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Mathematical Logic for Mathematicians, Part I
Joseph R. Mileti
September 20, 2007

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Contents
1 Introduction 7
1.1 The Nature of Mathematical Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.2 The Language of Mathematics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.3 Syntax and Semantics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.4 The Point of It All . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
1.5 Some Basic Terminology and Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
2 Induction and Recursion 15

2.1 Induction and Recursion on N . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
2.2 Generation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
2.2.1 From Above . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
2.2.2 From Below: Building by Levels . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2.3 From Below: Witnessing Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2.4 Equivalence of the Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.3 Step Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
2.4 Step Recursion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.5 An Illustrative Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5.1 Proving Freeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.5.2 The Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.5.3 An Alternate Syntax - Polish Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
3 Propositional Logic 29
3.1 The Syntax of Propositional Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.1.1 Standard Syntax . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
3.1.2 Polish Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
3.1.3 Official Syntax and Our Abuses of It . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
3.1.4 Recursive Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32
3.2 Truth Assignments and Semantic Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
3.3 Boolean Functions and Connectives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37
3.4 Syntactic Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4.2 Official Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.4.3 Examples Of Deductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.4.4 Theorems about ` . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.5 Soundness and Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.5.1 The Soundness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.5.2 The Completeness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.6 Compactness and Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
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4 CONTENTS
3.6.1 The Compactness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

3.6.2 Combinatorial Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.6.3 An Algebraic Application . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48
4 First-Order Logic : Syntax and Semantics 51

4.1 The Syntax of First-Order Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51
4.2 Structures: The Semantics of First-Order Logic . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.2.1 Structures: Definition and Satisfaction . . . . . . . . . . . . . . . . . . . . . . . . . . . 56
4.2.2 Elementary Classes of Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59
4.2.3 Definability in Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
4.2.4 Substitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.3 Relationships Between Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.3.1 Homomorphisms and Embeddings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.3.2 An Application To Definability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69
4.3.3 Substructures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70
4.3.4 Elementary Substructures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71
4.4 Changing the Language . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.4.1 Expansions and Restrictions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
4.4.2 Adding Constants to Name Elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5 Semantic and Syntactic Implication 75

5.1 Semantic Implication and Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.1.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.1.2 Finite Models of Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.1.3 Countable Models of Theories . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2 Syntactic Implication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2.1 Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.2.2 Some Fundamental Deductions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.2.3 Theorems About ` . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
6 Soundness, Completeness, and Compactness 85

6.1 Soundness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85
6.2 Prime Formulas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
6.3 Completeness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
6.3.1 Motivating the Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88
6.3.2 The Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89
6.4 Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97
6.5 Applications of Compactness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98
6.6 An Example: The Random Graph . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
7 Quantifier Elimination 103

7.1 Motivation and Definition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
7.2 What Quantifier Elimination Provides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104
7.3 Quantifier Manipulation Rules . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
7.4 Examples of Theories With QE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106
7.5 Algebraically Closed Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108
CONTENTS 5
8 Nonstandard Models of Arithmetic and Analysis 111

8.1 Nonstandard Models of Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111
8.2 The Structure of Nonstandard Models of Arithmetic . . . . . . . . . . . . . . . . . . . . . . . 113
8.3 Nonstandard Models of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116
6 CONTENTS
Chapter 1
Introduction
1.1 The Nature of Mathematical Logic

Mathematical logic originated as an attempt to codify and formalize
1. The language of mathematics.
2. The basic assumptions of mathematics.
3. The permissible rules of proof.
One successful result of such a program is that we can study mathematical language and reasoning using
mathematics. For example, we will eventually give a precise mathematical definition of a formal proof, and to
avoid confusion with your current intuitive understand of what a proof is, we’ll call these objects deductions.
You should think of this as analogous to giving a precise mathematical definition of continuity to replace the
fuzzy “a graph that can be drawn without lifting your pencil”. Once we’ve codified the notion in this way,
we’ll have turned deductions into mathematical objects and this allows use to prove mathematical theorems
about deductions using normal mathematical reasoning. Thus, we’ve opened up the possibility of proving
that there is no deduction of a certain statement.
Some newcomers to the subject find the whole enterprise perplexing. For instance, if you come to the
subject with the belief that the role of mathematical logic is to serve as a foundation to make mathematics
more precise and secure, then the above description probably seems a little circular and this will almost
certainly lead to a great deal of confusion. You may ask yourself:
Okay, we’ve just given a decent definition of a deduction. However, instead of proving things
about deductions following this formal definition, we’re proving things about deductions using
the usual informal proof style that I’ve grown accustomed to in other math courses. Why should
I trust these informal proofs about deductions? How can we formally prove things (using deduc-
tions) about deductions? Isn’t that circular? Is that why we’re only giving informal proofs? I
thought that I’d come away from this subject feeling better about the philosophical foundations
of mathematics, but we’ve just added a new layer to mathematics and we have both informal
proofs and deductions, making the whole thing even more dubious.
To others who begin a study of the subject, there is no problem. After all, mathematics is the most
reliable method we have to establish truth, and there was never any serious question as to its validity. Such
a person may react to the above thoughts as follows:
7
8 CHAPTER 1. INTRODUCTION
We gave a mathematical definition of a deduction, so what’s wrong with using mathematics to

prove things about deductions? There’s obviously a “real world” of true mathematics, and we’re
just working in that world to build a certain model of mathematical reasoning which is susceptible
to mathematical analysis. It’s quite cool, really, that we can subject mathematical proofs to a
mathematical study by building this internal model. All of this philosophical speculation and
worry about secure foundations is tiresome and in the end probably meaningless. Let’s get on
with the subject!
Should we be so dismissive of the first philosophically inclined student? The answer, or course, depends
on your own philosophical views, but I’ll try to give my own views as a mathematician specializing in logic
with a definite interest in foundational questions. It is my firm belief that you should put all philosophical
questions out of your mind during a first reading of the material (and perhaps forever, if you’re so inclined),
and come to the subject with a point of view which accepts an independent mathematical reality susceptible
to the mathematical analysis you’ve grown accustomed to. In your mind, you should keep a careful distinction
between normal “real” mathematical reasoning and the formal precise model of mathematical reasoning we
are developing. Some people like to give this distinction a name by calling the normal mathematical realm
we’re working in the metatheory.
To those who are interested, we’ll eventually be able to give reasonable answers to the first student and
provide other respectable philosophical accounts of the nature of mathematics, but this should wait until
we’ve developed the necessary framework. Once we’ve done so, we can give examples of formal theories,
such as first-order set theory, which are able to support the entire enterprise of mathematics including
mathematical logic. This is of great philosophical interest, because this makes it possible to carry out
(nearly) all of mathematics inside this formal theory.
The ideas and techniques that were developed with philosophical goals in mind have now found application
in other branches of mathematics and in computer science. The subject, like all mature areas of mathematics,
has also developed its own very interesting internal questions which are often (for better or worse) divorced
from its roots. Most of the subject developed after the 1930’s has been concerned with these internal and
tangential questions, along with applications to other areas, and now foundational work is just one small
(but still important) part of mathematical logic. Thus, if you have no interest in the more philosophical
aspects of the subject, there remains an impressive, beautiful, and mathematically applicable theory which
is worth your attention.
1.2 The Language of Mathematics

Our first and probably most important task in providing a mathematical model of mathematics is to deal with
the language of mathematics. In this section, we sketch the basic ideas and motivation for the development
of a language, but we will leave precise detailed definitions until later.
The first important point is that we should not use English (or any other natural language) because
it’s constantly changing, often ambiguous, and allows the construction of statements that are certainly not
mathematical and arguably express very subjective sentiments. Once we’ve thrown out natural language,
our only choice is to invent our own formal language. This seems quite daunting. How could we possibly
write down one formal language which encapsulates geometry, algebra, analysis, and every other field of
mathematics, not to mention those we haven’t developed yet, without using natural language? Our approach
to this problem will be to avoid (consciously) doing it all at once.
Instead of starting from the bottom and trying to define primitive mathematical statements which can’t
be broken down further, let’s first think about how to build new mathematical statements from old ones. The
simplest way to do this is take already established mathematical statements and put them together using
and, or, not, and implies. To keep a careful distinction between English and our language, we’ll introduce
symbols for each of these, and we’ll call these symbols connectives.
1.2. THE LANGUAGE OF MATHEMATICS 9
1. ∧ will denote and.
2. ∨ will denote or.
3. ¬ will denote not.
4. → will denote implies.
In order to ignore the nagging question of what constitutes a primitive statement, our first attempt will be to
simply take an arbitrary set whose elements we think of as the primitive statements and put them together
in all possible ways using the connectives.
For example, suppose we start with the set P = {A, B, C}. We think of A, B, and C as our primitive
statements, and we may or may not care what they might express. We now want to put together the elements
of P using the connectives, perhaps repeatedly, but to avoid ambiguity we should be careful. Should the
“meaning” of A ∧ B ∨ C be “A holds, and either B holds or C holds”, corresponding to A ∧ (B ∨ C), or should
it be “Either both A and B holds, or C holds”, corresponding to (A ∧ B) ∨ C? We need some way to avoid
this ambiguity. Probably the most natural way to achieve this is to insert parentheses to make it clear how to
group terms (although we’ll see another natural way later). We now describe the formulas of our language,
denoted by F ormP . First, we put every element of P in F ormP , and then we generate other formulas using
the following rules.
1. If ϕ and ψ are in F ormP , then (ϕ ∧ ψ) is in F ormP ..
2. If ϕ and ψ are in F ormP , then (ϕ ∨ ψ) is in F ormP .
3. If ϕ is in F ormP , then (¬ϕ) is in F ormP .
4. If ϕ and ψ are in F ormP , then (ϕ → ψ) is in F ormP .
Thus, the following is an element of F ormP :
((¬(B ∨ ((¬A) → C))) ∨ A)
This simple setup, called propositional logic, is a drastic simplification of the language of mathematics,
but there are already many interesting questions and theorems that arise from a careful study. We’ll spend
some time on it in Chapter 3.
Of course, mathematical language is much more rich and varied than what we can get using propositional
logic. One important way to make more complicated and interesting mathematical statements is to make
use of the quantifiers for all and there exists which we’ll denote using the symbols ∀ and ∃. In order to do
so, we will need variables to act as something to quantify over. We’ll denote variables by letters like x, y, z,
etc. Once we’ve come this far, however, we’ll have have to refine our naive notion of primitive statements
above because it’s unclear how to interpret a statement like ∀xB without knowledge of the role of x “inside”
B.
Let’s think a little about our primitive statements. As we mentioned above, it seems daunting to come
up with primitive statements for all areas of mathematics at once, so let’s think of the areas in isolation.
For instance, take group theory. A group is a set G equipped with a binary operation · (that is, · takes in
two elements x, y ∈ G and produces a new element of G denoted by x · y) and an element e such satisfying
1. Associativity: For all x, y, z ∈ G, we have (x · y) · z = x · (y · z).
2. Identity: For all x ∈ G, we have x · e = x = e · x.
3. Inverses: For all x ∈ G, there exists y ∈ G such that x · y = e = y · x.

Although it is customary and certainly easier on the eyes to put · between two elements of the group, let’s
instead use the standard function notation in order to make the mathematical notation uniform across fields.
In this setting, a group is a set G equipped with a function f : G × G → G and an element e satisfying
1. For all x, y, z ∈ G, we have f (f (x, y), z) = f (x, f (y, z)).
2. For all x ∈ G, we have f (x, e) = x = f (e, x).
3. For all x ∈ G, there exists y ∈ G such that f (x, y) = e = f (y, x).
In order to allow our language to make statement about groups, we introduce a function symbol which we
denote by f to represent the group operation, and a constant symbol which we denote by e to represent the
group identity. Now the group operation is supposed to take in two elements of the group, so if x and y are
variables, then we should allow the formation of f(x, y) which should denote an element of the group (once
we’ve assigned elements of the group to x and y). Also, we should allow the constant symbol to be used
in this way, allowing us to form things like f(x, e). Once we’ve formed these, we should be allowed to use
them like variables in more complicated expressions such as f(f(x, e), y). Each of these expressions formed
by putting together, perhaps repeatedly, variables and the constant symbol e using the function symbol f is
called a term. Intuitively, a term will name a certain element of the group once we’ve assigned elements to
the variables.
With a way to name group elements in hand, we’re now in position to say what out primitive statements
are. The most basic thing that we can say about two group elements is whether or not they are equal, so
we introduce a new equality symbol, which we will denote by the customary =. Given two terms t1 and t2 ,
we call the expression (t1 = t2 ) an atomic formula. These are our primitive statements.
With atomic formulas in hand, we can use the old connectives and the new quantifiers to make new
statements. This puts us in a position to define formulas. First off, all atomic formulas are formulas. Given
formulas we already know, we can put them together using the above connectives. Also, if ϕ is a formula
and x is a variable then each of the following are formulas:
1. ∀xϕ
2. ∃xϕ
Perhaps without realizing it, we’ve described quite a powerful language which can make many nontrivial
statements. For instance, we can write formulas in this language which express the axioms for a group:
1. ∀x∀y∀z(f(f(x, y), z) = f(x, f(y, z)))
2. ∀x((f(x, e) = x) ∧ (f(e, x) = x))
3. ∀x∃y((f(x, y) = e) ∧ (f(y, x) = e))
We can also write a statement saying that the group is abelian:
∀x∀y(f(x, y) = f(y, x))
or that the center is nontrivial:

∃x(¬(x = e) ∧ ∀y(f(x, y) = f(y, x)))
Perhaps unfortunately, we can write syntactically correct formulas which express things nobody would ever
utter, such as:
∀x∃y∃x(¬(e = e))
What if you want to do an area different from group theory? Commutative ring theory doesn’t pose
much of a problem, so long as we’re allowed to alter the number of function symbols and constant symbols.
1.2. THE LANGUAGE OF MATHEMATICS 11
We can simply have two function symbols a and m which take two arguments (a to represent addition and
m to represent multiplication) and two constant symbols 0 and 1 (0 to represent the additive identity and 1
to represent the multiplicative identity). Writing the axioms for commutative rings in this language is fairly
straightforward.
To take something fairly different, what about the theory of partially ordered sets? Recall that a partially
ordered set is a set P equipped with a subset ≤ of P × P , where we write x ≤ y to mean than (x, y) is an
element of this subset, satisfying
1. Reflexive: For all x ∈ P , we have x ≤ x.
2. Antisymmetric: If x, y ∈ P are such that x ≤ y and y ≤ x, then x = y.
3. Transitivity: If x, y, z ∈ P are such that x ≤ y and y ≤ z, then x ≤ z.
Similar to how we handled the group operation, we’ll use notation which puts the ordering in front of the
two arguments. This may seem odd at this point, given how we’re putting equality in the middle, but we’ll
see that this provides a unifying notation for other similar objects. We thus introduce a relation symbol R,
and we keep the equality symbol =, but we no longer have a need for constant symbols or function symbols.
In this setting without constant or function symbols, the only terms that we have (i.e. the only names for
elements of the partially ordered set) are the variables. However, our atomic formulas are more interesting
because now there are two basic things we can say about elements of the partial ordering: whether they are
equal and whether they are related by the ordering. Thus, our atomic formulas are things of the form t1 = t2
and R(t1 , t2 ) where t1 and t2 are terms. From these atomic formulas, we build up all our formulas as above.
Similar to the situation for groups, we can know write formulas expressing the axioms of partial orderings:
1. ∀xR(x, x)
2. ∀x∀y((R(x, y) ∧ R(y, x)) → (x = y))
3. ∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z))
We can also write a statement saying that the partial ordering is a linear ordering:
∀x∀y(R(x, y) ∨ R(y, x))
or that there exists a maximal element:

∃x∀y(¬R(x, y))
The general idea is that by leaving flexibility in the types and number of constant symbols, relation
symbols, and function symbols, we’ll be able to handle many areas of mathematics. We call this setup
first-order logic. An analysis of first-order logic will consume the vast majority of our time.
Now we don’t claim that first-order logic allows us to do and express everything in mathematics, nor do
we claim that each of the setups above allow us to do and express everything of importance in that particular
field. For example, take the group theory setting above. We can express that every nonidentity element has
order two with:
∀x(f(x, x) = e)
but it’s unclear how to say that every element of the group has finite order. The natural guess is:
∀x∃n(xn = e)
but this poses a problem for two reasons. The first is that our variables are supposed to quantify over
elements of the group in question, not the natural numbers. The second is that we put no construction in
our language to allow us to write something like xn . For each fixed n, we can express it (for example, for
n = 3, we can write f(f(x, x), x) and for n = 4, we can write f(f(f(x, x), x), x)), but it’s not clear how to write
it in a general way that would even allow quantification over the natural numbers.
Another example is trying to express that a group is simple (i.e. has no nontrivial normal subgroups).
The natural instinct is to quantify over all subsets of the group H, and say that if it so happens that H
is a normal subgroup, then H is either trivial or everything. However, we have no way to quantify over
subsets. It’s certainly possible to allow such constructions, and this gives second-order logic. If you allow
quantifications over sets of subsets (for example one way of expressing that a ring is Noetherian is to say
that every nonempty set of ideals has a minimal element), you get third-order logic, etc.
Newcomers to the field often find it strange that we focus primarily on first-order logic. There are many
reasons to give special attention to first-order logic that will be developed throughout our study, but for
now you should think of it as providing a simple example of a language which is capable of expressing many
important aspects of various branches of mathematics.
1.3 Syntax and Semantics

Notice that in the above discussion, we introduced symbols to denote certain concepts (such as using ∧ in
place of “and”, ∀ in place of “for all”, and a function symbol f in place of the group operation f ). Building
and maintaining a careful distinction between formal symbols and how to interpret them is a fundamental
aspect of mathematical logic.
The basic structure of the formal statements that we write down using the symbols, connectives, and
quantifiers is known as the syntax of the logic that we’re developing. This corresponds to the grammar
of the language in question with no thought given to meaning. Imagine an English instructor who cared
nothing for the content of your writings, but only that the it was grammatically correct. That is exactly
what the syntax of a logic is all about. Syntax is combinatorial in nature and is based on rules which provide
admissible ways to manipulate symbols devoid of meaning.
The manner in which we are permitted (or forced) to interpret the symbols, connectives, and quantifiers is
known as the semantics of the the logic that we’re developing. In a logic, some symbols are to be interpreted
in only one way. For instance, in the above examples, we interpret the symbol ∧ to mean and. In the
propositional logic setting, this doesn’t settle how to interpret a formula because we haven’t said how to
interpret the elements of P . We have some flexibility here, but once we assert that we should interpret
certain elements of P as true and the others as false, our formulas express statements that are either true
or false.
The first-order logic setting is more complicated. Since we have quantifiers, the first thing that must be
done in order to interpret a formula is to fix a set X which will act as the set of objects over which the
quantifiers will range. Once this is done, we can interpret any function symbol f taking k arguments as any
actual function f : X k → X, any relation R symbol taking k arguments as a subset of X k , and any constant
symbol c as an element of X. Once we’ve fixed what we’re talking about by provided such interpretations,
we can view them as expressing something meaningful. For example, if we’ve fixed a group G interpreted f
as the group operation and e as the identity, the formula
∀x∀y(f(x, y) = f(y, x))
is either true or false, according to whether G is abelian or not.

Always keep the distinction between syntax and semantics clear in your mind. The basic theorems of the
subject involve the interplay between syntax and semantics. For example, in the logics we discuss, we will
have two types of implication between formulas. Let Γ be a set of formulas and let ϕ be a formula. One way
of saying that the formulas in Γ imply ϕ is semantic: whenever we provide an interpretation which makes all
of the formulas of Γ true, it happens that ϕ is also true. For instance, if we’re working in propositional logic
and we have Γ = {((A ∧ B) ∨ C)} and ϕ = (A ∨ C), then Γ implies ϕ in this sense because no matter how we
1.4. THE POINT OF IT ALL 13
assign true/false values to A, B, and C that make the formulas in Γ true, it happens that ϕ will also be true.
Another approach that we’ll develop is syntactic. We’ll define deductions which are “formal proofs” built
from certain permissible syntactic manipulations, and Γ will imply ϕ in this sense if there is a witnessing
deduction. The Soundness Theorem and the Completeness Theorem for first-order logic (and propositional
logic) says that the semantic version and syntactic version are the same. This result amazingly allows one
to mimic mathematical reasoning with syntactic manipulations.
1.4 The Point of It All

One important aspect, often mistaken as the only aspect, of mathematical logic is that it allows us to study
mathematical reasoning. A prime example of this is given by the last sentence of the previous section.
The Completeness Theorem says that we can capture the idea of one mathematical statement following
from other mathematical statements with nothing more than syntactic rules on symbols. This is certainly
computationally, philosophically, and foundationally interesting, but it’s much more than that. A simple
consequence of this result is the Compactness Theorem, which says something very deep about mathematical
reasoning and has many interesting applications in mathematics.
Although we’ve developed the above logics with modest goals of handling certain fields of mathematics,
it’s a wonderful and surprising fact that we can embed (nearly) all of mathematics in an elegant and natural
first-order system: first-order set theory. This opens the door to the possibility of proving that certain
mathematical statements are independent of our usual axioms. That is, that there are formulas ϕ such that
there is no deduction from the usual axioms of either ϕ or (¬ϕ). Furthermore, the field of set theory has
blossomed into an intricate field with its own deep and interesting questions.
Other very interesting and fundamental subjects arise when we ignore the foundational aspects and
deductions altogether, and simply look at what we’ve accomplished by establishing a precise language to
describe an area of mathematics. With a language in hand, we now have a way to say that certain objects
are definable in that language. For instance, take the language of commutative rings mentioned above. If we
fix a particular commutative ring, then the formula
∃y(m(x, y) = 1)
has a free variable x and “defines” the set of units in the ring. With this point of view, we’ve opened up
the possibility of proving lower bounds on the complexity of any definition of a certain object, or even of
proving that no such definition exists in the language.
Another, closely related, way to take our definitions of precise languages and run with it is the subject
of model theory. In group theory, you state some axioms and work from there in order to study all possible
realizations of the axioms, i.e. groups. However, as we saw above, the group axioms arise in one possible
language with one possible set of axioms. Instead, we can study all possible languages and all possible sets
of axioms and see what we can prove in general and how the realizations compare to each other. In this
sense, model theory is a kind of abstract abstract algebra.
Finally, although it’s probably far from clear how it fits in at this point, computability theory is intimately
related to the above subjects. To see the first glimmer of a connection, notice that computer programming
languages are also formal languages with a precise grammar and a clear distinction between syntax and
semantics. As we’ll see in time, however, the connection is much deeper.
1.5 Some Basic Terminology and Notation

Definition 1.5.1. We let N = {0, 1, 2, . . . } and we let N+ = N\{0}.
Definition 1.5.2. For each n ∈ N, we let [n] = {m ∈ N : m < n}.

