informal
Problems - Undeclared Introduction - Undeclared
3.6.2: Some Derived Rules - Undeclared 3.13.2: Soundness for Derivations- Formal Details
3.6.3: Logical Truth, Contradictions, - Undeclared
Inconsistency, and Logical Equivalence -
3.13.3: Completeness for Derivations- Informal 3.15.2: Soundness and Completeness for Tree -
Introduction - Undeclared Undeclared
3.13.4: Completeness for Derivations- Formal 3.15.3: Soundness for Predicate Logic Derivations
Details - Undeclared - Undeclared
3.14: Compactness, and Generalization to Infinite 3.15.4: Completeness for Predicate Logic
Sets of Premises - Undeclared Derivations - Undeclared
3.14.1: Koenig's Lemma - Undeclared 3.15.5: Compactness, Identity, and Functions -
3.14.2: Compactness and Infinite Sets of Premises Undeclared
- Undeclared 3.15.6: Conclusion - Undeclared
Back Matter - Undeclared
3.15: Interpretations, Soundness, and Completeness
for Predicate Logic - Undeclared Index - Undeclared
Glossary - Undeclared
3.15.1: Interpretations - Undeclared
Detailed Licensing - Undeclared

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A MODERN FORMAL

1: Preface- A Guide to the Primer

2: Volume I- Sentence Logic

3: Volume II- Predicate Logic

Contributors and Attributions

Contributors and Attributions

2.1: Basic Ideas and Tools

2.1.2: Sentences and Connectives

2.1.4: Truth Functions

2.1.5: Compounding Compound Sentences

2.1.6: Rules of Formation and Rules of Valuation

2.1.S: Basic Ideas and Tools (Summary)

Contributors and Attributions

Contributors and Attributions

To summarize this section:

Contributors and Attributions

Contributors and Attributions

[(A&(BvC ))] (2.1.5.3)

Sentence Main Connective Components

Truth table definition of '&'

Truth table definition of 'v'

Contributors and Attributions

Contributors and Attributions

2.2: Transciption Between English and Sentence Logic

2.2.2: GROUPING IN TRANSCRIPTION

2.2.3: Adequacy of Transcriptions

Contributors and Attributions

 First Tramscription Test

2- 1. Consider the sentence I

 CHAPTER SUMMARY EXERCISE

2.3: Logical Equivalence, Logical Truths, and Contradictions

2.3.2: Substitution of Logical Equivalents and Some More Laws

2.3.3: Logical Truths and Contradictions

2.3.4: Disjunctive Normal Form and the Sheffer Stroke

Contributors and Attributions

'Av~A' is true no matter what. Such a sentence is called a Logical Truth.

Such a sentence is called a Contradiction.

Show that '*' is expressively complete.

 Chapter Summary Exercise

2.4: Validity and Conditionals

2.4.2: Invalidity and Counter Examples

2.4.4: The Conditional

2.4.5: The Biconditional

Contributors and Attributions

The Law of Contraposition (CP): X⊃Y is logically equivalent to ~Y⊃~X.

 Chapter summary Exercise

2.5: Natural Deduction for Sentence Logic - Fundamentals

2.5.3: The Complete Rules of Inference

Contributors and Attributions

 chapter summary Exercise

2.6: Natural Deduction for Sentence Logic - Strategies

2.6.2: Recognizing the Main Connective

2.6.3: Derivations- Overview, Definitions, and Points to Watch out for

Contributors and Attributions

SENTENCE MAIN CONNECTIVE COMPONENT OR COMPONENTS

(AvB)⊃~(B&D) ⊃ AvB and ~(B&D)

[(A⊃B)v~D)v(~A≡D) v (A⊃B)v~D and ~A≡D

Give derivations which establish the validity of the following arguments:

 chapter summary Exercise

2.7.2: Argument by Cases