We will often find a need to work with finite sequences, so we establish notation here.
Definition 1.5.3. Let X be a set. Given n ∈ N, we call a function σ : [n] → X a finite sequence from X
of length n. We denote the set of all finite sequences from X of length n by X n . We use λ to denote the
unique sequence of length 0, so X 0 = {λ}.
Definition 1.5.4. Let X be a set. We let X ∗ = n∈N X n , i.e. X ∗ is the set of all finite sequences from X.
S
We denote finite sequences by simply listing the elements in order. For instance, if X = {a, b}, the
sequence aababbba is an element of X ∗ . Sometimes for clarity, we’ll insert commas and instead write
a, a, b, a, b, b, b, a.
Definition 1.5.5. If σ, τ ∈ X ∗ , we say that σ is an initial segment of τ , and write σ ⊆ τ , if σ = τ [n] for
some n. We say that σ is a proper initial segment of τ , and write σ ⊂ τ , if σ ⊆ τ and σ 6= τ .
Definition 1.5.6. If σ, τ ∈ X ∗ , we denote the concatenation of σ and τ by στ or σ ∗ τ .
Definition 1.5.7. If σ, τ ∈ X ∗ , we say that σ is a substring of τ if there exists θ, ρ ∈ X ∗ such that σ = θτ ρ.
Definition 1.5.8. A set A is countably infinite if there exists a bijection g : N → A. A set A is countable
if it is either finite or countably infinite.
Definition 1.5.9. Given a set A, we let P(A) be the set of all subsets of A, and we call P(A) the power set
of A.
Chapter 2
Induction and Recursion
The natural numbers are perhaps the only structure that you’ve had the pleasure of working with when doing
proofs by induction or definitions by recursion, but there are many more arenas in which variants of induction
and recursion apply. In fact, more delicate and exotic proofs by induction and definitions by recursion are
two central tools in mathematical logic. Once we get to set theory, we’ll see how to do transfinite induction
and recursion, and this tool is invaluable in set theory and model theory. In this section, we develop the
more modest tools of induction and recursion along structures which are generated by one-step processes,
like the natural numbers.
2.1 Induction and Recursion on N

We begin by compiling the basic facts about induction and recursion on the natural numbers. We don’t
seek to “prove” that proofs by induction or definitions by recursion on N are valid methods because these
are “obvious” from the normal mathematical perspective which we are adopting. Besides, in order to do so,
we would first have to fix a context in which we are defining N, which we will do much later in the context
of axiomatic set theory. Although you’re no doubt familiar with the intuitive content of the results here,
our goal here is simply to carefully codify these facts in more precise ways to ease the transition to more
complicated types of induction and recursion.
Definition 2.1.1. We define S : N → N by letting S(n) = n + 1 for all n ∈ N.
Induction is often stated in the form “If we know something holds of 0, and we know that it holds of S(n)
whenever it holds of n, then we know that it holds for all n ∈ N”. We state it in the following more precise
set-theoretic fashion (avoiding explicit mention of “somethings” or “properties”) because we can always form
the set X = {n ∈ N : something holds of n}.
Theorem 2.1.2 (Induction on N - Step Form). Suppose that X ⊆ N is such that 0 ∈ X and S(n) ∈ X
whenever n ∈ X. We then have X = N.
Definitions by recursion is usually referred to by saying that “When defining f (S(n)), you are allowed to
refer to the value of f (n)”. For instance, let f : N → N be the factorial function f (n) = n!. One usually sees
this defined in the following manner:
f (0) = 1
f (S(n)) = S(n) · f (n)
We aim to codify this idea a little more abstractly and rigorously in order to avoid the self-reference of f in
the definition and allowable rules so that we can generalize it to other situations.
15
16 CHAPTER 2. INDUCTION AND RECURSION
Suppose that X is a set and we’re trying to define f : N → X recursively. What do you need? Well, you
need to know f (0), and you need to have a method telling you how to define f (S(n)) from knowledge of n
and the value of f (n). If we want to avoid the self-referential reference to f in invoking the value of f (n),
what we need is a method which tells us what to do next regardless of the actual particular value of f (n).
That is, it needs to tell you what to do on any possible value, not just the one the ends up happening to be
f (n). Formally, this “method” can be given by a function g : N × X → X which tells you what to do at the
next step. Intuitively, this function acts as an iterator. That is, it says if the the last thing you were working
on was input n and it so happened that you set f (n) to equal x ∈ A, then you should define f (S(n)) to be
the value g(n, x).
With all this setup, we now state the theorem which says that no matter what value you want to assign
to f (0), and no matter what iterating function g : N × X → X you give, there exists a unique function
f : N → X obeying the rules.
Theorem 2.1.3 (Recursion on N - Step Form). Let X be a set, let y ∈ X, and let g : N × X → X. There
exists a unique function f : N → X such that
1. f (0) = y.
2. f (S(n)) = g(n, f (n)) for all n ∈ N.
In the case of the factorial function, we have X = N, y = 1, and g : N×N → N defined by g(n, x) = S(n)·x.
The above theorem implies that there is a unique function f : N → N such that
1. f (0) = y = 1.
2. f (S(n)) = g(n, f (n)) = S(n) · f (n) for all n ∈ N.
Notice how we’ve avoiding a self-reference in the definition. Our definition of g is given nonrecursively, and
then the theorem states the existence and uniqueness of a function which behaves properly.
There is another version of induction on N, sometimes called “strong induction”, which appeals to the
ordering of the natural numbers rather than the stepping of the successor function.
Theorem 2.1.4 (Induction on N - Order Form). Suppose that X ⊆ N is such that n ∈ X whenever m ∈ X
for all m < n. We then have X = N.
Notice that there is no need to deal with the “base case” of n = 0, because this is handled vacuously due
to the fact that there is no m < 0.
Theorem 2.1.5 (Recursion on N - Order Form). Let X be a set and let g : X ∗ → X. There exists a unique
function f : N → X such that
f (n) = g(f [n])
for all n ∈ N.
2.2 Generation
There are many situations throughout mathematics when we want to look at what a certain subset “gen-
erates”. For instance, you have a subset of a group (vector space, ring), and you want to consider the
subgroup (subspace, ideal) that they generate. Another example is you have a subset of a graph, and you
want to consider the set of vertices in the graph reachable from the subset. In the introduction, we talked
about generating all formulas from primitive formulas using certain connections. This situation will arise so
frequently in what follows that it’s a good idea to unify them all in a common framework.
2.2. GENERATION 17
Definition 2.2.1. Let A be a set and let k ∈ N+ . A function h : Ak → A is called a k-ary function on A.
We call k the arity of h. A 1-ary function is sometimes called unary and a 2-ary function is sometimes
called binary.
Definition 2.2.2. Suppose that A is a set, B ⊆ A and H is a collection of functions such that each h ∈ H is
a k-ary function on A for some k ∈ N+ . We call (A, B, H) a simple generating system. In such a situation,
for each k ∈ N+ , we denote the set of k-ary functions in H by Hk .
Examples.
1. Let G be a group and let B ⊆ G. We want the subgroup of G that B generates. The operations in
question here are the group operation and inversion, so we let H = {h1 , h2 } where
(a) h1 : G2 → G is given by h1 (x, y) = x · y for all x, y ∈ G.

(b) h2 : G → G is given by h2 (x) = x−1 for all x ∈ G.
(G, B, H) is a simple generating system.
2. Let V be a vector space over R and let B ⊆ V . We want the subspace of V that B generates. The
operations in question consist of vector addition and scalar multiplication, so we let H = {g} ∪ {hα :
α ∈ R} where
(a) g : V 2 → V is given by g(v, w) = v + w for all v, w ∈ V .

(b) For each α ∈ R, hα : V → V is given by hα (v) = α · v for all v ∈ V .
(V, B, H) is a simple generating system.
There are certain cases when the natural functions to put into H are not total or are “multi-valued”. For
instance, in the first example below, we’ll talk about the subfield generated by a certain subset of a field,
and we’ll want to include multiplicative inverses for all nonzero elements. When putting a corresponding
function in H, there is no obvious way to define it on 0. Also, if generating the vertices reachable from a
subset of a graph, we may want to throw many vertices in because a vertex can be linked to many others.
Definition 2.2.3. Let A be a set and let k ∈ N+ . A function h : Ak → P(A) is called a set-valued k-ary
function on A. We call k the arity of h. A 1-ary set-valued function is sometimes called unary and a 2-ary
set-valued function is sometimes called binary.
Definition 2.2.4. Suppose that A is a set, B ⊆ A and H is a collection of functions such that each h ∈ H
is a set-valued k-ary function on A for some k ∈ N+ . We call (A, B, H) a generating system. In such a
situation, for each k ∈ N+ , we denote the set of multi-valued k-ary functions in H by Hk .
Examples.
1. Let K be a field and let B ⊆ K. We want the subfield of K that B generates. The operations in
question here are addition, multiplication, and both additive and multiplicative inverses. We thus let
H = {h1 , h2 , h3 , h4 } where
(a) h1 : K 2 → P(K) is given by h1 (a, b) = {a + b} for all a, b ∈ K.

(b) h2 : K 2 → P(K) is given by h2 (a, b) = {a · b} for all a, b ∈ K.
(c) h3 : K → P(K) is given by h3 (a) = {−a} for all a ∈ K.
(d) h4 : K → P(K) is given by (

{a−1 } if a 6= 0
h4 (a) =
∅ if a = 0
(K, B, H) is a generating system.

2. Let G be a graph with vertex set V and edge set E, and let B ⊆ V . We want to consider the subset
of V reachable from B using edges from E. Thus, we want to say that if we’ve generated v ∈ V ,
and w ∈ V is connected to v via some edge, then we should generate w. We thus let H = {h} where
h : V → V is defined as follows:
h(v) = {u ∈ V : (v, u) ∈ E}
(V, B, H) is a generating system.
Notice that if we have a simple generating system (A, B, H), then we can associate to it the generating
system (A, B, H0 ) where H0 = {h0 : h ∈ H} where if h : Ak → A is an element of Hk , then h0 : Ak → P(A) is
defined by letting h0 (a1 , a2 , . . . , ak ) = {h(a1 , a2 , . . . , ak )}.
Given a generating system (A, B, H), we want to define the set of elements of A generated from B using
the functions in H. There are many natural ways of doing this. We discuss three different ways which divide
into approaches “from above” and approaches “from below”. Each of these descriptions can be slightly
simplified for simple generating systems, but it’s not much harder to handle the more general case.
2.2.1 From Above

Our first approach is a “top-down” approach.
Definition 2.2.5. Let (A, B, H) be a generating system, and let J ⊆ A. We say that J is inductive if
1. B ⊆ J.
2. If k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ J, then h(a1 , a2 , . . . , ak ) ⊆ J.
Given a generating system (A, B, H), we certainly have a candidate for an inductive set, namely A itself.
However, this set may be too big. For instance, consider the generating system A = R, B = {7}, and
H = {h} where h : R → R is the function h(x) = 2x. In this situation, each of the sets R, Z, N, and
{n ∈ N : n is a multiple of 7} is inductive, but they’re not what we want. The idea is to consider the smallest
inductive subset of A containing B. Of course, we need to prove that such a set exists.
Proposition 2.2.6. Let (A, B, H) be a generating system. There exists a unique inductive set I such that
I ⊆ J for every inductive set J.
Proof. We first prove existence. Let I be the intersection of all inductive sets, i.e. I = {a ∈ A : a ∈ J for
every inductive set J}. By definition, we have I ⊆ J for every inductive set J, so we need only show that
I is inductive. Since B ⊆ J for every inductive set J (by definition of inductive), it follows that B ⊆ I.
Suppose that k ∈ N+ , h ∈ Hk and a1 , a2 , . . . , ak ∈ I. For any inductive set J, we have a1 , a2 , . . . , ak ∈ J,
hence h(a1 , a2 , . . . , ak ) ⊆ J because J is inductive. Therefore, h(a1 , a2 , . . . , ak ) ⊆ J for every inductive set
J, hence h(a1 , a2 , . . . , ak ) ⊆ I. It follows that I is inductive.
To see uniqueness suppose that both I1 and I2 are inductive sets such that I1 ⊆ J and I2 ⊆ J for every
inductive set J. We then have I1 ⊆ I2 and I2 ⊆ I1 , hence I1 = I2 .
Definition 2.2.7. Let (A, B, H) be a generating system. We denote the unique set of the previous proposition
by I(A, B, H), or simply by I when the context is clear.
2.2. GENERATION 19
2.2.2 From Below: Building by Levels

The second idea is to make a system of levels, at each new level adding elements of A which are reachable
from elements already accumulated by applying an element of H.
Definition 2.2.8. Let (A, B, H) be a generating system. We define a sequence Vn (A, B, H), or simply Vn ,
recursively as follows.
V0 = B
Vn+1 = Vn ∪ {c ∈ A : There exists k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ Vn such that c ∈ h(a1 , a2 , . . . , ak )}
S
Let V (A, B, H) = V = Vn .
n∈N
The following remarks are immediate from our definition.

Remark 2.2.9. Let (A, B, H) be a generating system.
1. If m ≤ n, then Vm ⊆ Vn .
2. For all c ∈ V , either c ∈ B or there exists k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ V with c ∈
h(a1 , a2 , . . . , ak ).
2.2.3 From Below: Witnessing Sequences

The third method is to consider those elements of A which you are forced to put in because you see a
witnessing construction.
Definition 2.2.10. Let (A, B, H) be a generating system. A witnessing sequence is an element σ ∈ A∗ \{λ}
such that for all j < |σ|, either
1. σ(j) ∈ B
2. There exists k ∈ N, h ∈ Hk and i1 , i2 , . . . , ik < j such that σ(j) ∈ h(σ(i1 ), σ(i2 ), . . . , σ(ik )).
If σ is a witnessing sequence, we call it a witnessing sequence for σ(|σ| − 1).
Definition 2.2.11. Let (A, B, H) be a generating system. Set
W (A, B, H) = W = {a ∈ A : there exists a witnessing sequence for a}.
It sometimes useful to look only at those elements reachable which are witnessed by sequences of a bounded
length, so for each n ∈ N+ , set
Wn = {a ∈ A : there exists a witnessing sequence for a of length n}.
The first simple observation is that if we truncate a witnessing sequence, what remains is a witnessing
sequence.
Remark 2.2.12. If σ is a witnessing sequence and |σ| = n, then for all m ∈ N+ with m < n we have that
σ [m] is a witnessing sequence.
Another straightforward observation is that if we concatenate two witnessing sequences, the result is a
witnessing sequence.
Proposition 2.2.13. If σ and τ are witnessing sequences, then so is στ .
Finally, since we can always insert “dummy” elements from B (assuming it’s nonempty because otherwise
the result is trivial), we have the following observation.
Proposition 2.2.14. Let (A, B, H) be a generating system. If m ≤ n, then Wm ⊆ Wn .
2.2.4 Equivalence of the Definitions

Theorem 2.2.15. Let (A, B, H) be a generating system. We then have
I(A, B, H) = V (A, B, H) = W (A, B, H)
Proof. Let I = I(A, B, H), V = V (A, B, H), and W = W (A, B, H).

We first show that V is inductive, hence I ⊆ V . Notice first that B = V0 ⊆ V . Suppose now that k ∈ N+ ,
h ∈ Hk and a1 , a2 , . . . , ak ∈ V . For each i, fix ni such that ai ∈ Vni . Let m = max{n1 , n2 , . . . , nk }. We then
have ai ∈ Vm for all i, hence h(a1 , a2 , . . . , ak ) ⊆ Vm+1 ⊆ V . It follows that V is inductive.
We next show that W is inductive, hence I ⊆ W . Notice first that for every b ∈ B, the sequence b is
a witnessing sequence, so b ∈ W1 ⊆ W . Suppose now that k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ W . Let
c ∈ h(a1 , a2 , . . . , ak ). For each i, fix a witnessing sequence σi for ai . The sequence σ1 σ2 · · · σk c is a witnessing
sequence for c. Therefore. h(a1 , a2 , . . . , ak ) ⊆ W . It follows that W is inductive.
We next show that Vn ⊆ I by induction on n, and hence V ⊆ I. Notice first that V0 = B ⊆ I. Suppose
now that n ∈ N and Vn ⊆ I. Fix k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ Vn . Since Vn ⊆ I, we have
a1 , a2 , . . . , ak ∈ I, hence h(a1 , a2 , . . . , ak ) ⊆ I because I is inductive. It follows that Vn+1 ⊆ I. By induction,
Vn ⊆ I for every n ∈ N, hence V ⊆ I.
We next show that Wn ⊆ I by induction on n ∈ N+ , and hence W ⊆ I. Notice first that W1 = B ⊆ I.
Suppose now that n ∈ N+ and Wn ⊆ I. Let σ be an witnessing sequence of length n + 1. We then have that
that σ [m + 1] is a witnessing sequence of length m + 1 for all m < n, hence σ(m) ∈ Wm ⊆ Wn ⊆ I for
all m < n. Now either σ(n) ∈ B or there exists i1 , i2 , . . . , ik < n such that σ(n) = h(σ(i1 ), σ(i2 ), . . . , σ(ik )).
In either case, σ(n) ∈ I because I is inductive. It follows that Wn+1 ⊆ I. By induction, Wn ⊆ I for every
n ∈ N+ , hence W ⊆ I.
Definition 2.2.16. Let (A, B, H) be a generating system. We denote the common value of I, V, W by
G(A, B, H) or simply G.
The nice thing about having multiple equivalent definitions for the same concept is that we can use the
most convenient one when proving a theorem. For example, using (2) of Remark 2.2.9, we get the following
corollary.
Corollary 2.2.17. Let (A, B, H) be a generating system. For all c ∈ G, either c ∈ B or there exists k ∈ N+ ,
h ∈ Hk , and a1 , a2 , . . . , ak ∈ G with c ∈ h(a1 , a2 , . . . , ak ).
2.3 Step Induction

Here’s a simple example of using the I definition to prove that we can argue by induction.
Proposition 2.3.1 (Step Induction). Let (A, B, H) be a generating system. Suppose that X ⊆ A satisfies
1. B ⊆ X.
2. h(a1 , a2 , . . . , ak ) ⊆ X whenever k ∈ N+ , h ∈ Hk , and a1 , a2 , . . . , ak ∈ X.
We then have that G ⊆ X. Thus, if X ⊆ G, we have X = G.
Proof. Our assumption simply asserts that X is inductive, hence G = I ⊆ X.
The next example illustrates how we can sometimes identify G explicitly. Notice that we use 2 different
types of induction in the argument. One direction uses induction on N and the other uses induction on G
as just described.
2.4. STEP RECURSION 21
Example 2.3.2. Consider the following simple generating system. Let A = R, B = {7}, and H = {h}
where h : R → R is the function h(x) = 2x. Determine G explicitly.
Proof. Intuitively, we want the set {7, 14, 28, 56, . . . }, which we can write more formally as {7 · 2n : n ∈ N}.
Let X = {7 · 2n : n ∈ N}
We first show that X ⊆ G by showing that 7 · 2n ∈ G for all n ∈ N by induction (on N). We have
7 · 20 = 7 · 1 = 7 ∈ G because B ⊆ G as G is inductive. Suppose that n ∈ N is such that 7 · 2n ∈ G. Since
G is inductive, it follows that h(7 · 2n ) = 2 · 7 · 2n = 7 · 2n+1 ∈ G. Therefore, 7 · 2n ∈ G for all n ∈ N by
induction, hence X ⊆ G.
We now show that G ⊆ X by induction (on G). Notice that B ⊆ X because 7 = 7 · 1 = 7 · 20 ∈ X.
Suppose now that x ∈ X and fix n ∈ N with x = 7 · 2n . We then have h(x) = 2 · x = 7 · 2n+1 ∈ X. Therefore
G ⊆ X by induction.
It follows that X = G.
In many cases, it’s very hard to give a simple explicit description of the set G. This is where induction
really shines, because it allows us to prove something about all elements of G despite the fact that we have
a hard time getting a handle on what exactly the elements of G look like. Here’s an example.
Example 2.3.3. Consider the following simple generating system. Let A = Z, B = {6, 183}, and H = {h}
where h : A3 → A is given by h(k, m, n) = k · m + n. Every element of G is divisible by 3.
Proof. Let X = {n ∈ Z : n is divisible by 3}. We prove by induction that G ⊆ X. We first handle the bases
case. Notice that 6 = 3 · 2 and 183 = 3 · 61, so B ⊆ X.
We now do the inductive step. Suppose that k, m, n ∈ X, and fix `1 , `2 , `3 ∈ Z with k = 3`1 , m = 3`2 ,
and n = 3`3 . We then have
h(k, m, n) = k · m + n
= (3`1 ) · (3`2 ) + 3`3
= 9`1 `2 + 3`3
= 3(3`1 `2 + `3 )
hence h(k, m, n) ∈ X.
It follows by induction that G ⊆ X, i.e. that every element of G is divisible by 3.
2.4 Step Recursion

In this section, we restrict attention to simple generating systems for simplicity (and also because all examples
that we’ll need which support definition by recursion will be simple). Naively, one might expect that a
straightforward analogue of Step Form of Recursion on N will carry over to recursion on generated sets. The
hope would be the following.
Hope 2.4.1. Suppose that (A, B, H) is a simple generating system and X is a set. Suppose also that
ι : B → X and that for every h ∈ Hk , we have a function gh : (A × X)k → X. There exists a unique function
f : G → X such that
1. f (b) = ι(b) for all b ∈ B.
2. f (h(a1 , a2 , . . . , ak )) = gh (a1 , f (a1 ), a2 , f (a2 ), . . . , ak , f (ak )) for all a1 , a2 , . . . , ak ∈ G.
Unfortunately, this hope is too good to be true. Intuitively, we may generate an element a of A in many
very different ways, and our different iterating functions conflict on what values we should assign to a. Here’s
a simple example to see what can go wrong.
Example 2.4.2. Consider the following simple generating system. Let A = {1, 2}, B = {1}, and H = {h}
where h : A → A is given by h(1) = 2 and h(2) = 1. Let X = N. Define ι : B → N by letting ι(1) = 1 and
define gh : A × N → N by letting gh (a, n) = n + 1. There is no function f : G → N such that
2. f (h(a)) = gh (a, f (a)) for all a ∈ G.
Proof. Notice first that G = {1, 2}. Suppose that f : G → N satisfies (1) and (2) above. Since f satisfies (1),
we must have f (1) = ι(1) = 1. By (2), we then have that
f (2) = f (h(1)) = gh (1, f (1)) = f (1) + 1 = 1 + 1 = 2.
By (2) again, it follows that
f (1) = f (h(2)) = gh (2, f (2)) = f (2) + 2 = 1 + 2 = 3,
contradicting the fact that f (1) = 1.
To get around this problem, we want a definition of a “nice” simple generating system. Intuitively, we
want to say something like “every element of G is generated in a unique way”. The following definition is a
relatively straightforward way to formulate this.
Definition 2.4.3. A simple generating system (A, B, H) is free if
1. ran(h Gk ) ∩ B = ∅ whenever h ∈ Hk .
2. h Gk is injective for every h ∈ Hk .
3. ran(h1 Gk ) ∩ ran(h2 G` ) = ∅ whenever h1 ∈ Hk and h2 ∈ H` with h1 6= h2 .
Here’s a simple example which will play a role for us in Section 2.5. We’ll see more subtle and important
examples when we come to Propositional Logic and First-Order Logic.
Example 2.4.4. Let X be a set. Consider the following simple generating system. Let A = X ∗ , let B = X,
and let H = {hx : x ∈ X} where hx : X ∗ → X ∗ is the function hx (σ) = xσ. We then have that G = X ∗ \{λ}
and that (A, B, H) is free.
Proof. First notice that X ∗ \{λ} is inductive because λ ∈

/ B and hx (σ) 6= λ for all σ ∈ X ∗ . Next, a simple
induction on n shows that X ⊆ G for all n ∈ N . It follows that G = X ∗ \{λ}.
n +
We now show that (A, B, H) is free. First notice that for any x ∈ X, we have that ran(hx G) ∩ X = ∅
because every elemnts of ran(hx G) has length at least 2 (because λ ∈
/ G).
Now for any x ∈ X, we have that hx G is injective because if hx (σ) = hx (τ ), then xσ = xτ , and hence
σ = τ.
Finally, notice that if x, y ∈ X with x 6= y, we have that ran(hx G) ∩ ran(hy G) = ∅ because every
elements of ran(hx G) begins with x while every element of ran(hy G) begins with y.
On to the theorem which says that if a simple generating system is free, then we can perform recursive
definitions on it.
Theorem 2.4.5. Suppose that the simple generating system (A, B, H) is free and X is a set. Suppose also
that ι : B → X and that for every h ∈ Hk , we have a function gh : (A × X)k → X. There exists a unique
function f : G → X such that

2.4. STEP RECURSION 23
2. f (h(a1 , a2 , . . . , ak )) = gh (a1 , f (a1 ), a2 , f (a2 ), . . . , ak , f (ak )) for all h ∈ Hk and all a1 , a2 , . . . , ak ∈ G.

Proof. We first prove the existence of an f by using a fairly slick argument. The basic idea is to build a new
simple generating system whose elements are pairs (a, x) where a ∈ A and x ∈ X. Intuitively, we want to
generate the pair (a, x) if something (either ι or one the gh functions) tells us that we’d better set f (a) = x
if we want to satisfy the above the conditions. We then go on to prove (by induction on G) that for every
a ∈ A, there exists a unique x ∈ X such that (a, x) is in our new generating system. Thus, there are no
conflicts, so we can use this to define our function.
Now for the details. Let A0 = A × X, B 0 = {(b, ι(b)) : b ∈ B} ⊆ A0 , and H0 = {gh0 : h ∈ H} where for
each h ∈ Hk , the function gh0 : (A × X)k → A × X is given by
gh0 (a1 , x1 , a2 , x2 , . . . , ak , xk ) = (h(a1 , a2 , . . . , ak ), gh (a1 , x1 , a2 , x2 , . . . , ak , xk )).
Let G0 = G(A0 , B 0 , H0 ). A simple induction (on G0 ) shows that if (a, x) ∈ G0 , then a ∈ G. Let
Z = {a ∈ G : there exists a unique x ∈ X such that (a, x) ∈ G0 }
We prove by induction (on G) that Z = G.
Base Case: Notice that for each b ∈ B, we have (b, ι(b)) ∈ B 0 ⊆ G0 , hence there exists x ∈ X such
that (b, x) ∈ G0 . Fix b ∈ B and suppose that y ∈ X is such that (b, y) ∈ G0 and y 6= ι(b). We then have
/ B 0 , hence by Corollary 2.2.17 there exists h ∈ Hk and (a1 , x1 ), (a2 , x2 ), . . . , (ak , xk ) ∈ G0 such that
(b, y) ∈
(b, y) = gh0 (a1 , x1 , a2 , x2 , . . . , ak , xk )
= (h(a1 , a2 , . . . , ak ), gh (a1 , x1 , a2 , x2 , . . . , ak , xk )).
Since a1 , a2 , . . . , ak ∈ G, this contradicts the fact that ran(h Gk ) ∩ B = ∅. Therefore, for every b ∈ B,
there exists a unique x ∈ X, namely ι(b), such that (b, x) ∈ G0 . Thus, B ⊆ Z.
Inductive Step: Fix h ∈ Hk , and suppose that a1 , a2 , . . . , ak ∈ Z. For each i, let xi be the unique element
of X with (ai , xi ) ∈ G0 . Notice that
(h(a1 , a2 , . . . , ak ), gh (a1 , x1 , a2 , x2 , . . . , ak , xk )) = gh0 (a1 , x1 , a2 , x2 , . . . , ak , xk ) ∈ G0
hence there exists x ∈ X such that (h(a1 , a2 , . . . , ak ), x) ∈ G0 . Suppose now that y ∈ X is such that
(h(a1 , a2 , . . . , ak ), y) ∈ G0 . We have (h(a1 , a2 , . . . , ak ), y) ∈
/ B 0 because ran(h Gk ) ∩ B = ∅, so there exists
0
ĥ ∈ H` together with (c1 , z1 ), (c2 , z2 ), . . . , (c` , z` ) ∈ G such that
(h(a1 , a2 , . . . , ak ), y) = (ĥ(c1 , c2 , . . . , c` ), gĥ (c1 , z1 , c2 , z2 , . . . , c` , z` )).
Since c1 , c2 , . . . , c` ∈ G, it follows that h = ĥ because ran(h Gk ) ∩ ran(ĥ G` ) = ∅ if h 6= ĥ, and

hence k = `. Also, since h Gk is injective, it follows that ai = ci for all i. We therefore have y =
gh (a1 , x1 , a2 , x2 , . . . , ak , xk ). Therefore, there exists a unique x ∈ X, namely gh (a1 , x1 , a2 , x2 , . . . , ak , xk ),
such that (h(a1 , a2 , . . . , ak ), x) ∈ G0 . It now follows by induction that Z = G.
Define f : G → X by letting f (a) be the unique x ∈ X such that (a, x) ∈ G0 . We need to check that f
satisfies the needed conditions. As stated above, for each b ∈ B, we have (b, ι(b)) ∈ G0 , so f (b) = ι(b). Thus,
f satisfies condition (1). Suppose now that h ∈ Hk and all a1 , a2 , . . . , ak ∈ G. We have (ai , f (ai )) ∈ G0 for
all i, hence
(h(a1 , a2 , . . . , ak ), gh (a1 , f (a1 ), a2 , f (a2 ), . . . , ak , f (ak ))) ∈ G0
by the above comments. It follows that f (h(a1 , a2 , . . . , ak )) = gh (a1 , f (a1 ), a2 , f (a2 ), . . . , ak , f (ak )). Thus,
f also satisfies condition (2).
Finally, we need show that f is unique. Suppose that f1 , f2 : G → X satisfy the conditions (1) and (2).
Let Y = {a ∈ G : f1 (a) = f2 (a)} We show that Y = G by induction on G. First notice that for any b ∈ B
we have
f1 (b) = ι(b) = f2 (b)
hence b ∈ Y . It follows that B ⊆ Y . Suppose now that h ∈ Hk and a1 , a2 , . . . , ak ∈ Y . Since ai ∈ Y for each
i, we have f1 (ai ) = f2 (ai ) for each i, and hence
f1 (h(a1 , a2 , . . . , ak )) = gh (a1 , f1 (a1 ), a2 , f1 (a2 ), . . . , ak , f1 (ak ))

= gh (a1 , f2 (a1 ), a2 , f2 (a2 ), . . . , ak , f2 (ak ))
= f2 (h(a1 , a2 , . . . , ak ))
Thus, h(a1 , a2 , . . . , ak ) ∈ Y . It follows by induction that Y = G, i.e. f1 (a) = f2 (a) for all a ∈ G.
2.5 An Illustrative Example

We now embark on a careful formulation and proof of the statement: If f : A2 → A is associative, i.e.
f (a, f (b, c)) = f (f (a, b), c) for all a, b, c ∈ A, then any “grouping” of terms which preserves the ordering of
the elements inside the grouping gives the same value. In particular, if we are working in a group A, then
we can write things like acabba without parentheses because any allowable insertion of parentheses gives the
same value.
Throughout this section, let A be a set not containing the symbols [, ], or ?. Let SymA = A ∪ {[, ], ?}.
Definition 2.5.1. Define a binary function h : (Sym∗A )2 → Sym∗A by letting h(σ, τ ) be the sequence [σ ? τ ].
Let V alidExpA = G(Sym∗A , A, {h}) (viewed as a simple generating system).
For example, suppose that A = {a, b, c}. Typical elements of G(Sym∗A , A, {h}) are c, [b ? [a ? c]] and
[c ? [[c ? b] ? a]. The idea now is that if we have a particular function f : A2 → A, we can intrepret ? as
application of the function, and then this should give us a way to “make sense of”, that is evaluate, any
element of V alidExpA .
2.5.1 Proving Freeness

Proposition 2.5.2. The simple generating system (Sym∗A , A, {h}) is free.
Definition 2.5.3. Define K : Sym∗A → Z as follows. We first define w : SymA → Z by letting w(a) = 0 for
∗
Pw([) = −1, and letting w(]) =∗ 1. We then define K : SymA → Z by letting K(λ) = 0 and
all a ∈ A, letting
letting K(σ) = i<|σ| w(σ(i)) for all σ ∈ SymA \{λ}.
Remark 2.5.4. If σ, τ ∈ Sym∗A , then K(στ ) = K(σ) + K(τ ).
Proposition 2.5.5. If ϕ ∈ V alidExpA , then K(ϕ) = 0.
Proof. The proof is by induction on ϕ. In other words, we let X = {ϕ ∈ V alidExpA : K(ϕ) = 0}, and we
prove by induction that X = V alidExpA . Notice that for every a ∈ A, we have that K(a) = 0. Suppose
that ϕ, ψ ∈ V alidExpA are such that K(ϕ) = 0 = K(ψ). We then have that
K([ϕ ? ψ]) = K([) + K(ϕ) + +K(?) + K(ψ) + K(])

= −1 + 0 + 0 + 0 + 1
= 0.
The result follows by induction.
Proposition 2.5.6. If ϕ ∈ V alidExpA and σ ⊂ ϕ with σ 6= λ, then K(σ) ≤ −1.

2.5. AN ILLUSTRATIVE EXAMPLE 25
Proof. In other words, we let X = {ϕ ∈ V alidExpA : whenever σ ⊂ ϕ and σ 6= λ, we have that K(σ) ≤ −1},
and we prove by induction that X = V alidExpA .
For every a ∈ A, this is trivial because there is no σ 6= λ with σ ⊂ a.
Suppose that ϕ, ψ ∈ V alidExpA and the result holds for ϕ and ψ. We prove the result for [ϕ ? ψ].
Suppose that σ ⊂ [ϕ ? ψ] and σ 6= λ. If σ is [, then K(σ) = −1. If σ is [τ where τ 6= λ and τ ⊂ ϕ, then
K(σ) = −1 + K(τ )
≤ −1 − 1 (by induction)
≤ −1.
If σ is [ϕ or [ϕ?, then
K(σ) = −1 + K(ϕ)
= −1 + 0 (by Proposition 2.5.5)
= −1.
If σ is [ϕ ? τ , where τ 6= λ and τ ⊂ ϕ, then
K(σ) = −1 + K(ϕ) + K(τ )

= −1 + 0 + K(τ ) (by Proposition 2.5.5)
≤ −1 + 0 − 1 (by induction)
≤ −1.
Otherwise, σ is [ϕ ? ψ, and
K(σ) = −1 + K(ϕ) + K(ψ)

= −1 + 0 + 0 (by Proposition 2.5.5)
= −1.
Thus, the result holds for [ϕ ? ψ].
Corollary 2.5.7. If ϕ, ψ ∈ V alidExpA , then ϕ 6⊂ ψ.
Proof. This follows by combining Proposition 2.5.5 and Proposition 2.5.6, along with noting that λ ∈
/
V alidExpA (which follows by a trivial induction).
Theorem 2.5.8. The generating system (Sym∗A , A, {h}) is free.
Proof. First notice that ran(h (V alidExpA )2 ) ∩ A = ∅ because all elements of ran(h) begin with [.
Suppose that ϕ1 , ϕ2 , ψ1 , ψ2 ∈ V alidExpA and h(ϕ1 , ψ1 ) = h(ϕ2 , ψ2 ). We then have [ϕ1 ? ψ1 ] = [ϕ2 ? ψ2 ],
hence ϕ1 ? ψ1 = ϕ2 ? ψ2 . Since ϕ1 ⊂ ϕ2 and ϕ2 ⊂ ϕ1 are both impossible by Corollary 2.5.7, it follows that
ϕ1 = ϕ2 . Therefore, ?ψ1 = ?ψ2 , and so ψ1 = ψ2 . It follows that h (V alidExpA )2 is injective.
2.5.2 The Result

Since we have established freeness, we can define functions recursively. The first such function we define is
the “evaluation” function.
Definition 2.5.9. We define a function Evf : V alidExpA → A recursively by letting
• Evf (a) = a for all a ∈ A.

• Evf ([ϕ ? ψ]) = f (Evf (ϕ), Evf (ψ)) for all ϕ, ψ ∈ V alidExpA .
Formally, we use freeness to justify this definition as follows. Let ι : A → A be the identity map, and let
gh : (Sym∗A × A)2 → A be the function defined by letting gh ((ϕ, a), (ψ, b)) = f (a, b). By freeness, there is a
unique function Evf : V alidExpA → A such that
1. Evf (a) = ι(a) for all a ∈ A.
2. Evf (h(ϕ, ψ)) = gh ((ϕ, Evf (ϕ)), (ψ, Evf (ψ))) for all ϕ, ψ ∈ V alidExpA .
which, unravelling definitions, is exactly what we wrote above.
We now define the function which elimanates all mention of parentheses and ?. Thus, it produces the
sequence of elements of A occuring in the given sequence in order.
Definition 2.5.10. Define a function D : V alidExpA → A∗ recursively by letting
• D(a) = a for all a ∈ A.
• D([ϕ ? ψ]) = D(ϕ) ∗ D(ψ) for all ϕ, ψ ∈ V alidExpA .
With these definitions in hand, we can now precisely state our theorem.
Theorem 2.5.11. Suppose that f : A2 → A is associative, i.e. f (a, f (b, c)) = f (f (a, b), c) for all a, b, c ∈ A.
For all ϕ, ψ ∈ V alidExpA with D(ϕ) = D(ψ), we have Evf (ϕ) = Evf (ψ).
In order to prove our theorem, we’ll make use of the following function. Intuitively, it takes a sequence
such as cabc and “associates to the right” to produce [c ? [a ? [b ? c]]]. Thus, it provides a canonical way to
put together the elements of the sequence into something we can evaluate.
To make the recursive definition precise, consider the simple generating system (A∗ , A, {ha : a ∈ A})
where ha : A∗ → A∗ is defined by ha (σ) = aσ. As shown in Example 2.4.4, we know that (A∗ , A, {ha : a ∈ A})
is free and we have that G = A∗ \{λ}.
Definition 2.5.12. We define R : A∗ \{λ} → Sym∗A recursively by letting R(a) = a for all a ∈ A, and letting
R(aσ) = [a ? R(σ)] for all a ∈ A and all σ ∈ A∗ \{λ}.
In order to prove our theorem, we will show that Evf (ϕ) = Evf (R(D(ϕ))) for all ϕ ∈ V alidExpA ,
i.e. that we can take any ϕ ∈ V alidExpA , rip it apart so that we see the elements of A in order, and then
associate to the right, without affecting the result of the evaluation. We first need the following lemma.
Lemma 2.5.13. Evf ([R(σ) ? R(τ )]) = Evf (R(στ )) for all σ, τ ∈ A∗ \{λ}.
Proof. Fix τ ∈ A∗ \{λ}. We prove the result for this fixed τ by induction on A∗ \{λ}. That is, we let
X = {σ ∈ A∗ \{λ} : Evf ([R(σ) ? R(τ )]) = Evf (R(στ ))} and prove by induction on (A∗ , A, {ha : a ∈ A})
that X = A∗ \{λ}. Suppose first that a ∈ A. We then have
Evf ([R(a) ? R(τ )]) = Evf ([a ? R(τ )]) (by definition of R)
= Evf (R(aτ )) (by definition of R)
so a ∈ X. Suppose now that σ ∈ X and that a ∈ A. We show that aσ ∈ X. We have
Evf ([R(aσ) ? R(τ )]) = Evf ([[a ? R(σ)] ? R(τ )]) (by definition of R)
= f (Evf ([a ? R(σ)]), Evf (R(τ ))) (by definition of Evf )
= f (f (a, Evf (R(σ))), Evf (R(τ ))) (by definition of Evf using Evf (a) = a)
= f (a, f (Evf (R(σ)), Evf (R(τ )))) (since f is associative)
= f (a, Evf ([R(σ) ? R(τ )])) (by definition of Evf )
= f (a, Evf (R(στ ))) (since σ ∈ X)
= Evf ([a ? R(στ )]) (by definition of Evf using Evf (a) = a)
= Evf (R(aστ )) (by definition of R)
2.5. AN ILLUSTRATIVE EXAMPLE 27
so aσ ∈ X. The result follows by induction.
Lemma 2.5.14. Evf (ϕ) = Evf (R(D(ϕ))) for all ϕ ∈ V alidExpA .
Proof. By induction on V alidExpA . If a ∈ A, this is trivial because R(D(a)) = R(a) = a. Suppose that
ϕ, ψ ∈ V alidExpA and the result holds for ϕ and ψ.
Evf ([ϕ ? ψ]) = f (Evf (ϕ), Evf (ψ)) (by definition of Evf )
= f (Evf (R(D(ϕ))), Evf (R(D(ψ)))) (by induction)
= Evf ([R(D(ϕ)) ? R(D(ψ))]) (by definition of Evf )
= Evf (R(D(ϕ) ∗ D(ψ))) (by Lemma 2.5.13)
= Evf (R(D([ϕ ? ψ]))) (by definition of D)
Proof of Theorem 2.5.11. Suppose that ϕ, ψ ∈ V alidExpA are such that D(ϕ) = D(ψ). We then have that
Evf (σ) = Evf (R(D(σ))) (by Lemma 2.5.14)

= Evf (R(D(τ ))) (since D(σ) = D(τ ))
= Evf (τ ) (by Lemma 2.5.14)
2.5.3 An Alternate Syntax - Polish Notation

Throughout this section, let A be a set not containing the symbol ?. Let SymA = A ∪ {?}.
Definition 2.5.15. Define a binary function h : (Sym∗A )2 → Sym∗A by letting h(σ, τ ) be the sequence ?στ .
Let P olishExpA = G(Sym∗A , A, {h}) (viewed as a simple generating system).
Proposition 2.5.16. The simple generating system (Sym∗A , A, {h}) is free.
Definition 2.5.17. Define K : Sym∗A → Z as follows. We first define w : SymA → Z by letting w(a) = 1
for all aP∈ A and letting w(?) = −1. We then define K : Sym∗A → Z by letting K(λ) = 0 and letting
K(σ) = i<|σ| w(σ(i)) for all σ ∈ Sym∗A \{λ}.
Remark 2.5.18. If σ, τ ∈ Sym∗A , then K(στ ) = K(σ) + K(τ ).
Proposition 2.5.19. If ϕ ∈ P olishExpA , then K(ϕ) = 1.
Proof. The proof is by induction on ϕ. Notice that for every a ∈ A, we have that K(a) = 1. Suppose that
ϕ, ψ ∈ P olishExpA are such that K(ϕ) = 1 = K(ψ). We then have that
K(?ϕψ) = K(?) + K(ϕ) + K(ψ)

= K(ϕ)
= 1.
Proposition 2.5.20. If ϕ ∈ P olishExpA and σ ⊂ ϕ, then K(σ) ≤ 0.

Proof. The proof is by induction on ϕ. For every a ∈ A, this is trivial because the only σ ⊂ A is σ = λ, and
we have K(λ) = 0.
Suppose that ϕ, ψ ∈ P olishExpA and the result holds for ϕ and ψ. We prove the result for ?ϕψ. Suppose
that σ ⊂ ?ϕψ. If σ = λ, then K(σ) = 0. If σ is ?τ for some τ ⊂ ϕ, then
K(σ) = K(?) + K(τ )

≤ −1 + 0 (by induction)
≤ −1.
Otherwise, σ is ?ϕτ for some τ ⊂ ψ, in which case
K(σ) = K(?) + K(ϕ) + K(τ )

≤ −1 + 0 + 0 (by induction)
≤ −1.
Thus, the result holds for ?ϕψ.

Corollary 2.5.21. If ϕ, ψ ∈ P olishExpA , then ϕ 6⊂ ψ.
Proof. This follows by combining Proposition 2.5.19 and Proposition 2.5.20.

Theorem 2.5.22. The generating system (Sym∗A , A, H) is free.
Proof. First notice that ran(h (P olishExpA )2 ) ∩ A = ∅ because all elements of ran(h) begin with ?.
Suppose that ϕ1 , ϕ2 , ψ1 , ψ2 ∈ P olishExpA and that h(ϕ1 , ψ1 ) = h(ϕ2 , ψ2 ). We then have ?ϕ1 ψ1 = ?ϕ2 ψ2 ,
hence ϕ1 ψ1 = ϕ2 ψ2 . Since ϕ1 ⊂ ϕ2 and ϕ2 ⊂ ϕ1 are both impossible by Corollary 2.5.21, it follows that
ϕ1 = ϕ2 . Therefore, ψ1 = ψ2 . It follows that h (P olishExpA )2 is injective.
Chapter 3
Propositional Logic
3.1 The Syntax of Propositional Logic

3.1.1 Standard Syntax
Definition 3.1.1. Let P be a nonempty set not containing the symbols (, ), ¬, ∧, ∨, and →. Let SymP =
P ∪ {(, ), ¬, ∧, ∨, →}. Define a unary function h¬ and binary functions h∧ , h∨ , and h→ on Sym∗P as follows.
h¬ (σ) = (¬σ)
h∧ (σ, τ ) = (σ ∧ τ )
h∨ (σ, τ ) = (σ ∨ τ )
h→ (σ, τ ) = (σ → τ )
Definition 3.1.2. Fix P . Let F ormP = G(Sym∗P , P, H) where H = {h¬ , h∧ , h∨ , h→ }.
Definition 3.1.3. Define K : Sym∗P → Z as follows. We first define w : SymP → Z by letting w(A) = 0
for all A ∈ P , letting w(3) = 0 for all 3 ∈ {¬, ∧, ∨, →}, letting
Pw(() = −1, and letting w()) = 1. We then
define K : Sym∗P → Z by letting K(λ) = 0 and letting K(σ) = i<|σ| w(σ(i)) for all σ ∈ Sym∗P \{λ}.
Remark 3.1.4. If σ, τ ∈ Sym∗P , then K(στ ) = K(σ) + K(τ ).
Proposition 3.1.5. If ϕ ∈ F ormP , then K(ϕ) = 0.
Proof. A simple induction as above.
Proposition 3.1.6. If ϕ ∈ F ormP and σ ⊂ ϕ with σ 6= λ, then K(σ) ≤ −1.
Proof. A simple induction as above.
Corollary 3.1.7. If ϕ, ψ ∈ F ormP , then ϕ 6⊂ ψ.
Proof. This follows by combining Proposition 3.1.5 and Proposition 3.1.6, along with noting that λ ∈
/ F ormP
(which follows by a simple induction).
Theorem 3.1.8. The generating system (Sym∗P , P, H) is free.
29
30 CHAPTER 3. PROPOSITIONAL LOGIC
Proof. First notice that ran(h¬ F ormP ) ∩ P = ∅ because all elements of ran(h¬ ) begin with (. Similarly,
for any 3 ∈ {∧, ∨, →}, we have ran(h3 F orm2P ) ∩ P = ∅ since all elements of ran(h3 ) begin with (.
Suppose that ϕ, ψ ∈ F ormP and h¬ (ϕ) = h¬ (ψ). We then have (¬ϕ) = (¬ψ), hence ϕ = ψ. Therefore,
h¬ F ormP is injective. Fix 3 ∈ {∧, ∨, →}. Suppose that ϕ1 , ϕ2 , ψ1 , ψ2 ∈ F ormP and that h3 (ϕ1 , ψ1 ) =
h3 (ϕ2 , ψ2 ). We then have (ϕ1 3ψ1 ) = (ϕ2 3ψ2 ), hence ϕ1 3ψ1 = ϕ2 3ψ2 . Since ϕ1 ⊂ ϕ2 and ϕ2 ⊂ ϕ1 are
both impossible by Corollary 3.1.7, it follows that ϕ1 = ϕ2 . Therefore, 3ψ1 = 3ψ2 , and so ψ1 = ψ2 . It
follows that h3 F orm2P is injective.
Let 3 ∈ {∧, ∨, →}. Suppose that ϕ, ψ1 , ψ2 ∈ F ormP and h¬ (ϕ) = h3 (ψ1 , ψ2 ). We then have (¬ϕ) =
(ψ1 3ψ2 ), hence ¬ϕ = ψ1 3ψ2 , contradicting the fact that no element of F ormP begins with ¬ (by a simple
induction). Therefore, ran(h¬ F ormP ) ∩ ran(h3 F orm2P ) = ∅.
Suppose now that 31 , 32 ∈ {∧, ∨, →} with 31 6= 32 . Suppose that ϕ1 , ϕ2 , ψ1 , ψ2 ∈ F ormP and
h31 (ϕ1 , ψ1 ) = h32 (ϕ2 , ψ2 ). We then have (ϕ1 31 ψ1 ) = (ϕ2 32 ψ2 ), hence ϕ1 31 ψ1 = ϕ2 32 ψ2 . Since ϕ1 ⊂ ϕ2
and ϕ2 ⊂ ϕ1 are both impossible by Corollary 3.1.7, it follows that ϕ1 = ϕ2 . Therefore, 31 = 32 , a
contradiction. It follows that ran(h31 F orm2P ) ∩ ran(h32 F orm2P ) = ∅.
3.1.2 Polish Notation

Definition 3.1.9. Let P be a set not containing the symbols ¬, ∧, ∨, and →. Let SymP = P ∪ {¬, ∧, ∨, →}.
Define a unary function h¬ and binary functions h∧ , h∨ , and h→ on Sym∗P as follows.
h¬ (σ) = ¬σ
h∧ (σ, τ ) = ∧στ
h∨ (σ, τ ) = ∨στ
h→ (σ, τ ) = → στ
Definition 3.1.10. Fix P . Let F ormP = G(Sym∗P , P, H) where H = {h¬ , h∧ , h∨ , h→ }.

Definition 3.1.11. Define K : Sym∗P → Z as follows. We first define w : SymP → Z by letting w(A) = 1 for
P = −1 for all 3 ∈ {¬, ∧, ∨,∗→}. We then define K : SymP → Z
∗
all A ∈ P , letting w(¬) = 0, and letting w(3)
by letting K(λ) = 0 and letting K(σ) = i<|σ| w(σ(i)) for all σ ∈ SymP \{λ}.
Remark 3.1.12. If σ, τ ∈ Sym∗P , then K(στ ) = K(σ) + K(τ ).

Proposition 3.1.13. If ϕ ∈ F ormP , then K(ϕ) = 1.
Proof. The proof is by induction on ϕ. Notice that for every A ∈ P , we have that K(A) = 1. Suppose that
ϕ ∈ F ormP is such that K(ϕ) = 1. We then have that
K(¬ϕ) = 0 + K(ϕ)
= K(ϕ)
= 1.
Suppose now that ϕ, ψ ∈ F ormP are such that K(ϕ) = 1 = K(ψ), and 3 ∈ {∧, ∨, →}. We then have that
K(3ϕψ) = −1 + K(ϕ) + K(ψ)

= −1 + 1 + 1
= 1.

Proposition 3.1.14. If ϕ ∈ F ormP and σ ⊂ ϕ, then K(σ) ≤ 0.
3.1. THE SYNTAX OF PROPOSITIONAL LOGIC 31
Proof. The proof is by induction on ϕ. For every A ∈ P , this is trivial because the only σ ⊂ A is σ = λ, and
we have K(λ) = 0.
Suppose that ϕ ∈ F ormP and the result holds for ϕ. We prove the result for ¬ϕ. Suppose that σ ⊂ ¬ϕ.
If σ = λ, then K(σ) = 0. Otherwise, σ is ¬τ for some τ ⊂ ϕ, in which case
K(σ) = 0 + K(τ )
≤0+0 (by induction)
≤ 0.
Thus, the result holds for ¬ϕ.

Suppose that ϕ, ψ ∈ F ormP and the result holds for ϕ and ψ. Let 3 ∈ {∧, ∨, →}. We prove the result
for 3ϕψ. Suppose that σ ⊂ 3ϕψ. If σ = λ, then K(σ) = 0. If σ is 3τ for some τ ⊂ ϕ, then
K(σ) = −1 + K(τ )
≤ −1 + 0 (by induction)
≤ −1.
Otherwise, σ is 3ϕτ for some τ ⊂ ψ, in which case
K(σ) = −1 + K(ϕ) + K(τ )

≤ −1 + 0 + 0 (by induction)
≤ −1.
Thus, the result holds for 3ϕψ.
Corollary 3.1.15. If ϕ, ψ ∈ F ormP , then ϕ 6⊂ ψ.
Theorem 3.1.16. The generating system (Sym∗P , P, H) is free.
Proof. First notice that ran(h¬ F ormP ) ∩ P = ∅ because all elements of ran(h¬ ) begin with ¬. Similarly,
for any 3 ∈ {∧, ∨, →}, we have ran(h3 F orm2P ) ∩ P = ∅ since all elements of ran(h3 ) begin with 3.
Suppose that ϕ, ψ ∈ F ormP and h¬ (ϕ) = h¬ (ψ). We then have ¬ϕ = ¬ψ, hence ϕ = ψ. Therefore,
h¬ F ormP is injective. Fix 3 ∈ {∧, ∨, →}. Suppose that ϕ1 , ϕ2 , ψ1 , ψ2 ∈ F ormP and that h3 (ϕ1 , ψ1 ) =
h3 (ϕ2 , ψ2 ). We then have 3ϕ1 ψ1 = 3ϕ2 ψ2 , hence ϕ1 ψ1 = ϕ2 ψ2 . Since ϕ1 ⊂ ϕ2 and ϕ2 ⊂ ϕ1 are both
impossible by Corollary 3.1.15, it follows that ϕ1 = ϕ2 . Therefore, ψ1 = ψ2 . It follows that h3 F orm2P is
injective.
For any 3 ∈ {∧, ∨, →}, we have ran(h¬ F ormP )∩ran(h3 F orm2P ) = ∅ because all elements of ran(h¬ )
begin with ¬ and all elements of ran(h3 ) begin with 3. Similarly, if 31 , 32 ∈ {∧, ∨, →} with 31 6= 32 , we
have ran(h31 F orm2P ) ∩ ran(h32 F orm2P ) = ∅ because all elements of ran(h31 ) begin with 31 and all
elements of ran(h32 ) begin with 32 .
3.1.3 Official Syntax and Our Abuses of It

Since we should probably fix an official syntax, let’s agree to use Polish notation because it’s simpler in
many aspects and it will be natural to generalize when we talk about the possibility of other connectives
and when we discuss first-order logic. However, as with many official definitions in mathematics, we’ll ignore
and abuse this convention constantly in the interest of readability. For example, we’ll often write things in
standard syntax or in more abbreviated forms. For example, we’ll write A ∧ B instead of ∧AB (or (A ∧ B)
in the original syntax) . We’ll also write something like
A1 ∧ A2 ∧ · · · ∧ An−1 ∧ An
or
n
^
Ai
i=1
instead of (A1 ∧ (A2 ∧ (· · · (An−1 ∧ An ) · · · ))) in standard syntax or ∧A1 ∧ A2 · · · ∧ An−1 An in Polish notation
(which can be precisely defined in a similar manner as R in Section 2.5). In general, when we string together
multiple applications of an operation (such as ∧) occur in order, we always associate to the right.
When it comes to mixing symbols, let’s agree to the following conventions about “binding” in a similar
fashion to how we think of · as more binding than + (so that 3·5+2 is read as (3·5)+2). We think of ¬ as the
most binding, so we read ¬A ∧ B as ((¬A) ∧ B). After that, we consider ∧ and ∨ as the next most binding,
and → has the least binding. We’ll insert parentheses when we wish to override this binding. For example,
A ∧ ¬B → C ∨ D is really ((A ∧ (¬B)) → (C ∨ D)) while A ∧ (¬B → C ∨ D) is really (A ∧ ((¬B) → (C ∨ D))).
3.1.4 Recursive Definitions

Since we’ve shown that our generating system is free, we can define functions recursively. It is possible
to avoid using recursion on F ormP to define some of functions. In such cases, you may wonder why we
bother. Since our only powerful way to prove things about the set F ormP is by induction, and definitions
of functions by recursion are well-suited to induction, it’s simply the easiest way ahead.
Definition 3.1.17. If X is a set, we denote by P(X) the set of all subsets of X. Thus P(X) = {Z : Z ⊆ X}.
We call P(X) the power set of X.
Definition 3.1.18. We define a function OccurP rop : F ormP → P(P ) recursively as follows.
• OccurP rop(A) = {A} for all A ∈ P .
• OccurP rop(¬ϕ) = OccurP rop(ϕ).
• OccurP rop(3ϕψ) = OccurP rop(ϕ) ∪ OccurP rop(ψ) for each 3 ∈ {∧, ∨, →}.
If you want to be precise in the previous definition, we’re defining functions ι : P → P(P ), gh¬ : Sym∗P ×
P(P ) → P(P ) and gh3 : (Sym∗P × P(P ))2 → P(P ) for each 3 ∈ {∧, ∨, →} as follows.
• ι(A) = {A} for all A ∈ P .
• gh¬ (σ, Z) = Z.
• gh3 (σ1 , Z1 , σ2 , Z2 ) = Z1 ∪ Z2 for each 3 ∈ {∧, ∨, →}.
and we’re using our result on freeness to assure that there is a unique function OccurP rop : F ormP → P(P )
which satisfy the associated requirements. Of course, this method is more precise, but it’s hardly more
intuitive to use. It’s a good exercise to make sure that you can translate a few more informal recursive
definitions in this way, but once you understand how it works you can safely keep the formalism in the back
of your mind.
Here’s a somewhat trivial example of using induction to prove a result based on a recursive definition.
Proposition 3.1.19. Suppose that Q ⊆ P . We then have that F ormQ ⊆ F ormP .
Proof. A trivial induction on ϕ ∈ F ormQ .
3.1. THE SYNTAX OF PROPOSITIONAL LOGIC 33
Proposition 3.1.20. Fix P . For any ϕ ∈ F ormP , we have ϕ ∈ F ormOccurP rop(ϕ) .
Proof. The proof is by induction on ϕ ∈ F ormP . Suppose first that A ∈ P . Since OccurP rop(A) = {A} and
A ∈ F ormA , we have A ∈ F ormOccurP rop(A) .
Suppose that ϕ ∈ F ormP and that the result holds for ϕ, i.e. we have ϕ ∈ F ormOccurP rop(ϕ) . Since
OccurP rop(¬ϕ) = OccurP rop(ϕ), it follows that ϕ ∈ F ormOccurP rop(¬ϕ) . Hence, ¬ϕ ∈ F ormOccurP rop(¬ϕ) .
Suppose that ϕ, ψ ∈ F ormP , that 3 ∈ {∧, ∨, →}, and that the result holds for ϕ and ψ, i.e. we have
ϕ ∈ F ormOccurP rop(ϕ) and ψ ∈ F ormOccurP rop(ψ) . Since
OccurP rop(ϕ) ⊆ OccurP rop(3ϕψ) and OccurP rop(ψ) ⊆ OccurP rop(3ϕψ)
it follows from Proposition 3.1.19 that ϕ, ψ ∈ F ormOccurP rop(3ϕψ) . Therefore, 3ϕψ ∈ F ormOccurP rop(3ϕψ) .
On to some more important recursive definitions.
Definition 3.1.21. We define a function Depth : F ormP → N recursively as follows.
• Depth(A) = 0 for all A ∈ P .
• Depth(¬ϕ) = Depth(ϕ) + 1.
• Depth(3ϕψ) = max{Depth(ϕ), Depth(ψ)} + 1 for each 3 ∈ {∧, ∨, →}.
Example 3.1.22. Depth(∨ ∧ AB ∨ CD) = 2.
Definition 3.1.23. We define a function Subf orm : F ormP → P(F ormP ) recursively as follows.
• Subf orm(A) = {A} for all A ∈ P .
• Subf orm(¬ϕ) = {¬ϕ} ∪ Subf orm(ϕ).
• Subf orm(3ϕψ) = {3ϕψ} ∪ Subf orm(ϕ) ∪ Subf orm(ψ) for each 3 ∈ {∧, ∨, →}.
Example 3.1.24. Subf orm(∧¬AB) = {∧¬AB, ¬A, A, B}.
Definition 3.1.25. Let θ, γ ∈ F ormP . We define a function Substθγ : F ormP → F ormP recursively as
follows.
(
θ θ if γ = A
• Substγ (A) =
A otherwise
(
θ if γ = ¬ϕ
• Substθγ (¬ϕ) =
¬Substθγ (ϕ) otherwise
(
θ if γ = 3ϕψ
• Substθγ (3ϕψ) =
3Substθγ (ϕ)Substθγ (ψ) otherwise
for each 3 ∈ {∧, ∨, →}.
Example 3.1.26. Subst∧AB

¬C (→ ∨¬CA¬C) = → ∨ ∧ ABA ∧ AB.
3.2 Truth Assignments and Semantic Implication

Definition 3.2.1. A function v : P → {0, 1} is called a truth assignment on P .
Definition 3.2.2. Let v : P → {0, 1} be a truth assignment. We denote by v the unique function v : F ormP →
{0, 1} such that
• v(A) = v(A) for all A ∈ P .
(
1 if v(ϕ) = 0
• v(¬ϕ) =
0 if v(ϕ) = 1

0


if v(ϕ) = 0 and v(ψ) = 0
0 if v(ϕ) = 0 and v(ψ) = 1
• v(∧ϕψ) =
0
 if v(ϕ) = 1 and v(ψ) = 0

1 if v(ϕ) = 1 and v(ψ) = 1



 0 if v(ϕ) = 0 and v(ψ) = 0

1 if v(ϕ) = 0 and v(ψ) = 1
• v(∨ϕψ) =


 1 if v(ϕ) = 1 and v(ψ) = 0
1 if v(ϕ) = 1 and v(ψ) = 1



 1 if v(ϕ) = 0 and v(ψ) = 0

1 if v(ϕ) = 0 and v(ψ) = 1
• v(→ ϕψ) =
0
 if v(ϕ) = 1 and v(ψ) = 0

1 if v(ϕ) = 1 and v(ψ) = 1

Before moving on, we should a couple of things about what happens when we shrink/enlarge the set P .
Intuitively, if ϕ ∈ F ormQ and Q ⊆ P , then we can extend the truth assigment from Q to P arbitrarily
without affecting the value of v(ϕ). Here is the precise statement.
Proposition 3.2.3. Suppose that Q ⊆ P and that v : P → {0, 1} is a truth assignment on P . We then have
that v(ϕ) = (v Q)(ϕ) for all ϕ ∈ F ormQ .
Proof. A trivial induction on ϕ ∈ F ormQ .
Proposition 3.2.4. Suppose ϕ ∈ F ormP . Whenever v1 and v2 are truth assignments on P such that
v1 (A) = v2 (A) for all A ∈ OccurP rop(ϕ), we have v 1 (ϕ) = v 2 (ϕ).
Proof. Let Q = OccurP rop(ϕ). We then have that ϕ ∈ F ormQ by Proposition 3.1.20. Since v1 Q = v2 Q,
we have
v 1 (ϕ) = (v1 Q)(ϕ) = (v2 Q)(ϕ) = v 2 (ϕ)
With a method of assigning true/false values to formulas in hand (once we’ve assigned them to P ), we’re
now in position to use our semantic definitions to given a precise meaning to “The set of formulas Γ implies
the formula ϕ”.
Definition 3.2.5. Let P be given Let Γ ⊆ F ormP and let ϕ ∈ F ormP . We write Γ P ϕ, or simply Γ ϕ
if P is clear, to mean that whenever v is a truth assignment on P such that v(γ) = 1 for all γ ∈ Γ, we have
v(ϕ) = 1. We pronounce Γ ϕ as “Γ semantically implies ϕ”.
3.2. TRUTH ASSIGNMENTS AND SEMANTIC IMPLICATION 35
We also have a semantic way to say that a set of formulas is not contradictory.
Definition 3.2.6. Γ is satisfiable if there exists a truth assignment v : P → {0, 1} such that v(γ) = 1 for all
γ ∈ Γ. Otherwise, we say that Γ is unsatisfiable.
Example 3.2.7. Let P = {A, B, C}. We have {A ∨ B, ¬(A ∧ (¬C))} B ∨ C.
Proof. Let v : P → {0, 1} be a truth assignment such that v(A ∨ B) = 1 and v(¬(A ∧ (¬C))) = 1. We need to
show that v(B ∨ C) = 1. Suppose not. We would then have that v(B) = 0 and v(C) = 0. Since v(A ∨ B) = 1,
this implies that v(A) = 1. Therefore, v(A ∧ (¬C)) = 1, so v(¬(A ∧ (¬C))) = 0, a contradiction.
Example 3.2.8. Let P be given. For any ϕ, ψ ∈ F ormP , we have {ϕ → ψ, ϕ} ψ
Proof. Let v : P → {0, 1} be a truth assignment and suppose that v(ϕ → ψ) = 1 and v(ϕ) = 1. If v(ψ) = 0,
it would follows that v(ϕ → ψ) = 0, a contradiction. Thus, v(ψ) = 1.
Notation 3.2.9.
1. If Γ = ∅, we write ϕ instead of ∅ ϕ.
2. If Γ = {γ}, we write γ ϕ instead of {γ} ϕ.
Definition 3.2.10.
1. Let ϕ ∈ F ormP . We say that ϕ is a tautology if ϕ.
2. If ϕ ψ and ψ ϕ, we say that ϕ and ψ are semantically equivalent.
Remark 3.2.11. Notice that ϕ and ψ are semantically equivalent if and only if for all truth assigments
v : P → {0, 1}, we have v(ϕ) = v(ψ).
Example 3.2.12. ϕ ∨ ¬ϕ is a tautology for any ϕ ∈ F ormP .
Proof. Fix ϕ ∈ F ormP . Let v : P → {0, 1} be a truth assignment. If v(ϕ) = 1, then v(ϕ ∨ ¬ϕ) = 1.
Otherwise, we have v(ϕ) = 0, in which case v(¬ϕ) = 1, and hence v(ϕ ∨ ¬ϕ) = 1. Therefore, v(ϕ ∨ ¬ϕ) = 1
for all truth assignments v : P → {0, 1}, hence ϕ ∨ ¬ϕ is a tautology.
Example 3.2.13. ϕ is semantically equivalent to ¬¬ϕ for any ϕ ∈ F ormP .
Proof. Fix ϕ ∈ F ormP . We need to show that for any truth assignment v : P → {0, 1}, we have v(ϕ) = 1 if
and only if v(¬¬ϕ) = 1. We have
v(ϕ) = 1 ⇐⇒ v(¬ϕ) = 0
⇐⇒ v(¬¬ϕ) = 1
Example 3.2.14. ϕ → ψ and ¬ϕ ∨ ψ are semantically equivalent for any ϕ, ψ ∈ F ormP .

Proof. Fix ϕ, ψ ∈ F ormP . We need to show that for any truth assignment v : P → {0, 1}, we have v(ϕ →
ψ) = 1 if and only if v(¬ϕ ∨ ψ) = 1. Let v : P → {0, 1} be a truth assignment. We have
v(ϕ → ψ) = 1 ⇐⇒ v(ϕ) = 0 or v(ψ) = 1

⇐⇒ v(¬ϕ) = 1 or v(ψ) = 1
⇐⇒ v(¬ϕ ∨ ψ) = 1
Definition 3.2.15. Define OccurP rop : P(F ormP ) → P(P ) by letting
OccurP rop(Γ) = {A ∈ P : A ∈ OccurP rop(γ) for some γ ∈ Γ}.
Proposition 3.2.16. Suppose that Q ⊆ P , that Γ ⊆ F ormQ , and that ϕ ∈ F ormQ . We then have that
Γ P ϕ if and only if Γ Q ϕ.
Proof. First notice that Γ ⊆ F ormP and ϕ ∈ F ormP by Proposition 3.1.19 and Proposition 3.1.20.
Suppose first that Γ Q ϕ. Let v : P → {0, 1} be a truth assignment such that v(γ) = 1 for all γ ∈ Γ.
We then have that (v Q)(γ) = 1 for all γ ∈ Γ, hence (v Q)(ϕ) = 1 because Γ Q ϕ. Therefore, v(ϕ) = 1.
It follows that Γ P ϕ.
Suppose then that Γ P ϕ. Let v : Q → {0, 1} be a truth assignment such that v(γ) = 1 for all γ ∈ Γ.
Define a truth assignment w : P → {0, 1} by letting w(A) = v(A) for all A ∈ Q and letting w(A) = 0 for all
A ∈ P \Q. Since w Q = v, we have w(γ) = v(γ) = 1 for all γ ∈ Γ. Since Γ P ϕ, it follows that w(ϕ) = 1,
hence v(ϕ) = 1. Therefore, Γ Q ϕ.
Suppose that Γ is finite, and we want to determine whether or not Γ ϕ. By the previous proposition,
instead of examining all truth assignments v : P → {0, 1} on P , we need only consider truth assignments
v : OccurP rop(Γ ∪ {ϕ}) → {0, 1}. Now OccurP rop(Γ ∪ {ϕ}) is a finite set, so there are only finitely many
possibilities. Thus, one way of determining whether Γ ϕ is simply to check all of them. If |OccurP rop(Γ ∪
{ϕ})| = n, then there are 2n different truth assignments. We can systematically arrange them in a table
like below, where we ensure we put the the elements of OccurP rop(Γ ∪ {ϕ}) in the first columns, and put
all elements of Γ ∪ {ϕ} in later columns. We also ensure that if ψ is in a column, then all subformulas of ψ
appear in an earlier column. This allows us to fill in the table one column at a time. This simple-minded
exhaustive technique is called the method of truth tables.
Example 3.2.17. Show that {(A ∨ B) ∧ C, A → (¬C)} (¬C) → B.
Proof.
A B C A∨B (A ∨ B) ∧ C ¬C A → (¬C) (¬C) → B
0 0 0 0 0 1 1 0
0 0 1 0 0 0 1 1
0 1 0 1 0 1 1 1
0 1 1 1 1 0 1 1
1 0 0 1 0 1 1 0
1 0 1 1 1 0 0 1
1 1 0 1 0 1 1 1
1 1 1 1 1 0 0 1
Notice that every row in which both of the (A ∨ B) ∧ C column and the A → (¬C) column have a 1, namely
just the row beginning with 011, we have that the entry under the (¬C) → B column is a 1. Therefore,
{(A ∨ B) ∧ C, A → (¬C)} (¬C) → B.
Example 3.2.18. Show that ¬(A ∧ B) is semantically equivalent to ¬A ∨ ¬B.
Proof.
A B A∧B ¬(A ∧ B) ¬A ¬B ¬A ∨ ¬B
0 0 0 1 1 1 1
0 1 0 1 1 0 1
1 0 0 1 0 1 1
1 1 1 0 0 0 0
3.3. BOOLEAN FUNCTIONS AND CONNECTIVES 37
Notice that the rows in which the ¬(A ∧ B) column has a 1 are exactly the same as the rows in which the
¬A ∨ ¬B column has a 1. Therefore, ¬(A ∧ B) is semantically equivalent to ¬A ∨ ¬B.
3.3 Boolean Functions and Connectives

It’s natural to wonder if our choice of connectives is the “right” one. For example, why didn’t we introduce a
new connective ↔, allowing ourselves to form the formulas ϕ ↔ ψ (or ↔ ϕψ in Polish notation) and extend
our definition of v so that 
1 if v(ϕ) = 0 and v(ψ) = 0


0 if v(ϕ) = 0 and v(ψ) = 1
v(↔ ϕψ) =


 0 if v(ϕ) = 1 and v(ψ) = 0
1 if v(ϕ) = 1 and v(ψ) = 1

The idea is that there’s no real need to introduce this connective because for any ϕ, ψ ∈ F ormP we would
have that ϕ ↔ ψ is semantically equivalent to (ϕ → ψ) ∧ (ψ → ϕ).
Perhaps we could be more exotic and introduce a new connective with takes three formulas allowing
us to form the formulas ϕψθ (here’s an instance when Polish notation becomes important), and extend
our definition of v so that
(
1 if at least two of v(ϕ) = 1, v(ψ) = 1, v(θ) = 1
v(ϕψθ) =
0 otherwise
It’s not hard (and a good exercise) to show that for any ϕ, ψ, θ ∈ F ormP , there exists α ∈ F ormP such that
ϕψθ is semantically equivalent to α. We want a general theorem which says that no matter how exotic a
connective one invents, it’s always possible to find an element of F ormP which is semantically equivalent,
and thus our choice of connectives is sufficient to express everything we’d ever want.
Rather than deal with arbitrary connectives, the real issue here is whether we can express any possible
function taking k true/false values to true/false values.
Definition 3.3.1. Let k ∈ N+ . A function f : {0, 1}k → {0, 1} is called a boolean function of arity k.
Definition 3.3.2. Suppose that P = {A0 , A1 , . . . , Ak−1 }. Given ϕ ∈ F ormP , we define a boolean function
Bϕ : {0, 1}k → {0, 1} as follows. Given σ ∈ {0, 1}k , define a truth assignment v : P → {0, 1} by letting
v(Ai ) = σ(i) for all i, and set Bϕ (σ) = v(ϕ).
Theorem 3.3.3. Fix k ∈ N+ , and let P = {A0 , A1 , . . . , Ak−1 }. For any boolean function f : {0, 1}k → {0, 1}
of arity k, there exists ϕ ∈ F ormP such that f = Bϕ .
In fact, we’ll prove a stronger theorem below which says that we may assume that our formula ϕ is in a
particularly simple form.
Let’s look at an example before we do the proof. Suppose that f : {0, 1}3 → {0, 1} is given by
0 0 0 1
0 0 1 0
0 1 0 1
0 1 1 0
1 0 0 0
1 0 1 0
1 1 0 1
1 1 1 1
Suppose we wanted to come up with a formula ϕ such that f = Bϕ . One option is to use a lot of thought
to come up with an elegant solution. Another is simply to think as follows. Since f (000) = 1, perhaps we
should put
¬A0 ∧ ¬A1 ∧ ¬A2
into the formula somewhere. Similarly, since f (010) = 1, perhaps we should put
¬A0 ∧ A1 ∧ ¬A2
into the formula somewhere. If we do the same to the other lines which have value 1, we can put all of these
pieces together in a manner which makes them all play nice by connecting them with ∨. Thus, our formula
is
(¬A0 ∧ ¬A1 ∧ ¬A2 ) ∨ (¬A0 ∧ A1 ∧ ¬A2 ) ∨ (A0 ∧ A1 ∧ ¬A2 ) ∨ (A0 ∧ A1 ∧ A2 )
We now give the general proof.
Definition 3.3.4. A literal is a element of P ∪ {¬A : A ∈ P }. We denote the set of literals by LitP .
Definition 3.3.5.
• Let ConjP = G(Sym∗P , LitP , {h∧ }). We call the elements of ConjP conjunctive formulas.
• Let DisjP = G(Sym∗P , LitP , {h∨ }). We call the elements of DisjP disjunctive formulas.
Definition 3.3.6.
• Let DN FP = G(Sym∗P , ConjP , {h∨ }). We say that an element of DN FP is in disjunctive normal
form.
• Let CN FP = G(Sym∗P , DisjP , {h∧ }). We say that an element of CN FP is in conjunctive normal
form.
Theorem 3.3.7. Fix k ∈ N+ , and let P = {A0 , A1 , . . . , Ak−1 }. For any boolean function f : {0, 1}k → {0, 1}
of arity k, there exists ϕ ∈ DN FP such that f = Bϕ .
Proof. Let T = {σ ∈ {0, 1}k : f (σ) = 1}. If T = ∅, we may let ϕ be A0 ∧ (¬A0 ). Suppose then that T 6= ∅.
For each σ ∈ T , let
k−1
^
ψσ = θi
i=0
where (
Ai if σ(i) = 1
θi =
¬Ai if σ(i) = 0
For each σ ∈ T , notice that ψσ ∈ ConjP because θi ∈ LitP for all i. Finally, let
_
ϕ= ψσ
σ∈T
and notice that ϕ ∈ DN FP .
Since DN FP formulas suffice, we don’t even need → if all we want to do is have the ability to express
all boolean functions. In fact we can also get rid of one of ∧ or ∨ as well (think about why).
3.4. SYNTACTIC IMPLICATION 39
3.4 Syntactic Implication

3.4.1 Motivation
We now seek to define a different notion of implication which is based on syntactic manipulations instead
of a detour through truth assignments and other semantic notions. We will do this by setting up a “proof
system” which gives rules on how to transform certain implications to other implications. There are many
many ways to do this. Some approaches pride themselves on being minimalistic by using a minimal number
of axioms and rules, often at the expense of making the system unnatural to actually work with. We’ll take
a different approach and set down our rules and axioms based on the types of steps in a proof that are used
naturally throughout mathematics.
We begin with a somewhat informal description of what we plan to do. The objects that will manipulate
are pairs, where the first component is a set of formulas and the second is a formula. Given Γ ⊆ F ormP
and ϕ ∈ F ormP , we write Γ ` ϕ to intuitively mean that there is a proof of ϕ from the assumptions Γ. We
begin with the most basic proofs. If ϕ ∈ Γ, i.e. if ϕ is one of your assumptions, then you’re permitted to
assert that Γ ` ϕ.
Basic Proofs: Γ ` ϕ if ϕ ∈ Γ.
Rules for ∧: We have two rules for ∧-elimination and one for ∧-introduction.
Γ`ϕ∧ψ Γ`ϕ∧ψ Γ`ϕ Γ`ψ
(∧EL) (∧ER) (∧I)
Γ`ϕ Γ`ψ Γ`ϕ∧ψ
Rules for ∨: We have two rules for introducing ∨.
Γ`ϕ Γ`ψ
(∨IL) (∨IR)
Γ`ϕ∨ψ Γ`ϕ∨ψ
Rules for →:
Γ`ϕ→ψ Γ ∪ {ϕ} ` ψ
(→ E) (→ I)
Γ ∪ {ϕ} ` ψ Γ`ϕ→ψ
Rules for proofs by cases:
Γ ∪ {ϕ} ` θ Γ ∪ {ψ} ` θ Γ ∪ {ψ} ` ϕ Γ ∪ {¬ψ} ` ϕ

(∨P C) (¬P C)
Γ ∪ {ϕ ∨ ψ} ` θ Γ`ϕ
Rule for proof by contradiction:
Γ ∪ {¬ϕ} ` ψ Γ ∪ {¬ϕ} ` ¬ψ
(Contr)
Γ`ϕ
3.4.2 Official Definitions

Definition 3.4.1. Let LineP = P(F ormP ) × F ormP .
Definition 3.4.2. Let AssumeP = {(Γ, ϕ) ∈ LineP : ϕ ∈ Γ}.
We need to define functions corresponding to various rules. For example, we define h∧EL : LineP →
P(LineP ) by letting
(
{(Γ, ϕ)} if α = ϕ ∧ ψ where ϕ, ψ ∈ F ormP
h∧EL (Γ, α) =
∅ otherwise
h∧EL is similar, and we define h∧I : (LineP )2 → P(LineP ) by

(
{(Γ1 , ϕ1 ∧ ϕ2 )} if Γ1 = Γ2
h∧I ((Γ1 , ϕ1 ), (Γ2 , ϕ2 )) =
∅ otherwise
For the ∨IL rule, we have the function h∨IL : LineP → P(LineP ) by
h∨IL (Γ, ϕ) = {(Γ, ϕ ∨ ψ) : ψ ∈ F ormP }
Similarly, we define functions h∨IR .

We let H be the collection of all of these functions.
Definition 3.4.3. A deduction is a witnessing sequence in (LineP , AssumeP , H).
Definition 3.4.4. Let Γ ⊆ F ormP and let ϕ ∈ F ormP . We write Γ `P ϕ, or simply Γ ` ϕ if P is clear, to
mean that
(Γ, ϕ) ∈ G(LineP , AssumeP , H).
We pronounce Γ ` ϕ as “Γ syntactically implies ϕ”.
Notation 3.4.5.
1. If Γ = ∅, we write ` ϕ instead of ∅ ` ϕ.
2. If Γ = {γ}, we write γ ` ϕ instead of {γ} ` ϕ.
Definition 3.4.6. Γ is inconsistent if there exists θ ∈ F ormP such that Γ ` θ and Γ ` ¬θ. Otherwise, we
say that Γ is consistent.
3.4.3 Examples Of Deductions

Proposition 3.4.7. A ∧ B ` A ∨ B.
Proof.
{A ∧ B} ` A ∧ B (AssumeP ) (1)
{A ∧ B} ` A (∧EL on 1) (2)
{A ∧ B} ` A ∨ B (∨I on 2) (3)
Proposition 3.4.8. ¬¬ϕ ` ϕ for all ϕ ∈ F ormP .
Proof.
{¬¬ϕ, ¬ϕ} ` ¬ϕ (AssumeP ) (1)

{¬¬ϕ, ¬ϕ} ` ¬¬ϕ (AssumeP ) (2)
{¬¬ϕ} ` ϕ (Contr on 1 and 2) (3)
Proposition 3.4.9. ` ϕ ∨ ¬ϕ for all ϕ ∈ F ormP .

Proof.
{ϕ} ` ϕ (AssumeP ) (1)

{ϕ} ` ϕ ∨ ¬ϕ (∨IL on 1) (2)
{¬ϕ} ` ¬ϕ (AssumeP ) (3)
{¬ϕ} ` ϕ ∨ ¬ϕ (∨IR on 3) (4)
∅ ` ϕ ∨ ¬ϕ (¬P C on 2 and 4) (5)
Proposition 3.4.10. {¬ϕ, ϕ ∨ ψ} ` ψ for all ϕ, ψ ∈ F ormP .
Proof.
{¬ϕ, ϕ, ¬ψ} ` ϕ (AssumeP ) (1)

{¬ϕ, ϕ, ¬ψ} ` ¬ϕ (AssumeP ) (2)
{¬ϕ, ϕ} ` ψ (Contr on 1 and 2) (3)
{¬ϕ, ψ} ` ψ (AssumeP ) (4)
{¬ϕ, ϕ ∨ ψ} ` ψ (∨P C on 3 and 4) (5)
3.4.4 Theorems about `

Proposition 3.4.11. If Γ ` ϕ and Γ0 ⊇ Γ, then Γ0 ` ϕ.
Proof. The proof is by induction. We let X = {(Γ, ϕ) ∈ G : Γ0 ` ϕ for all Γ0 ⊇ Γ} and we show by induction
on G that X = G. We begin by noting that if ϕ ∈ Γ, then for every Γ0 ⊇ Γ, we have ϕ ∈ Γ0 and hence
Γ0 ` ϕ. Therefore, (Γ, ϕ) ∈ X for all (Γ, ϕ) ∈ AssumeP .
We first handle the ∧EL rule. Suppose that (Γ, ϕ ∧ ψ) ∈ X. We need to show that (Γ, ϕ) ∈ X. Suppose
that Γ0 ⊇ Γ. We then have that Γ0 ` ϕ ∧ ψ by induction (i.e. since (Γ, ϕ ∧ ψ) ∈ X), hence Γ0 ` ϕ by the
∧EL rule. Therefore, (Γ, ϕ) ∈ X. The other ∧ rules and the ∨ rules are similar.
We now handle → E rule. Suppose that (Γ, ϕ → ψ) ∈ X. We need to show that (Γ ∪ {ϕ}, ψ). Suppose
that Γ0 ⊇ Γ ∪ {ϕ}. We then have that Γ0 ⊇ Γ, hence Γ0 ` ϕ → ψ by induction, and so Γ0 ∪ {ϕ} ` ψ by the
→ E rule. However, Γ0 ∪ {ϕ} = Γ0 because ϕ ∈ Γ0 , so Γ0 ` ψ. Therefore, (Γ ∪ {ϕ}, ψ) ∈ X.
We now handle → I rule. Suppose that (Γ ∪ {ϕ}, ψ) ∈ X. We need to show that (Γ, ϕ → ψ). Suppose
that Γ0 ⊇ Γ. We then have that Γ0 ∪ {ϕ} ⊇ Γ ∪ {ϕ}, hence Γ0 ∪ {ϕ} ` ψ by induction, and so Γ0 ` ϕ → ψ
by the → I rule. Therefore, (Γ, ϕ → ψ) ∈ X.
Let’s go for the ∨P C rule. Suppose that (Γ ∪ {ϕ}, θ) ∈ X and (Γ ∪ {ψ}, θ) ∈ X. We need to show
that (Γ ∪ {ϕ ∨ ψ}, θ) ∈ X. Suppose that Γ0 ⊇ Γ ∪ {ϕ ∨ ψ}. We then have that Γ0 ∪ {ϕ} ⊇ Γ ∪ {ϕ} and
Γ0 ∪ {ψ} ⊇ Γ ∪ {ψ}, hence Γ0 ∪ {ϕ} ` θ and Γ0 ∪ {ψ} ` θ by induction, and so Γ0 ∪ {ϕ ∨ ψ} ` θ by the ¬P C
rule. However, Γ0 ∪ {ϕ ∨ ψ} = Γ0 because ϕ ∨ ψ ∈ Γ0 , so Γ0 ` θ. Therefore, (Γ ∪ {ϕ ∨ ψ}, θ) ∈ X.
Let’s next attack the ¬P C rule. Suppose that (Γ ∪ {ψ}, ϕ) ∈ X and (Γ ∪ {¬ψ}, ϕ) ∈ X. We need to show
that (Γ, ϕ) ∈ X. Suppose that Γ0 ⊇ Γ. We then have that Γ0 ∪ {ψ} ⊇ Γ ∪ {ψ} and Γ0 ∪ {¬ψ} ⊇ Γ ∪ {¬ψ},
hence Γ0 ∪ {ψ} ` ϕ and Γ0 ∪ {¬ψ} ` ϕ by induction, and so Γ0 ` ϕ by the ¬P C rule. Therefore, (Γ, ϕ) ∈ X.
We finish off with the Contr rule. Suppose that (Γ ∪ {¬ϕ}, ψ) ∈ X and (Γ ∪ {¬ϕ}, ¬ψ) ∈ X. We need to
show that (Γ, ϕ) ∈ X. Suppose that Γ0 ⊇ Γ. We then have that Γ0 ∪ {¬ϕ} ⊇ Γ ∪ {¬ϕ}, hence Γ0 ∪ {¬ϕ} ` ψ
and Γ0 ∪ {¬ϕ} ` ¬ψ by induction, and so Γ0 ` ϕ by the Contr rule. Therefore, (Γ, ϕ) ∈ X.
Proposition 3.4.12. If Γ is inconsistent, then Γ ` ϕ for all ϕ ∈ F ormP .
Proof. Fix θ such that Γ ` θ and Γ ` ¬θ, and fix ϕ ∈ F ormP . We have that Γ ∪ {¬ϕ} ` θ and Γ ∪ {¬ϕ} ` ¬θ
by Proposition 3.4.11. Therefore, Γ ` ϕ by using the Contr rule.
Proposition 3.4.13.
1. If Γ ∪ {ϕ} is inconsistent, then Γ ` ¬ϕ.
2. If Γ ∪ {¬ϕ} is inconsistent, then Γ ` ϕ.
Proof.
1. Since Γ ∪ {ϕ} is inconsistent, we know that Γ ∪ {ϕ} ` ¬ϕ by Proposition 3.4.12. Since we also have
that Γ ∪ {¬ϕ} ` ¬ϕ by AssumeP , it follows that Γ ` ¬ϕ by the ¬P C rule.
2. Since Γ ∪ {¬ϕ} is inconsistent, we know that Γ ∪ {¬ϕ} ` ϕ by Proposition 3.4.12. Since we also have
that Γ ∪ {ϕ} ` ϕ by AssumeP , it follows that Γ ` ϕ by the ¬P C rule.
Corollary 3.4.14. If Γ ⊆ F ormP is consistent and ϕ ∈ F ormP , then either Γ ∪ {ϕ} is consistent or
Γ ∪ {¬ϕ} is consistent.
Proof. If both Γ ∪ {ϕ} and Γ ∪ {¬ϕ} are inconsistent, then both Γ ` ¬ϕ and Γ ` ϕ by Proposition 3.4.13,
so Γ is inconsistent.
Proposition 3.4.15.
1. If Γ ` ϕ and Γ ∪ {ϕ} ` ψ, then Γ ` ψ.
2. If Γ ` ϕ and Γ ` ϕ → ψ, then Γ ` ψ.
Proof.
1. Since Γ ` ϕ, it follows from Proposition 3.4.11 that Γ ∪ {¬ϕ} ` ϕ. Since we also have Γ ∪ {¬ϕ} ` ¬ϕ
by AssumeP , we may conclude that Γ ∪ {¬ϕ} is inconsistent. Therefore, by Proposition 3.4.12, we
have that Γ ∪ {¬ϕ} ` ψ. Now we also have Γ ∪ {ϕ} ` ψ by assumption, so the ¬P C rule gives that
Γ ` ψ.
2. Since Γ ` ϕ → ψ, we can conclude that Γ ∪ {ϕ} ` ψ by rule → E. The result follows from part 1.
Proposition 3.4.16. Γ ` ϕ if and only if there is a finite Γ0 ⊆ Γ such that Γ0 ` ϕ.
Proof. The proof is by induction. We let X = {(Γ, ϕ) ∈ G : there exists a finite Γ0 ⊆ Γ such that Γ0 ` ϕ}
and we show by induction on G that X = G. We begin by noting that if ϕ ∈ Γ, then we have {ϕ} is a finite
subset of Γ and {ϕ} ` ϕ. Therefore, (Γ, ϕ) ∈ X for all (Γ, ϕ) ∈ AssumeP .
We first handle the ∧EL rule. Suppose that (Γ, ϕ ∧ ψ) ∈ X. We need to show that (Γ, ϕ) ∈ X. By
induction (i.e. since (Γ, ϕ ∧ ψ) ∈ X), we may fix a finite Γ0 ⊆ Γ such that Γ0 ` ϕ ∧ ψ. We then have that
Γ0 ` ϕ by the ∧EL rule. Therefore, (Γ, ϕ) ∈ X. The other ∧ER rule and the ∨ rules are similar.
We now handle the ∧I rule. Suppose that (Γ, ϕ) ∈ X and (Γ, ψ) ∈ X. We need to show that (Γ, ϕ ∧ ψ) ∈
X. By induction, we may fix a finite Γ0 ⊆ Γ such that Γ0 ` ϕ and we may fix a finite Γ1 ⊆ Γ such that
Γ1 ` ψ. We then have that Γ0 ∪ Γ1 ` ϕ and Γ0 ∪ Γ1 ` ψ by Proposition 3.4.11, hence Γ0 ∪ Γ1 ` ϕ ∧ ψ by the
∧I rule. Therefore, (Γ, ϕ ∧ ψ) ∈ X because Γ0 ∪ Γ1 is a finite subset of Γ.
3.5. SOUNDNESS AND COMPLETENESS 43
We now handle → E rule. Suppose that (Γ, ϕ → ψ) ∈ X. We need to show that (Γ ∪ {ϕ}, ψ) ∈ X. By
induction, we may fix a finite Γ0 ⊆ Γ such that Γ0 ` ϕ → ψ. We then have that Γ0 ∪ {ϕ} ` ψ by the → E
rule. Therefore, (Γ ∪ {ϕ}, ψ) ∈ X because Γ0 ∪ {ϕ} is a finite subset of Γ ∪ {ϕ}.
We now handle → I rule. Suppose that (Γ ∪ {ϕ}, ψ) ∈ X. We need to show that (Γ, ϕ → ψ) ∈ X. By
induction, we may fix a finite Γ0 ⊆ Γ ∪ {ϕ} such that Γ0 ` ψ. Let Γ00 = Γ0 − {ϕ}. We then have that
Γ0 ⊆ Γ00 ∪ {ϕ}, hence Γ00 ∪ {ϕ} ` ψ by Proposition 3.4.11, and so Γ00 ` ϕ → ψ by the → I rule. Therefore,
(Γ, ϕ → ψ) ∈ X because Γ00 is a finite subset of Γ.
The other rules are exercises. The result follows by induction.
Corollary 3.4.17. If every finite subset of Γ is consistent, then Γ is consistent.
Proof. Suppose that Γ is inconsistent, and fix θ ∈ F ormP such that Γ ` θ and Γ ` ¬θ. By Proposition
3.4.16, there exists finite sets Γ0 ⊆ Γ and Γ1 ⊆ Γ such that Γ0 ` θ and Γ1 ` ¬θ. Using Proposition 3.4.11,
it follows that Γ0 ∪ Γ1 ` θ and Γ0 ∪ Γ1 ` ¬θ, so Γ0 ∪ Γ1 is a finite inconsistent subset of Γ.
3.5 Soundness and Completeness

3.5.1 The Soundness Theorem
Theorem 3.5.1 (Soundness Theorem).
1. If Γ ` ϕ, then Γ ϕ.
2. Every satisfiable set of formulas is consistent.
Proof.
1. The proof is by induction. We let X = {(Γ, ϕ) ∈ G : Γ ϕ} and we show by induction on G that
X = G. We begin by noting that if ϕ ∈ Γ, then Γ ϕ because if v : P → {0, 1} is such that v(γ) = 1
for all γ ∈ Γ, then v(ϕ) = 1 simply because ϕ ∈ Γ. Therefore, (Γ, ϕ) ∈ X for all X ∈ AssumeP .
We first handle the ∧EL rule. Suppose that Γ ϕ ∧ ψ. We need to show that Γ ϕ. However, this is
straightforward because if v : P → {0, 1} is such that v(γ) = 1 for all γ ∈ Γ, then v(ϕ ∧ ψ) = 1 because
Γ ϕ ∧ ψ, hence v(ϕ) = 1. Therefore, Γ ϕ. The other ∧ rules and the ∨ rules are similar.
We now handle → E rule. Suppose that Γ ϕ → ψ. We need to show that Γ ∪ {ϕ} ψ. Let
v : P → {0, 1} be such that v(γ) = 1 for all γ ∈ Γ ∪ {ϕ}. Since Γ ϕ → ψ, we have v(ϕ → ψ) = 1.
Since v(ϕ) = 1, it follows that v(ψ) = 1. Therefore, Γ ∪ {ϕ} ψ. The → I rule is similar.
Let’s next attack the ¬P C rule. Suppose that Γ ∪ {ψ} ϕ and Γ ∪ {¬ψ} ϕ. We need to show that
Γ ϕ. Let v : P → {0, 1} be such that v(γ) = 1 for all γ ∈ Γ. If v(ψ) = 1, then v(ϕ) = 1 because
Γ ∪ {ψ} ϕ. Otherwise, we have v(ψ) = 0, hence v(¬ψ) = 1, and thus v(ϕ) = 1 because Γ ∪ {¬ψ} ϕ.
Therefore, Γ ϕ. The ∨P C rule is similar.
We finish off with the Contr rule. Suppose that Γ ∪ {¬ϕ} ψ and Γ ∪ {¬ϕ} ¬ψ. We need to show
that Γ ϕ. Let v : P → {0, 1} be such that v(γ) = 1 for all γ ∈ Γ. Suppose that v(ϕ) = 0. We then
have v(¬ϕ) = 1, and so both v(ψ) = 1 and v(¬ψ) = 1 because Γ ∪ {¬ϕ} ψ and Γ ∪ {¬ϕ} ¬ψ. This
is a contradiction, so we may conclude that v(ϕ) = 1. Therefore, Γ ϕ.
2. Let Γ be a satisfiable set of formulas. Fix a truth assignment v : P → {0, 1} such that v(γ) = 1 for
all γ ∈ Γ. Suppose that Γ is inconsistent, and fix θ ∈ F ormP such that Γ ` θ and Γ ` ¬θ. We then
have Γ θ and Γ ¬θ by part 1, hence v(θ) = 1 and v(¬θ) = 1, a contradiction. It follows that Γ is
consistent.
3.5.2 The Completeness Theorem

Our first task is to show that every consistent set of formulas is satisfiable. From here, the Completeness
Theorem follows easily, as we’ll see below.
Definition 3.5.2. Γ is complete if for all ϕ ∈ F ormP , either ϕ ∈ Γ or ¬ϕ ∈ Γ.
Proposition 3.5.3. Suppose that P is countable. If Γ is consistent, then there exists a set ∆ ⊇ Γ which is
consistent and complete.
Proof. Since P is countable, it follows that F ormP is countable. List F ormP as ψ1 , ψ2 , ψ3 , . . . . We define a
sequence of sets Γ0 , Γ1 , Γ2 , . . . recursively as follows. Let Γ0 = Γ. Suppose that n ∈ N and we have defined
Γn . Let (
Γn ∪ {ψn } if Γn ∪ {ψn } is consistent
Γn+1 =
Γn ∪ {¬ψn } otherwise
S
Using induction and Corollary 3.4.14, it follows that Γn is consistent for all n ∈ N. Let ∆ = n∈N Γn .
We first argue that ∆ is consistent. For any finite subset ∆0 of ∆, there exists an n ∈ N such that
∆0 ⊆ Γn , and so ∆0 is consistent because every Γn is consistent. Therefore, ∆ is consistent by Proposition
3.4.17. We end by arguing that ∆ is complete. Fix ϕ ∈ F ormP , and fix n ∈ N+ such that ϕ = ψn . By
construction, we either have ϕ ∈ Γn ⊆ ∆ or ¬ϕ ∈ Γn ⊆ ∆. Therefore, ∆ is complete.
Definition 3.5.4. ∆ is maximal consistent if ∆ is consistent and there is no ∆0 ⊃ ∆ which is consistent.
Proposition 3.5.5. ∆ is maximal consistent if and only if ∆ is consistent and complete.
Proof. Suppose that ∆ is maximal consistent. We certainly have that ∆ is consistent. Fix ϕ ∈ F ormP . By
Corollary 3.4.14, either ∆ ∪ {ϕ} is consistent or ∆ ∪ {¬ϕ} is consistent. If ∆ ∪ {ϕ} is consistent, then ϕ ∈ ∆
because ∆ is maximal consistent. Similarly, If ∆ ∪ {¬ϕ} is consistent, then ¬ϕ ∈ ∆ because ∆ is maximal
consistent. Therefore, either ϕ ∈ ∆ or ¬ϕ ∈ ∆.
Suppose that ∆ is consistent and complete. Suppose that ∆0 ⊃ ∆ and fix ϕ ∈ ∆0 − ∆. Since ∆ is
complete and ϕ ∈ / ∆, we have ¬ϕ ∈ ∆. Therefore, ∆0 ` ϕ and ∆0 ` ¬ϕ, so ∆0 is inconsistent. It follows that
∆ is maximal consistent.
Proposition 3.5.6. If Γ is consistent, then there exists a set ∆ ⊇ Γ which is consistent and complete.
Proof. Let S = {Φ ⊆ F ormP : Γ ⊆ Φ and Φ is consistent}, and S order S be ⊆. Notice that S is nonempty
because Γ ∈ S. Suppose that C ⊆ S is a chain in S. Let Ψ = C = {ψ ∈ F ormP : ψ ∈ Φ for some Φ ∈ C}.
We need to argue that Ψ is consistent. Suppose that Ψ0 is a finite subset of Ψ, say Ψ0 = {ψ1 , ψ2 , . . . , ψn }.
For each ψi , fix Φi ∈ C with ψi ∈ Φi . Since C is a chain, there exists j such that Φj ⊇ Φi for all i. Now
Φj ∈ C ⊆ S, so Φj is consistent, and hence Ψ0 is consistent. Therefore, Ψ is consistent by Proposition 3.4.17.
It follows that Ψ ∈ S and using the fact that Φ ⊆ Ψ for all Φ ∈ C, we may conclude that C has an upper
bound.
Therefore, by Zorn’s Lemma, S has a maximal element ∆. Notice that ∆ is maximal consistent, hence
∆ is complete and consistent by Proposition 3.5.5.
Lemma 3.5.7. Suppose that ∆ is consistent and complete. If ∆ ` ϕ, then ϕ ∈ ∆.
Proof. Suppose that ∆ ` ϕ. Since ∆ is complete, we have that either ϕ ∈ ∆ or ¬ϕ ∈ ∆. Now if ¬ϕ ∈ ∆,

then ∆ ` ¬ϕ, hence ∆ is inconsistent contradicting our assumption. It follows that ϕ ∈ ∆.
Lemma 3.5.8. Suppose that ∆ is consistent and complete. We have
1. ¬ϕ ∈ ∆ if and only if ϕ ∈
/ ∆.
3.5. SOUNDNESS AND COMPLETENESS 45
2. ϕ ∧ ψ ∈ ∆ if and only if ϕ ∈ ∆ and ψ ∈ ∆.
3. ϕ ∨ ψ ∈ ∆ if and only if either ϕ ∈ ∆ or ψ ∈ ∆.
4. ϕ → ψ ∈ ∆ if and only if either ϕ ∈

/ ∆ or ψ ∈ ∆.
Proof.
1. If ¬ϕ ∈ ∆, then ϕ ∈
/ ∆ because otherwise ∆ ` ϕ and so ∆ would be inconsistent.
Conversely, if ϕ ∈
/ ∆, then ¬ϕ ∈ ∆ because ∆ is complete.
2. Suppose first that ϕ ∧ ψ ∈ ∆. We then have that ∆ ` ϕ ∧ ψ, hence ∆ ` ϕ by the ∧EL rule and ∆ ` ψ
by the ∧ER rule. Therefore, ϕ ∈ ∆ and ψ ∈ ∆ by Lemma 3.5.7.
Conversely, suppose that ϕ ∈ ∆ and ψ ∈ ∆. We then have ∆ ` ϕ and ∆ ` ψ, hence ∆ ` ϕ ∧ ψ by the
∧I rule. Therefore, ϕ ∧ ψ ∈ ∆ by Lemma 3.5.7.
3. Suppose first that ϕ ∨ ψ ∈ ∆. Suppose that ϕ ∈

/ ∆. Since ∆ is complete, we have that ¬ϕ ∈ ∆. From
Proposition 3.4.10, we know that {¬ϕ, ϕ ∨ ψ} ` ψ, hence ∆ ` ψ by Proposition 3.4.11. Therefore,
ψ ∈ ∆ by Lemma 3.5.7. It follows that either ϕ ∈ ∆ or ψ ∈ ∆.
Conversely, suppose that either ϕ ∈ ∆ or ψ ∈ ∆.
Case 1: Suppose that ϕ ∈ ∆. We have ∆ ` ϕ, hence ∆ ` ϕ ∨ ψ by the ∨IL rule. Therefore, ϕ ∨ ψ ∈ ∆
by Lemma 3.5.7.
Case 2: Suppose that ψ ∈ ∆. We have ∆ ` ψ, hence ∆ ` ϕ ∨ ψ by the ∨IR rule. Therefore, ϕ ∨ ψ ∈ ∆
by Lemma 3.5.7.
4. Suppose first that ϕ → ψ ∈ ∆. Suppose that ϕ ∈ ∆. We then have that ∆ ` ϕ and ∆ ` ϕ → ψ,

hence ∆ ` ψ by Proposition 3.4.15. Therefore, ψ ∈ ∆ by Lemma 3.5.7. It follows that either ϕ ∈
/ ∆ or
ψ ∈ ∆.
Conversely, suppose that either ϕ ∈
/ ∆ or ψ ∈ ∆.
Case 1: Suppose that ϕ ∈
/ ∆. We have ¬ϕ ∈ ∆ because ∆ is complete, hence ∆ ∪ {ϕ} is inconsistent
(as ∆ ∪ {ϕ} ` ϕ and ∆ ∪ {¬ϕ} ` ¬ϕ). It follows that ∆ ∪ {ϕ} ` ψ by Proposition 3.4.12, hence
∆ ` ϕ → ψ by the → I rule. Therefore, ϕ → ψ ∈ ∆ by Lemma 3.5.7.
Case 2: Suppose that ψ ∈ ∆. We have ψ ∈ ∆ ∪ {ϕ}, hence ∆ ∪ {ϕ} ` ψ, and so ∆ ` ϕ → ψ by the
→ I rule. Therefore, ϕ → ψ ∈ ∆ by Lemma 3.5.7.
Proposition 3.5.9. If ∆ is consistent and complete, then ∆ is satisfiable.
Proof. Suppose that ∆ is complete and consistent. Define v : P → {0, 1} by

(
1 if A ∈ ∆
v(A) =
0 if A ∈ /∆
We prove by induction on ϕ that ϕ ∈ ∆ if and only if v(ϕ) = 1. For any A ∈ P , we have
A ∈ ∆ ⇔ v(A) = 1 ⇔ v(A) = 1
by our definition of v.
Suppose that the result holds for ϕ. We have
¬ϕ ∈ ∆ ⇔ ϕ ∈
/∆ (by Lemma 3.5.8)
⇔ v(ϕ) = 0 (by induction)
⇔ v(¬ϕ) = 1
Suppose that the result holds for ϕ and ψ. We have
ϕ ∧ ψ ∈ ∆ ⇔ ϕ ∈ ∆ and ψ ∈ ∆ (by Lemma 3.5.8)

⇔ v(ϕ) = 1 and v(ψ) = 1 (by induction)
⇔ v(ϕ ∧ ψ) = 1
and
ϕ ∨ ψ ∈ ∆ ⇔ ϕ ∈ ∆ or ψ ∈ ∆ (by Lemma 3.5.8)

⇔ v(ϕ) = 1 or v(ψ) = 1 (by induction)
⇔ v(ϕ ∨ ψ) = 1
and finally
ϕ→ψ∈∆⇔ϕ∈
/ ∆ or ψ ∈ ∆ (by Lemma 3.5.8)
⇔ v(ϕ) = 0 or v(ψ) = 1 (by induction)
⇔ v(ϕ → ψ) = 1
Therefore, by induction, we have ϕ ∈ ∆ if and only if v(ϕ) = 1. In particular, we have v(ϕ) = 1 for all
ϕ ∈ ∆, hence ∆ is satisfiable.
Theorem 3.5.10 (Completeness Theorem). (Suppose that P is countable.)
1. Every consistent set of formulas is satisfiable.
2. If Γ ϕ, then Γ ` ϕ.
Proof.
1. Suppose that Γ is consistent. By Proposition 3.5.6, we may fix ∆ ⊇ Γ which is consistent and complete.
Now ∆ is satisfiable by Proposition 3.5.9, so we may fix v : P → {0, 1} such that v(δ) = 1 for all δ ∈ ∆.
Since Γ ⊆ ∆, it follows that v(γ) = 1 for all γ ∈ Γ, hence Γ is satisfiable.
2. Suppose that Γ ϕ. We then have that Γ ∪ {¬ϕ} is unsatisfiable, hence Γ ∪ {¬ϕ} is inconsistent by
part 1. It follows from Proposition 3.4.13 that Γ ` ϕ.
3.6 Compactness and Applications

3.6.1 The Compactness Theorem
Corollary 3.6.1 (Compactness Theorem).
1. If Γ ϕ, then there exists a finite Γ0 ⊆ Γ such that Γ0 ϕ.

3.6. COMPACTNESS AND APPLICATIONS 47
2. If every finite subset of Γ is satisfiable, then Γ is satisfiable.

Proof. We first prove 1. Suppose that Γ ϕ. By the Completeness Theorem, we have Γ ` ϕ. Using
Proposition 3.4.16, we may fix a finite Γ0 ⊆ Γ such that Γ0 ` ϕ. By the Soundness Theorem, we have
Γ0 ϕ.
We now prove 2. If every finite subset of Γ is satisfiable, then every finite subset of Γ is consistent by the
Soundness Theorem, hence Γ is consistent by Corollary 3.4.17, and so Γ is satisfiable by the Completeness
Theorem.
3.6.2 Combinatorial Applications

Definition 3.6.2. Let G = (V, E) be a graph, and let k ∈ N+ . A k-coloring of G is a function f : V → [k]
such that for all u, v ∈ V which are linked by an edge in E, we have f (u) 6= f (v).
Proposition 3.6.3. Let G = (V, E) be a (possibly infinite) graph and let k ∈ N+ . If every finite subgraph
of G is k-colorable, then G is k-colorable.
Proof. Let P = {Au,i : u ∈ V and i ∈ [k]}. Let
k−1
_
Γ={ Au,i : u ∈ V } ∪ {¬(Au,i ∧ Au,j ) : u ∈ V and i, j ∈ [k] with i 6= j}
i=0
∪ {¬(Au,i ∧ Aw,i ) : u and w are linked by an edge in E and i ∈ [k]}}
We use the Compactness Theorem to show that Γ is satisfiable. Suppose that Γ0 ⊆ Γ is finite. Let
{u1 , u2 , . . . , un } be all of the elements u ∈ V such that Au,i occurs in some element of Γ0 for some i. Since
every finite subgraph of G is k-colorable, we may fix a k-coloring f : {u1 , u2 , . . . , un } → [k] such that whenever
u` and um are linked by an edge of E, we have f (u` ) 6= f (um ). If we define a truth assignment v : P → {0, 1}
by (
1 if w = u` and f (u` ) = i
v(Aw,i ) =
0 otherwise
we see that v(ϕ) = 1 for all ϕ ∈ Γ0 . Thus, Γ0 is satisfiable. Therefore, Γ is satisfiable by the Compactness
Theorem.
Fix a truth assignment v : P → {0, 1} such that v(ϕ) = 1 for all ϕ ∈ Γ. Notice that for each u ∈ V ,
there exists a unique i such that v(Au,i ) = 1 because of the first two sets in the definition of Γ. If we define
f : V → {1, 2, . . . , k} by letting f (u) be the unique i such that v(Au,i ) = 1, then for all u, w ∈ V linked by
an edge in E we have that f (u) 6= f (w) (because of the third set in the definition of Γ). Therefore, G is
k-colorable.
Corollary 3.6.4. Every (possibly infinite) planar graph is 4-colorable.
Definition 3.6.5. A set T ⊆ {0, 1}∗ is called a tree if whenever σ ∈ T and τ ⊆ σ, we have τ ∈ T .
Theorem 3.6.6 (Weak König’s Lemma). Suppose that T ⊆ {0, 1}∗ is a tree which is infinite. There exists
an f : N → {0, 1} such that f [n] ∈ T for all n ∈ N.
Proof. Let P = {Aσ : σ ∈ T }. For each n ∈ N, let Tn = {σ ∈ T : |σ| = n} and notice that Tn 6= ∅ for all
n ∈ N because T is infinite. Let
_
Γ={ Aσ : n ∈ N} ∪ {¬(Aσ ∧ Aτ ) : σ, τ ∈ Tn and σ 6= τ }
σ∈Tn
∪ {(Aσ → Aτ ) : σ, τ ∈ T, τ ⊆ σ}
We use the Compactness Theorem to show that Γ is satisfiable. Suppose that Γ0 ⊆ Γ is finite. Let
Σ = {σ1 , σ2 , . . . , σk } be all of the elements σ ∈ {0, 1}∗ such that Aσ occurs in some element of Γ0 . Let
n = max{|σ1 |, |σ2 |, . . . , |σk |}. Since Tn 6= ∅, we may fix τ ∈ Tn . If we define a truth assignment v : P → {0, 1}
by (
1 if σ ⊆ τ
v(Aσ ) =
0 otherwise
Theorem.
Fix a truth assignment v : P → {0, 1} such that v(ϕ) = 1 for all ϕ ∈ Γ. Notice that for each n ∈ N+ ,
there exists a unique σ ∈ Tn such that v(Aσ ) = 1 because of the first two sets in the definition of Γ. For
each n, denote the unique such σ by ρn and notice that ρm ⊆ ρn whenver m ≤ n. Define f : N → {0, 1} by
letting f (n) = ρn+1 (n). We then have that f [n] = ρn ∈ T for all n ∈ N.
3.6.3 An Algebraic Application

Definition 3.6.7. An ordered abelian group is an abelian group (A, +, 0) together with a relation ≤ on A2
such that
1. ≤ is a linear ordering on A, i.e. we have
• For all a ∈ A, we have a ≤ a.
• For all a, b ∈ A, either a ≤ b or b ≤ a.
• If a ≤ b and b ≤ a, then a = b.
• If a ≤ b and b ≤ c, then a ≤ c.
2. If a ≤ b and c ≤ d, then a + c ≤ b + d.
Example 3.6.8. (Z, +, 0) with its usual order is an ordered abelian group.
Example 3.6.9. Define ≤ on Zn using the lexicographic order. In other words, given distinct elements
~a = (a1 , a2 , . . . , an ) and ~b = (b1 , b2 , . . . , bn ) in Zn , let i be least such that ai 6= bi , and set ~a < ~b if ai <Z bi ,
and ~b < ~a if bi <Z ai . With this order, (Zn , +, 0) is an ordered abelian group.
Proposition 3.6.10. Suppose that (A, +, 0) is an ordered abelian group, and define < be letting a < b if
a ≤ b and a 6= b. We the have
1. For all a, b ∈ A, exactly one of a < b, a = b, or b < a holds.
2. If a < b and b ≤ c, then a < c.
3. If a ≤ b and b < c, then a < c.
Proof.
1. Let a, b ∈ A. We first show that at least one happens. Suppose then that a 6= b. We either have a ≤ b
or b ≤ a. If a ≤ b, we then have a < b, while if b ≤ a, we then have b < a.
We now show that at most one occurs. Clearly, we can’t have both a < b and a = b, nor can we have
both a = b and b < a. Suppose then that we have both a < b and b < a. We would then have both
a ≤ b and b ≤ a, hence a = b, a contradiction.
2. Since a ≤ b and b ≤ c, we have a ≤ c. If a = c, it would follow that a ≤ b and b ≤ a, hence a = b, a
contradiction.
3.6. COMPACTNESS AND APPLICATIONS 49
3. Since a ≤ b and b ≤ c, we have a ≤ c. If a = c, it would follow that c ≤ b and b ≤ c, hence b = c, a

contradiction.
Proposition 3.6.11. Suppose that (A, +, 0) is an ordered abelian group.
1. If a < b and c ∈ A, then a + c < b + c.
2. If a < b and c ≤ d, then a + c < b + d.
3. If a > 0, then −a < 0.
4. If a < 0, then −a > 0.
Proof.
1. Since a < b and c ≤ c, we have a + c ≤ b + c. If a + c = b + c, then we would have a = b, a contradiction.

Therefore, a + c < b + c.
2. We have a + c < b + c and b + c ≤ b + d, hence a + c < b + d by the previous proposition.
3. We have a 6= 0, hence −a 6= 0. Suppose that −a > 0. We would then have a + (−a) > 0, hence 0 > 0,
a contradiction.
4. Similar to 3.
Definition 3.6.12. An abelian group (A, +, 0) is torsion-free if every nonzero element of A has infinite
order.
Proposition 3.6.13. Every ordered abelian group is torsion-free.
Proof. Let (A, +, 0) be an ordered abelian group. Let a ∈ A. If a > 0, then we have n · a > 0 for every
n ∈ N+ by induction. If a < 0, then we have n · a < 0 for every n ∈ N+ by induction.
Theorem 3.6.14. Every torsion-free abelian group can be ordered.
Proof. First notice that every finitely generated torsion-free abelian group is isomorphic to Zn for some n,
which we can order lexicographically from above. We can transer this ordering across the isomorphism to
order our finitely generated abelian group.
Suppose now that A is an arbitrary torsion-free abelian group. Let P be the set {La,b : a, b ∈ A} and let
Γ be the union of the sets
• {La,a : a ∈ A}.
• {La,b ∨ Lb,a : a, b ∈ A}.
• {¬(La,b ∧ Lb,a ) : a, b ∈ A, a 6= b}
• {(La,b ∧ Lb,c ) → La,c : a, b, c ∈ A}.
• {(La,b ∧ Lc,d ) → La+c,b+d : a, b, c, d ∈ A}

We show that Γ is satisfiable. By Compactness, it suffices to show that any finite subset of Γ is satisfiable.
Suppose that Γ0 ⊆ Γ is finite, and let S the finite subset of A consisting of all elements of A which appear
as a subscript of a symbol occuring in Γ0 . Let B be the subgroup of A generated by S. We then have that
B is a finitely generated torsion-free abelian group, so from above we may fix an order ≤ on it. If we define
a truth assignment v : P → {0, 1} by
(
1 if a ≤ b
v(La,b ) =
0 otherwise
Theorem.
Fix a truth assignment v : P → {0, 1} such that v(γ) = 1 for all γ ∈ Γ. Define ≤ on A2 by letting a ≤ b
if and only if v(La,b ) = 1. We then have that ≤ orders A. Therefore, A can be ordered.
Chapter 4
First-Order Logic : Syntax and

Semantics
Now that we’ve succeeded in giving a decent analysis of propositional logic, together with proving a few
nontrivial theorems, it’s time to move on to a much more substantial and important logic: first-order logic.
As summarized in the introduction, the general idea is as follows. Many areas of mathematics deal with
mathematical structures consisting of special constants, relations, and functions, together with certain axioms
that these objects obey. We want our logic to be able to handle many different types of situations, so we
allow ourselves to vary the number and types of these symbols. Any such choice gives rise to a language,
and once we’ve fixed such a language, we can build up formulas which will express something meaningful
once we’ve decided on an interpretation of the symbols.
4.1 The Syntax of First-Order Logic

Since our logic will have quantifiers, the first thing that we need is a collection of variables.
Definition 4.1.1. Fix a countably infinite set V ar called variables.
Definition 4.1.2. A first-order language, or simply a language, consists of the following:
1. A set C of constant symbols.
2. A set F of function symbols together with a function ArityF : F → N+ .
3. A set R of relation symbols together with a function ArityR : R → N+ .
We also assume that C, R, F, V ar, and {∀, ∃, =, ¬, ∧, ∨, →} are pairwise disjoint. For each k ∈ N+ , we let
Fk = {f ∈ F : ArityF (f) = k}
and we let
Rk = {R ∈ R : ArityR (R) = k}
Definition 4.1.3. Let L be a language. We let SymL = C ∪ R ∪ F ∪ V ar ∪ {∀, ∃, =, ¬, ∧, ∨, →}.
Now that we’ve described all of the symbols that are available once we’ve fixed a language, we need to
talk about how to build up formulas. Before doing this, however, we need a way to name objects. Intuitively,
our constant symbols and variables name objects once we’ve fixed an interpretation. From here, we can get
new objects by applying, perhaps repeatedly, interpretations of function symbols. This is starting to sound
like a recursive definition.
51
52 CHAPTER 4. FIRST-ORDER LOGIC : SYNTAX AND SEMANTICS
Definition 4.1.4. Let L be a language. For each f ∈ Fk , define hf : (Sym∗L )k → Sym∗L by letting
hf (σ1 , σ2 , . . . , σk ) = fσ1 σ2 · · · σk
Let
T ermL = G(Sym∗L , C ∪ V ar, {hf : f ∈ F})
Now that we have terms which intuitively name elements once we’ve fixed an interpretation, we need to
say what our atomic formulas are. The idea is that the most basic things we can say are whether or not two
objects are equal or whether or not a k-tuple is in the interpretation of some relation symbol R ∈ Rk .
Definition 4.1.5. Let L be a language. We let
AtomicF ormL = {Rt1 t2 · · · tk : k ∈ N+ , R ∈ Rk , and t1 , t2 , . . . , tk ∈ T ermL } ∪ {= t1 t2 : t1 , t2 ∈ T ermL }
From here, we can build up all formulas.
Definition 4.1.6. Let L be a language. Define a unary function h¬ and binary functions h∧ , h∨ , and h→
on Sym∗L as follows.
h¬ (σ) = ¬σ
h∧ (σ, τ ) = ∧στ
h∨ (σ, τ ) = ∨στ
h→ (σ, τ ) =→ στ
Also, for each x ∈ V ar, define two unary functions h∀,x and h∃,x on Sym∗L as follows
h∀,x (σ) = ∀xσ
h∃,x (σ) = ∃xσ
Let
F ormL = G(Sym∗L , AtomicF ormL , {h¬ , h∧ , h∨ , h→ } ∪ {h∀,x , h∃,x : x ∈ V ar})
As with propositional logic, we’d like to be able to define things recursively, so we need to check that our
generating systems are free. Notice that in the construction of formulas, we have two generating systems
around. We first generate all terms. With terms taken care of, we next describe the atomic formulas, and
from them we generate all formulas. Thus, we’ll need to prove that two generating systems are free. The
general idea is to make use of the insights gained by proving the corresponding result for Polish notation in
propositional logic.
Definition 4.1.7. Let L be a language. Define K : Sym∗L → Z as follows. We first define w : SymL → Z
as follows.
w(c) = 1 for all c ∈ C
w(f) = 1 − k for all f ∈ Fk
w(R) = 1 − k for all R ∈ Rk
w(x) = 1 for all x ∈ V ar
w(=) = −1
w(Q) = −1 for all Q ∈ {∀, ∃}
w(¬) = 0
w(3) = −1 for all 3 ∈ {∧, ∨, →}
Sym∗L
P
We then define K on all of by letting K(λ) = 0 and letting K(σ) = i<|σ| w(σ(i)) for all σ ∈
Sym∗L \{λ}.
4.1. THE SYNTAX OF FIRST-ORDER LOGIC 53
Remark 4.1.8. If σ, τ ∈ Sym∗L , then K(στ ) = K(σ) + K(τ ).
Proposition 4.1.9. If t ∈ T ermL , then K(t) = 1.
Proof. The proof is by induction on t. Notice first that K(c) = 1 for all c ∈ C and K(x) = 1 for all x ∈ V ar.
Suppose that k ∈ N+ , f ∈ Fk , and t1 , t2 , . . . , tk ∈ T ermL are such that K(ti ) = 1 for all i. We then have
that
K(ft1 t2 · · · tk ) = K(f) + K(t1 ) + K(t2 ) + · · · + K(tk )

= (1 − k) + 1 + 1 + · · · + 1 (by induction)
= 1.
Proposition 4.1.10. If t ∈ T ermL and σ ⊂ t, then K(σ) ≤ 0.
Proof. The proof is by induction on t. For every c ∈ C, this is trivial because the only σ ⊂ c is σ = λ and
we have K(λ) = 0. Similarly, for every x ∈ V ar, the only σ ⊂ x is σ = λ and we have K(λ) = 0.
Suppose that k ∈ N+ , f ∈ Fk , and t1 , t2 , . . . , tk ∈ T ermL are such that the result holds for each ti . We
prove the result for ft1 t2 · · · tk . Suppose that σ ⊂ ft1 t2 · · · tk . If σ = λ, then K(σ) = 0. Otherwise, there
exists i < k and τ ⊂ ti such that σ = ft1 t2 · · · ti−1 τ , in which case
K(σ) = K(f) + K(t1 ) + K(t2 ) + · · · + K(ti−1 ) + K(τ )

= (1 − k) + 1 + 1 + · · · + 1 + K(τ ) (by Proposition 4.1.9)
= (1 − k) + i + K(τ )
≤ (1 − k) + i + 0 (by induction)
= 1 + (i − k)
≤ 0. (since i < k)
Thus, the result holds for ft1 t2 · · · tk .
Corollary 4.1.11. If t, u ∈ T ermL , then t 6⊂ u.
Theorem 4.1.12. The generating system (Sym∗L , C ∪ V ar, {hf : f ∈ F}) is free.
Proof. First notice that for all f ∈ F, we have that ran(hf (T ermL )k ) ∩ (C ∪ V ar) = ∅ because all elements
of ran(hf ) begin with f and we know that f ∈ / C ∪ V ar.
Fix f ∈ Fk . Suppose that t1 , t2 , . . . , tk , u1 , u2 , . . . , uk ∈ T ermL and hf (t1 , t2 , . . . , tk ) = hf (u1 , u2 , . . . , uk ).
We then have ft1 t2 · · · tk = fu1 u2 · · · uk , hence t1 t2 · · · tk = u1 u2 · · · uk . Since t1 ⊂ u1 and u1 ⊂ t1 are both
impossible by Corollary 4.1.11, it follows that t1 = u1 . Thus, t2 · · · tk = u2 · · · uk , and so t2 = u2 for the
same reason. Continuing in this fashion, we conclude that ti = ui for all i. It follows that hf (T ermL )k is
injective.
Finally notice that for any f ∈ Fk and any g ∈ F` with f 6= g, we have that ran(hf (T ermL )k ) ∩
ran(hg (T ermL )` ) = ∅ because all elements of ran(hf (T ermL )k ) begin with f while all elements of
ran(hg (T ermL )` ) begin with g.
Proposition 4.1.13. If ϕ ∈ F ormL , then K(ϕ) = 1.

Proof. The proof is by induction on ϕ. We first show that K(ϕ) = 1 for all ϕ ∈ AtomicF ormL . Suppose
that ϕ is Rt1 t2 · · · tk where R ∈ Rk and t1 , t2 , . . . , tk ∈ T ermL . We then have
K(Rt1 t2 · · · tk ) = K(R) + K(t1 ) + K(t2 ) + · · · + K(tk )

= (1 − k) + 1 + 1 + · · · + 1 (by Proposition 4.1.9)
= 1.
Suppose that ϕ is = t1 t2 where t1 , t2 ∈ T ermL . We then have
K(= t1 t2 ) = K(=) + K(t1 ) + K(t2 )

= −1 + 1 + 1 (by Proposition 4.1.9)
= 1.
Thus, K(ϕ) = 1 for all ϕ ∈ AtomicF ormL .

Suppose that ϕ ∈ F ormL is such that K(ϕ) = 1. We then have that
K(¬ϕ) = K(¬) + K(ϕ)

=0+1
= 1.
For any Q ∈ {∀, ∃} and any x ∈ V ar we also have
K(Qxϕ) = K(Q) + K(x) + K(ϕ)

= −1 + 1 + 1
= 1.
Suppose now that ϕ, ψ ∈ F ormL are such that K(ϕ) = 1 = K(ψ), and 3 ∈ {∧, ∨, →}. We then have that
K(3ϕψ) = −1 + K(ϕ) + K(ψ)

= −1 + 1 + 1
= 1.

Proposition 4.1.14. If ϕ ∈ F ormL and σ ⊂ ϕ, then K(σ) ≤ 0.
Proof. The proof is by induction on ϕ. We first show that the results holds for all ϕ ∈ AtomicF ormL .
Suppose that ϕ is Rt1 t2 · · · tk where R ∈ Rk and t1 , t2 , . . . , tk ∈ T ermL . Suppose that σ ⊂ Rt1 t2 · · · tk . If
σ = λ, then K(σ) = 0. Otherwise, there exists i < k and τ ⊂ ti such that σ is Rt1 t2 · · · ti−1 τ , in which case
K(σ) = K(R) + K(t1 ) + K(t2 ) + · · · + K(ti−1 ) + K(τ )

= (1 − k) + 1 + 1 + · · · + 1 + K(τ ) (by Proposition 4.1.9)
= (1 − k) + i + K(τ )
≤ (1 − k) + i + 0 (by induction)
= 1 + (i − k)
≤ 0. (since i < k)
Thus, the result holds for Rt1 t2 · · · tk . The same argument works for = t1 t2 where t1 , t2 ∈ T ermL , so the
result holds for all ϕ ∈ AtomicF ormL .
4.1. THE SYNTAX OF FIRST-ORDER LOGIC 55
Suppose that the result holds for ϕ ∈ F ormL . Suppose that σ ⊂ ¬ϕ. If σ = λ, then K(σ) = 0.
Otherwise, σ = ¬τ for some τ ⊂ ϕ, in which case
K(σ) = K(¬) + K(τ )

= 0 + K(τ )
≤ 0. (by induction)
Suppose now that Q ∈ {∀, ∃}, that x ∈ V ar, and that σ ⊂ Qxϕ. If σ = λ, then K(σ) = 0, and if σ = Q,
then K(σ) = −1. Otherwise, σ = Qxτ for some τ ⊂ ϕ, in which case
K(σ) = K(Q) + K(x) + K(τ )

= −1 + 1 + K(τ )
=0 (by induction)
Suppose now that the result holds for ϕ, ψ ∈ F ormL , and 3 ∈ {∧, ∨, →}. Suppose that σ ⊂ 3ϕψ. If σ = λ,
then K(σ) = 0. If σ is 3τ for some τ ⊂ ϕ, then
K(σ) = K(3) + K(τ )

= −1 + K(τ )
≤ −1. (by induction)
Otherwise, σ is 3ϕτ for some τ ⊂ ψ, in which case
K(σ) = K(3) + K(ϕ) + K(τ )

≤ −1. (by induction)
Thus, the result holds for 3ϕψ.
Corollary 4.1.15. If ϕ, ψ ∈ F ormL , then ϕ 6⊂ ψ.
Theorem 4.1.16. The generating system (Sym∗L , AtomicF ormL , {h¬ , h∧ , h∨ , h→ } ∪ {h∀,x , h∃,x : x ∈ V }) is
free.
Proof. Similar to the others.
With these freeness results, we are now able to define functions recursively on T ermL and F ormL . Since
we use terms in our definition of atomic formulas, which are the basic formulas, we will often need to make
two recursive definitions (on terms first, then on formulas) in order to define a function on formulas. Here’s
an example.
Definition 4.1.17. Let L be a language.
1. We define a function OccurV ar : T ermL → P(V ar) recursively as follows.
• OccurV ar(c) = ∅ for all x ∈ C.

• OccurV ar(x) = {x} for all x ∈ V ar.
• OccurV ar(ft1 t2 · · · tk ) = OccurV ar(t1 ) ∪ OccurV ar(t2 ) ∪ · · · ∪ OccurV ar(tk ) for all f ∈ Fk and
t1 , t2 , . . . , tk ∈ T ermL .
2. We define a function OccurV ar : F ormL → P(V ar) recursively as follows.

• OccurV ar(Rt1 t2 · · · tk ) = OccurV ar(t1 ) ∪ OccurV ar(t2 ) ∪ · · · ∪ OccurV ar(tk ) for all R ∈ Rk and
t1 , t2 , . . . , tk ∈ T ermL .
• OccurV ar(= t1 t2 ) = OccurV ar(t1 ) ∪ OccurV ar(t2 ) for all t1 , t2 ∈ T ermL .
• OccurV ar(¬ϕ) = OccurV ar(ϕ) for all ϕ ∈ F ormL .
• OccurV ar(3ϕψ) = OccurV ar(ϕ) ∪ OccurV ar(ψ) for each 3 ∈ {∧, ∨, →} and ϕ, ψ ∈ F ormL .
• OccurV ar(Qxϕ) = OccurV ar(ϕ) ∪ {x} for each Q ∈ {∀, ∃}, x ∈ V ar, and ϕ ∈ F ormL .
Definition 4.1.18. Let L be a language.
1. We define a function F reeV ar : F ormL → P(V ar) recursively as follows.
• F reeV ar(Rt1 t2 · · · tk ) = OccurV ar(t1 ) ∪ OccurV ar(t2 ) ∪ · · · ∪ OccurV ar(tk ) for all R ∈ Rk and
t1 , t2 , . . . , tk ∈ T ermL .
• F reeV ar(= t1 t2 ) = OccurV ar(t1 ) ∪ OccurV ar(t2 ) for all t1 , t2 ∈ T ermL .
• F reeV ar(¬ϕ) = F reeV ar(ϕ) for all ϕ ∈ F ormL .
• F reeV ar(3ϕψ) = F reeV ar(ϕ) ∪ F reeV ar(ψ) for each 3 ∈ {∧, ∨, →} and ϕ, ψ ∈ F ormL .
• F reeV ar(Qxϕ) = F reeV ar(ϕ)\{x} for each Q ∈ {∀, ∃}, x ∈ V ar, and ϕ ∈ F ormL .
2. We define a function BoundV ar : F ormL → P(V ar) recursively as follows.
• BoundV ar(Rt1 t2 · · · tk ) = ∅ for all R ∈ Rk and t1 , t2 , . . . , tk ∈ T ermL .
• BoundV ar(= t1 t2 ) = ∅ for all t1 , t2 ∈ T ermL .
• BoundV ar(¬ϕ) = BoundV ar(ϕ) for all ϕ ∈ F ormL .
• BoundV ar(3ϕψ) = BoundV ar(ϕ) ∪ BoundV ar(ψ) for each 3 ∈ {∧, ∨, →} and ϕ, ψ ∈ F ormL .
• BoundV ar(Qxϕ) = BoundV ar(ϕ) ∪ {x} for each Q ∈ {∀, ∃}, x ∈ V ar, and ϕ ∈ F ormL .
Definition 4.1.19. Let L be a language and let ϕ ∈ F ormL . We say that ϕ is an L-sentence, or simply a
sentence, if F reeV ar(ϕ) = ∅. We let SentL be the set of sentences.
4.2 Structures: The Semantics of First-Order Logic

4.2.1 Structures: Definition and Satisfaction
Up until this point, all that we’ve dealt with are sequences of symbols without meaning. Sure, our motivation
was to capture meaningful situations with our languages and the way we’ve described formulas, but all we’ve
done so far is describe the grammar. If we want our formulas to actually express something, we need to set
up a context in which to interpret them. Since we have quantifiers, the first thing we’ll need is a nonempty
set M to serve as the domain of objects that the quantifiers range over. Once we’ve fixed that, we need
to interpret the symbols of language as actual elements of our set (in the case of constant symbols), actual
k-ary relations on M (in the case of R ∈ Rk ), and actual k-ary functions on M (in the case of f ∈ Fk ).
Definition 4.2.1. Let L be a language. An L-structure, or simply a structure, is a set M = (M, gC , gF , gR )
where
1. M is a nonempty set called the universe of M.
2. gC : C → M .
4.2. STRUCTURES: THE SEMANTICS OF FIRST-ORDER LOGIC 57
3. gR is a function on R such that gR (R) is a subset of M k for all R ∈ Rk .
4. gF is a function on F such that gF (f) is a k-ary function on M for all f ∈ Fk .
We use the following notation.
1. For each c ∈ C, we use cM to denote gC (c).
2. For each R ∈ Rk , we use RM to denote gR (R).
3. For each f ∈ Fk , we use f M to denote gF (f).
Example 4.2.2. Let L = {R} where R is a binary relation symbol. Here are some examples of L-structures.
1. M = N and RM = {(m, n) ∈ M 2 : m | n}.
2. M = {0, 1}∗ and RM = {(σ, τ ) ∈ M 2 : σ ⊆ τ }.
3. M = R2 and RM = {((a1 , b1 ), (a2 , b2 )) ∈ M 2 : a1 = a2 }.
4. M = {0, 1, 2, 3, 4} and RM = {(0, 2), (3, 3), (4, 1), (4, 2), (4, 3)}.
Example 4.2.3. Let L = {c, f} where c is a constant symbol and f is a binary function symbol. Here are
some examples of L-structures.
1. M = Z, cM = 3 and f M is the addition function (m, n) 7→ m − n.
2. M = R, cM = π and f M is the function (a, b) 7→ sin(a · b).
3. For any group (G, e, ·), we get an L-structure by letting M = G, cM = e, and letting f M be the group
operation.
At first, it may appear than an L-structure provides a means to make sense out of any formula. However,
this is not the case, as you can see by looking at the formula x = y where x, y ∈ V ar. Until we provide a way
to interpret the elements of V ar, this formula is meaningless. This motivates the following definition.
Definition 4.2.4. Let M be an L-structure. A function s : V ar → M is called a variable assignment on

M.
Recall in propositional logic that every truth assignment v : P → {0, 1} gave rise to an extension
v : F ormP → {0, 1} telling us how to interpret every formula. In our case, every variable assignment
s : V ar → M gives rise to an extension s : T ermL → M telling us which element of M to assign to each
term.
Definition 4.2.5. Let M be an L-structure, and let s : V ar → M be a variable assignment. By freeness,

there exists a unique s : T ermL → M such that
• s(x) = s(x) for all v ∈ V ar.
• s(c) = cM for all c ∈ C.
• s(ft1 t2 · · · tk ) = f M (s(t1 ), s(t2 ), . . . , s(tk ))
We’re now in position to define the intuitive statement “ϕ holds in the L-structure M with variable
assignment s” recursively. We need the following definition.
Definition 4.2.6. Let M be an L-structure, and let s : V ar → M be a variable assignment. Given x ∈ V ar

and a ∈ M , we let s[x ⇒ a] denote the variable assignment
(
a if y = x
s[x ⇒ a](y) =
s(y) otherwise
Definition 4.2.7. Let M be an L-structure. We define a relation (M, s) ϕ (pronounced “ϕ holds in

(M, s)”, or “ϕ is true in (M, s)”, or “(M, s) models ϕ”) for all ϕ ∈ F ormL and all variable assignments
s by induction on ϕ.
• Suppose first that ϕ is an atomic formula.
– If ϕ is Rt1 t2 · · · tk ,, we have (M, s) ϕ if and only if (s(t1 ), s(t2 ), . . . , s(tk )) ∈ RM .
– If ϕ is = t1 t2 , we have (M, s) ϕ if and only if s(t1 ) = s(t2 ).
• For any s, we have (M, s) ¬ϕ if and only if (M, s) 6 ϕ.
• For any s, we have (M, s) ϕ ∧ ψ if and only if (M, s) ϕ and (M, s) ψ.
• For any s, we have (M, s) ϕ ∨ ψ if and only if either (M, s) ϕ or (M, s) ψ.
• For any s, we have (M, s) ϕ → ψ if and only if either (M, s) 6 ϕ or (M, s) ψ.
• For any s, we have (M, s) ∃xϕ if and only if there exists a ∈ M such that (M, s[x ⇒ a]) ϕ.
• For any s, we have (M, s) ∀xϕ if and only if for all a ∈ M , we have (M, s[x ⇒ a]) ϕ.
Comments. The above recursive definition takes a little explanation, because some recursive “calls” change
the variable assignment. Thus, we are not fixing an L-structurve M and a variable assignment s on M,
and then doing a recursive definition on ϕ ∈ F ormL . We can make the definition formal as follows. Fix an
L-structure M. Let V arAssignM be the set of all variable assignments on M. We then define a function
gM : F ormP → V arAssignM recursively using the above rules as guides, and we write (M, s) ϕ to mean
that s ∈ gM (ϕ).
Example. Let L = {R, f} where R is a unary relation symbol and f is a unary function symbol. Let M be
the following L-structure. We have M = {0, 1, 2, 3}, RM = {1, 3}, and f M : M → M is the function defined
by
f M (0) = 3 f M (1) = 1 f M (2) = 0 f M (3) = 3
1. For every s : V ar → M with s(x) = 2, we have (M, s) ¬Rx because
(M, s) ¬Rx ⇔ (M, s) 6 Rx
/ RM
⇔ s(x) ∈
/ RM
⇔2∈
which is true.
2. For every s, we have (M, s) ∃xRx. To see this, fix s : V ar → M . We have
(M, s) ∃xRx ⇔ There exists a ∈ M such that (M, s[x ⇒ a]) Rx
⇔ There exists a ∈ M such that s[x ⇒ a](x) ∈ RM
⇔ There exists a ∈ M such that s[x ⇒ a](x) ∈ RM
⇔ There exists a ∈ M such that a ∈ RM
with is true since 1 ∈ RM .

3. For every s, we have (M, s) ∀x(Rx → (fx = x)). To see this, fix s : V ar → M . We have
(M, s) ∀x(Rx → (f(x) = x)) ⇔ For all a ∈ M, we have (M, s[x ⇒ a]) (Rx → (fx = x))
⇔ For all a ∈ M, we have either
(M, s[x ⇒ a]) 6 Rx or (M, s[x ⇒ a]) (fx = x)
/ RM or s[x ⇒ a](fx) = s[x ⇒ a](x)
s[x ⇒ a](x) ∈
/ RM or f M (s[x ⇒ a](x)) = s[x ⇒ a](x)
s[x ⇒ a](x) ∈
/ RM or f M (s[x ⇒ a](x)) = s[x ⇒ a](x)
s[x ⇒ a](x) ∈
/ RM or f M (a) = a
⇔ For all a ∈ M, we have either a ∈
/ RM , f M (1) = 1, 2 ∈
which is true because 0 ∈ / RM , and f M (3) = 3.
In the above examples, it’s clear that only the value of s on the free variables in ϕ affect whether or not
(M, s) ϕ. The following precise statement of this fact follows by a straightforward induction.
Proposition 4.2.8. Let M be an L-structure. Suppose that t ∈ T ermL and s1 , s2 : V ar → M are two
variable assignments such that s1 (x) = s2 (x) for all x ∈ OccurV ar(t). We then have s1 (t) = s2 (t).
Proposition 4.2.9. Let M be an L-structure. Suppose that ϕ ∈ F ormL and s1 , s2 : V ar → M are two
variable assignments such that s1 (x) = s2 (x) for all x ∈ F reeV ar(ϕ). We then have
(M, s1 ) ϕ if and only if (M, s2 ) ϕ
Notation 4.2.10. Let L be a language.
1. If x1 , x2 , . . . , xn ∈ V ar are distinct, and we refer to a formula ϕ(x1 , x2 , . . . , xn ) ∈ F ormL we mean that

ϕ ∈ F ormL and F reeV ar(ϕ) ⊆ {x1 , x2 , . . . , xn }.
2. Suppose that M is an L-structure, ϕ(x1 , x2 , . . . , xn ) ∈ F ormL , and a1 , a2 , . . . , an ∈ M . We write

(M, a1 , a2 , . . . , an ) ϕ to mean that (M, s) ϕ for some (any) s : V ar → M with s(xi ) = ai for all i.
3. As a special case of 2, we have the following. Suppose that M is an L-structure and σ ∈ SentL . We
write M σ to mean that (M, s) σ for some (any) s : V ar → M .
4.2.2 Elementary Classes of Structures

As we’ve seen, once you fix a language L, an L-structure can fix any set M at all, interpret the elements
of C as arbitrary elements of M , interpret the elements of Rk as arbitrary subsets of M k , and interpret the
elements Fk as arbitrary k-ary functions on M . However, since we have a precise languague in hand, we
now carve out classes of structures which satisfy certain sentences of our language.
Definition 4.2.11. Let L be a language, and let Σ ⊆ SentL . We let M od(Σ) be the class of all L-structures
M such that M σ for all σ ∈ Σ. If σ ∈ SentL , we write M od(σ) instead of M od(Σ).
Definition 4.2.12. Let L be a language and let K be a class of L-structures.

1. K is an elementary class if there exists σ ∈ SentL such that K = M od(σ).

2. K is a weak elementary class if there exists Σ ⊆ SentL such that K = M od(Σ).
By taking conjunctions, we have the following simple proposition.
Proposition 4.2.13. Let L be a language and let K be a class of L-structures. K is an elementary class if
and only if there exists a finite Σ ⊆ SentL such that K = M od(Σ).
Examples. Let L = {R} where R is a binary relation symbol.
1. The class of partially ordered sets is an elementary class as we saw in the introduction. We may let Σ
be the following collection of sentences:
(a) ∀xRxx
(b) ∀x∀y((Rxy ∧ Ryx) → (x = y))
(c) ∀x∀y∀z((Rxy ∧ Ryz) → Rxz)
2. The class of equivalence relations is an elementary class. We may let Σ be the following collection of
sentences:
(a) ∀xRxx
(b) ∀x∀y(Rxy → Ryx)
(c) ∀x∀y∀z((Rxy ∧ Ryz) → Rxz)
3. The class of graphs is an elementary class. We may let Σ be the following collection of sentences:
(a) ∀x(¬Rxx)
(b) ∀x∀y(Rxy → Ryx)
Example. Let L be any language whatsoever, and let n ∈ N+ . The class of L-structures of cardinality at
least n is an elementary class as witnessed by the formula:
^
∃x1 ∃x2 · · · ∃xn ( 6 xj ))
(xi =
1≤i<j≤n
Furthermore, the class of L-structures of cardinality equal to n is an elementary class. Letting σn be the
above formula for n, we can see this by considering σn ∧ (¬σn+1 ).
Examples. Let L = {0, 1, +, ·} where 0, 1 are constant symbols and +, · are binary function symbols.
1. The class of fields is an elementary class. We may let Σ be the following collection of sentences:
(a) ∀x∀y∀z(x + (y + z) = (x + y) + z)
(b) ∀x((x + 0 = x) ∧ (0 + x = x))
(c) ∀x∃y((x + y = 0) ∧ (y + x = 0))
(d) ∀x∀y(x + y = y + x)
(e) ∀x∀y∀z(x · (y · z) = (x · y) · z)
(f) ∀x((x · 1 = x) ∧ (1 · x = x))
(g) ∀x(x 6= 0 → ∃y((x · y = 1) ∧ (y · x = 0)))
(h) ∀x∀y(x · y = y · x)
(i) ∀x∀y∀z(x · (y + z) = (x · y) + (x · z))
2. For each prime p > 0, the class of fields of characteristic p is an elementary class. Fix a prime p > 0,
and let Σp be the above sentences togheter with the sentence 1 + 1 + · · · + 1 = 0 (where there are p
1’s in the sum).
3. The class of fields of characteristic 0 is a weak elementary class. Let Σ be the above sentences together
with {τn : n ∈ N+ } where for each n ∈ N+ , we have τn = ¬(1 + 1 + · · · + 1 = 0) (where there are n 1’s
in the sum).
Example. Let F be a field, and let LF = {0, +} ∪ {hα : α ∈ F } where 0 is a constant symbol, + is binary
function symbol, and each hα is a unary function symbol. The class of vector spaces over F is a weak
elementary class. We may let Σ be the following collection of sentences:
1. ∀x∀y∀z(x + (y + z) = (x + y) + z)
2. ∀x((x + 0 = x) ∧ (0 + x = x))
3. ∀x∃y((x + y = 0) ∧ (y + x = 0))
4. ∀x∀y(x + y = y + x)
5. ∀x∀y(hα (x + y) = hα (x) + hα (y)) for each α ∈ F .
6. ∀x(hα+β (x) = (hα (x) + hβ (x))) for each α ∈ F .
7. ∀x(hα·β (x) = (hα (hβ (x))) for each α, β ∈ F .
8. ∀x(h1 (x) = x)
At this point, it’s often clear how to show that a certain class of structures is a (weak) elementary class:
simply exhibit the correct sentences. However, it may seem very difficult to show that a class is not a (weak)
elementary class. For example, is the class of fields of characteristic 0 an elementary class? Is the class of
finite groups a weak elementary class? There are no obvious ways to answer these questions affirmatively.
We’ll develop some tools later which will allow us to resolve these questions.
Another interesting case is that of Dedekind-complete ordered fields. Now the ordered field axioms are
easily written down in the first-order language L = {0, 1, <, +, ·}. In contrast, the Dedekind-completeness
axiom, which says that every nonempty subset which is bounded above has a least upper bound, can not
be directly translated in the language L because it involves quantifying over subsets instead of elements.
However, we are unable to immediately conclude that this isn’t due to a lack of cleverness on our part.
Perhaps there is an alternative approach which captures Dedekind-complete ordered fields in a first-order
way (by finding a clever equivalent first-order expression of Dedekind-completeness). More formally, the
precise question is whether the complete ordered fields are a (weak) elementary class in the language L.
We’ll be able to answer this question later.
4.2.3 Definability in Structures

Another wonderful side-effect of developing a formal language is the ability to talk about what objects we
can define using that language.
Definition 4.2.14. Let M be an L-structure. Suppose that k ∈ N+ and X ⊆ M k . We say that X is

definable in M if there exists ϕ(x1 , x2 , . . . , xk ) ∈ F ormL such that
X = {(a1 , a2 , . . . , ak ) ∈ M k : (M, a1 , a2 , . . . , ak ) ϕ}
Examples. Let L = {0, 1, +, ·} where 0 and 1 are constant symbols and + and · are binary function symbols.
1. The set X = {(m, n) ∈ N2 : m < n} is definable in (N, 0, 1, +, ·) as witnessed by the formula
∃z(z 6= 0 ∧ (x + z = y))
2. The set X = {n ∈ N : n is prime} is definable in (N, 0, 1, +, ·) as witnessed by the formula
¬(x = 1) ∧ ∀y∀z(x = y · z → (y = 1 ∨ z = 1))
3. The set X = {r ∈ R : r ≥ 0} is definable in (R, 0, 1, +, ·) as witnessed by the formula
∃y(y · y = x)
Example.
1. Let L = {<} where < is a binary relation symbol. For every n ∈ N, the set {n} is definable in (N, <).
To see this, first define ϕn (x) to be the formula
^ n
^
∃y1 ∃y2 · · · ∃yn ( (yi 6= yj ) ∧ (yi < x))
1≤i<j≤n i=1
Now notice that {0} is definable as witnessed by the formula
¬∃y(y < x)
and for each n ∈ N+ , the set {n} is definable as witnessed by the formula
ϕn (x) ∧ ¬ϕn+1 (x)
2. Let L = {e, f} where e is a constant symbol and f is a binary function symbol. Let (G, e, ·) be a group
interpreted as an L-stucture. The center of G is definable in (G, e, ·) as witnessed by the formula
∀y(f(x, y) = f(y, x))
Sometimes, there isn’t an obvious way to show that a set is definable, but some cleverness really pays off.
Examples. Let L = {0, 1, +, ·} where 0 and 1 are constant symbols and + and · are binary function symbols.
1. The set N is definable in (Z, 0, 1, +, ·) as witnessed by the formula:
∃y1 ∃y2 ∃y3 ∃y4 (x = y1 · y1 + y2 · y2 + y3 · y3 + y4 · y4 )
because by Lagrange’s Theorem, every natural number is the sum of four squares.
2. The set X = {(k, m, n) ∈ N3 : k m = n} is definable in (N, 0, 1, +, ·), as is the set {(m, n) ∈ N2 : m is

the nth digit in the decimal expansion of π}. These are nontrivial result we’ll prove later.
3. The set Z is definable in (Q, 0, 1, +, ·). This is a deep result of Julia Robinson.
4. Let (R, 0, 1, +, ·) be a commutative ring. The Jacobson radical of R is definable in (R, 0, 1, +, ·) as

witnessed by the formula
∀y∃z((x · y) · z = z + 1)
As for elementary classes, it’s clear how to attempt to show that something is definable (although as
we’ve seen this may require a great deal of cleverness). However, it’s not at all obvious how one could show
that a set is not definable. We’ll develop a few tools to do this in time.
4.2.4 Substitution
In time, we will see the need to “substitute” terms for variables. Roughly, you would think that if ∀xϕ was
true in some structure, then when you take any term and substitute it in for x in the formula ϕ, the resulting
formula would be true. We need a way to relate truth before substituting with truth after substituting. The
hope would be the following, where we use the notation ϕtx to intuitively mean that you substitute t for x:
Hope 4.2.15. Let M be an L-structure, let s : V ar → M , let t ∈ T ermL , and let x ∈ V ar. For all
ϕ ∈ F ormL , we have
(M, s) ϕtx if and only if (M, s[x ⇒ s(t)]) ϕ
In order to make this precise, we first need to define substitition. Even with the “correct” definition of
substitution, however, the above statement is not true. Let’s first define substitution for terms and show
that it behaves well.
Definition 4.2.16. Let x ∈ V ar and let t ∈ T ermL . We define a function Substtx : T ermL → T ermL
denoted by utx as follows.
1. ctx = c for all c ∈ C.

(
t t if y = x
2. yx =
y otherwise
for all y ∈ V ar.
3. (fu1 u2 . . . uk )tx = f(u1 )tx (u2 )tx · · · (uk )tx for all f ∈ Fk and all u1 , u2 , . . . , uk ∈ T ermL .
Here’s the key lemma that relates how to interpret a term before and after substitition.
Lemma 4.2.17. Let M be an L-structure, let s : V ar → M , let t ∈ T ermL , and let x ∈ V ar. For all
u ∈ T ermL , we have
s(utx ) = s[x ⇒ s(t)](u)
Proof. The proof is by induction on T ermL . For any c ∈ C, we have
s(ctx ) = s(c)
= cM
= s[x ⇒ s(t)](c)
= s[x ⇒ s(t)](c)
Suppose that u = x. We then have
s(xtx ) = s(t)
= s[x ⇒ s(t)](x)
= s[x ⇒ s(t)](x)
Suppose that u = y ∈ V ar and that y 6= x. We then have
s(yxt ) = y
= s[x ⇒ s(t)](y)
= s[x ⇒ s(t)](y)
Finally, suppose that f ∈ Fk and that the result holds for u1 , u2 , . . . , uk ∈ T ermL . We then have
s((fu1 u2 · · · uk )tx ) = s(f(u1 )tx (u2 )tx · · · (uk )tx )

= f M (s((u1 )tx ), s((u2 )tx ), . . . , s((uk )tx ))
= f M (s[x ⇒ s(t)](u1 ), s[x ⇒ s(t)](u2 ), . . . , s[x ⇒ s(t)](uk )) (by induction)
= s[x ⇒ s(t)](fu1 u2 · · · uk )
With subsitution in terms defined, we now move to define substitution in formals. The key fact about
this definition is that we only replace x by the term t for the free occurances of x because we certainly don’t
want to change ∀xϕ into ∀tϕ, nor do we want to mess with an x inside the scope of such a quantifier. We
thus make the following recursive definition.
Definition 4.2.18. We now define F reeSubstt,x : F ormL → F ormL , again denoted ϕtx , as follows.
1. (Ru1 u2 · · · uk )tx = R(u1 )tx (u2 )tx · · · (uk )tx for all R ∈ Rk and all u1 , u2 , . . . , uk ∈ T ermL .
2. (= u1 u2 )tx = “ = (u1 )tx (u2 )tx ” for all u1 , u2 ∈ T ermL .
3. (¬ϕ)tx = ¬(ϕtx ) for all ϕ ∈ F ormL .
4. (3ϕψ)tx = 3ϕtx ψxt for all ϕ, ψ ∈ F ormL and all 3 ∈ {∧, ∨, →}.
(
t Qyϕ if x = y
5. (Qyϕ)x =
Qy(ϕtx ) otherwise
for all ϕ ∈ F ormL , y ∈ V ar, and Q ∈ {∃, ∀}.
With the definition in hand, let’s analyze the above hope. Suppose that L = ∅, and consider the formula
ϕ(y) ∈ F ormL given by
∃x¬(x = y)
For any L-structure M and any s : V ar → M , we have (M, s) ϕ if and only if |M | ≥ 2. Now notice that
the formula ϕxy is
∃x¬(x = x)
so for any L-structure M and any s : V ar → M , we have (M, s) 6 ϕxy . Therefore, the above hope fails
whenever M is an L-stucture with |M | ≥ 2. The problem is that the term we substituted (in this case x)
had a variable which became “captured” by a quantifier, and thus the “meaning” of the formula became
transformed. We thus defined a function which indicates with this does not happen.
Definition 4.2.19. Let t ∈ T ermL and let x ∈ V ar. We define a function V alidSubsttx : F ormL → {0, 1}
as follows.
1. V alidSubsttx (ϕ) = 1 for all ϕ ∈ AtomicF ormL .
2. V alidSubsttx (¬ϕ) = V alidSubsttx (ϕ) for all ϕ ∈ F ormL .

(
t 1 if V alidSubsttx (ϕ) = 1 and V alidSubsttx (ψ) = 1
3. V alidSubstx (3ϕψ) =
0 otherwise
for all ϕ, ψ ∈ F ormL and all 3 ∈ {∧, ∨, →}.

1
 if x ∈
/ F reeV ar(Qyϕ)
t
4. V alidSubstx (Qyϕ) = 1 / OccurV ar(t) and V alidSubsttx (ϕ) = 1
if y ∈

0 otherwise

for all ϕ ∈ F ormL , x, y ∈ V ar, and Q ∈ {∀, ∃}.
Theorem 4.2.20 (Substitution Theorem). Let M be an L-structure, let s : V ar → M , let t ∈ T ermL , and
let x ∈ V ar. For all ϕ ∈ F ormL with V alidSubsttx (ϕ) = 1, we have
(M, s) ϕtx if and only if (M, s[x ⇒ s(t)]) ϕ
Proof. The proof is by induction on ϕ. We first handle the case when ϕ ∈ AtomicF ormL . Suppose that
R ∈ Rk and that u1 , u2 , . . . , uk ∈ T ermL . We then have
(M, s) (Ru1 u2 · · · uk )tx ⇔ (M, s) R(u1 )tx (u2 )tx · · · (uk )tx
⇔ (s((u1 )tx ), s((u2 )tx ), · · · , s((uk )tx )) ∈ RM
⇔ (s[x ⇒ s(t)](u1 ), s[x ⇒ s(t)](u2 ), · · · , s[x ⇒ s(t)](uk )) ∈ RM
⇔ (M, s[x ⇒ s(t)]) Ru1 u2 · · · uk
If u1 , u2 ∈ T ermL , we have
(M, s) (= u1 u2 )tx ⇔ (M, s) = (u1 )tx (u2 )tx

⇔ s((u1 )tx ) = s((u2 )tx )
⇔ s[x ⇒ s(t)](u1 ) = s[x ⇒ s(t)](u2 )
⇔ (M, s[x ⇒ s(t)]) = u1 u2
Suppose that the results holds for ϕ and that V alidSubsttx (¬ϕ) = 1. We then have that V alidSubsttx (ϕ) = 1,
and hence
(M, s) (¬ϕ)tx ⇔ (M, s) ¬(ϕtx )

⇔ (M, s) 6 ϕtx
⇔ (M, s[x ⇒ s(t)]) 6 ϕ (by induction)
⇔ (M, s[x ⇒ s(t)]) ¬ϕ
The connectives ∧,∨, and → are similarly uninteresting.

Suppose that the result holds for ϕ and that V alidSubsttx (∃yϕ) = 1. First, if x ∈
/ F reeV ar(∃yϕ), we have
(M, s) (∃yϕ)tx ⇔ (M, s) ∃yϕ

⇔ (M, s[x ⇒ s(t)]) ∃yϕ
Suppose then that x ∈ F reeV ar(∃yϕ), so in particular x 6= y. Since V alidSubsttx (∃yϕ) = 1, we have that
y∈/ OccurV ar(t), and also that V alidSubsttx (ϕ) = 1. Now using the fact that y ∈
/ OccurV ar(t), it follows
that s[y ⇒ a](t) = s(t) for every a ∈ M . Therefore,
(M, s) (∃yϕ)tx ⇔ (M, s) ∃y(ϕtx )

⇔ There exists a ∈ M such that (M, s[y ⇒ a]) ϕtx
⇔ There exists a ∈ M such that (M, (s[y ⇒ a])[x ⇒ s[y ⇒ a](t)]) ϕ (by induction)
⇔ There exists a ∈ M such that (M, (s[y ⇒ a])[x ⇒ s(t)]) ϕ
⇔ There exists a ∈ M such that (M, (s[x ⇒ s(t)])[y ⇒ a]) ϕ
⇔ (M, s[x ⇒ s(t)]) ∃yϕ
Suppose that the result holds for ϕ and that V alidSubsttx (∀yϕ) = 1. First, if x ∈
/ F reeV ar(∃yϕ), we have
(M, s) (∀yϕ)tx ⇔ (M, s) ∀yϕ

⇔ (M, s[x ⇒ s(t)]) ∀yϕ
Suppose then that x ∈ F reeV ar(∀yϕ), so in particular x 6= y. Since V alidSubsttx (∀yϕ) = 1, we have that
y∈/ OccurV ar(t) and also that V alidSubsttx (ϕ) = 1. Now using the fact that y ∈/ OccurV ar(t), it follows
that s[y ⇒ a](t) = s(t) for every a ∈ M . Therefore,
(M, s) (∀yϕ)tx ⇔ (M, s) ∀y(ϕtx )

⇔ For all a ∈ M , we have (M, s[y ⇒ a]) ϕtx
⇔ For all a ∈ M , we have (M, (s[y ⇒ a])[x ⇒ s[y ⇒ a](t)]) ϕ
⇔ For all a ∈ M , we have (M, (s[y ⇒ a])[x ⇒ s(t)]) ϕ
⇔ For all a ∈ M , we have (M, (s[x ⇒ s(t)])[y ⇒ a]) ϕ
⇔ (M, s[x ⇒ s(t)]) ∀yϕ
4.3 Relationships Between Structures

4.3.1 Homomorphisms and Embeddings
Definition 4.3.1. Let L be a language, and let M and N be L-structures.
4.3. RELATIONSHIPS BETWEEN STRUCTURES 67
1. A function h : M → N is called a homomorphism if
(a) For all c ∈ C, we have h(cM ) = cN

(b) For all R ∈ Rk and all a1 , a2 , . . . , ak ∈ M, we have
(a1 , a2 , . . . , ak ) ∈ RM if and only if (h(a1 ), h(a2 ), . . . , h(ak )) ∈ RN
(c) For all f ∈ Fk and all a1 , a2 , . . . , ak ∈ M, we have
h(f M (a1 , a2 , . . . , ak )) = f N (h(a1 ), h(a2 ), . . . , h(ak ))
2. A function h : M → N is called an embedding if it is an injective homomorphism.
3. A function h : M → N is called an isomorphism if it is a bijective homomorphism.
Notation 4.3.2. Let M and N be L-structures. If M and N are isomorphic, i.e. if there exists an
isomorphism h : M → N , then we write M ∼
= N.
Definition 4.3.3. Let M be an L-structure. An isomorphism h : M → M is called an automorphism.
Theorem 4.3.4. Let L be a language, and let M and N be L-structures. Suppose that h : M → N is a
homomorphism, and suppose that s : V ar → M is a variable assignment.
1. For any t ∈ T ermL , we have h(s(t)) = h ◦ s(t).
2. For every quantifier-free ϕ ∈ F ormL not containing the equality symbol, we have
(M, s) ϕ if and only if (N , h ◦ s) ϕ
3. If h is an embedding, then for every quantifier-free ϕ ∈ F ormL , we have
4. If h is an isomorphism, then for every ϕ ∈ F ormL , we have
Proof.
1. We prove this by induction on t. First, for any c ∈ C, we have
h(s(c)) = h(cM )
= cN (since h is a homomorphism)
= h ◦ s(c).
Now for x ∈ V ar, we have
h(s(x)) = h(s(x)))
= (h ◦ s)(x)
= h ◦ s(x)
Suppose now that f ∈ Fk , that t1 , t2 , . . . , tk ∈ T ermL , and the result holds for each ti . We then have
h(s(ft1 t2 · · · tk )) = h(f M (s(t1 ), s(t2 ), . . . , s(tk )))

= f M (h(s(t1 )), h(s(t2 )), . . . , h(s(tk ))) (since h is a homomorphism)
N
= f (h ◦ s(t1 ), h ◦ s(t2 ), . . . , h ◦ s(tk )) (by induction)
= h ◦ s(ft1 t2 · · · tk )
The result follows by induction
2. Suppose that h is an embedding. We prove the result by induction on ϕ. Suppose first that R ∈ Rk
and that t1 , t2 , . . . , tk ∈ T ermL . We then have
(M, s) Rt1 t2 · · · tk ⇔ (s(t1 ), s(t2 ), . . . , s(tk )) ∈ RM

⇔ (h(s(t1 )), h(s(t2 )), . . . , h(s(tk ))) ∈ RN (since h is a homomorphism)
N
⇔ (h ◦ s(t1 ), h ◦ s(t2 ), . . . , h ◦ s(tk )) ∈ R (by part 1)
⇔ (N , h ◦ s) Rt1 t2 · · · tk
Suppose that the result holds for ϕ. We prove it for ¬ϕ. We have
(M, s) ¬ϕ ⇔ (M, s) 6 ϕ
⇔ (N , h ◦ s) 6 ϕ (by induction)
⇔ (N , h ◦ s) ¬ϕ
(M, s) ϕ ∧ ψ ⇔ (M, s) ϕ and (M, s) ψ

⇔ (N , h ◦ s) ϕ and (N , h ◦ s) ϕ (by induction)
⇔ (N , h ◦ s) ϕ ∧ ψ
and similarly for ∨ and →. The result follows by induction.
3. In light of the proof of 2, we need only show that if ϕ is = t1 t2 where t1 , t2 ∈ T ermL , then (M, s) ϕ
if and only if (N , h ◦ s) ϕ. For any t1 , t2 ∈ T ermL , we have
(M, s) = t1 t2 ⇔ s(t1 ) = s(t2 )

⇔ h(s(t1 )) = h(s(t2 )) (since h is injective)
⇔ h ◦ s(t1 ) = h ◦ s(t2 ) (by part 1)
⇔ (N , h ◦ s) = t1 t2
4. Suppose that the result holds for ϕ and x ∈ V ar. We have
(M, s) ∃xϕ ⇔ There exists a ∈ M such that (M, s[x ⇒ a]) ϕ

⇔ There exists a ∈ M such that (N , h ◦ (s[x ⇒ a])) ϕ
⇔ There exists a ∈ M such that (N , (h ◦ s)[x ⇒ h(a)] ϕ
⇔ There exists b ∈ N such that (N , (h ◦ s)[x ⇒ b]) ϕ (since h is bijective)
⇔ (N , h ◦ s) ∃xϕ
and also
(M, s) ∀xϕ ⇔ For all a ∈ M , we have (M, s[x ⇒ a]) ϕ

⇔ For all a ∈ M , we have (N , h ◦ (s[x ⇒ a])) ϕ
⇔ For all a ∈ M , we have (N , (h ◦ s)[x ⇒ h(a)] ϕ
⇔ For all b ∈ N , we have (N , (h ◦ s)[x ⇒ b]) ϕ (since h is bijective)
⇔ (N , h ◦ s) ∀xϕ
Definition 4.3.5. Let L be a language, and let M and N be L-structures. We write M ≡ N , and say that
M and N are elementarily equivalent, if for all σ ∈ SentL , we have M σ if and only if N σ.
Corollary 4.3.6. Let L be a language, and let M and N be L-structures. If M ∼

= N , then M ≡ N .
4.3.2 An Application To Definability

Proposition 4.3.7. Suppose that M is an L-structure and k ∈ N+ . Suppose also that X ⊆ M k is definable
in M and that h : M → M is an automorphism. For every a1 , a2 , . . . , ak ∈ M , we have
(a1 , a2 , . . . , ak ) ∈ X if and only if (h(a1 ), h(a2 ), . . . , h(ak )) ∈ X
Proof. Fix ϕ(x1 , x2 , . . . , xk ) ∈ F ormL such that
X = {(a1 , a2 , . . . , ak ) ∈ M k : (M, a1 , a2 , . . . , ak ) ϕ}
By part 4 of Theorem 4.3.4, we know that for every a1 , a2 , . . . , ak ∈ M , we have
(M, a1 , a2 , . . . , ak ) ϕ if and only if (M, h(a1 ), h(a2 ), . . . , h(ak )) ϕ
Therefore, for every a1 , a2 , . . . , ak ∈ M , we have
(a1 , a2 , . . . , ak ) ∈ X if and only if (h(a1 ), h(a2 ), . . . , h(ak )) ∈ X
Corollary 4.3.8. Suppose that M is an L-structure and k ∈ N+ . Suppose also that X ⊆ M k is and that
h : M → M is an automorphism. If there exists a1 , a2 , . . . , ak ∈ M such that exactly one of the following
holds:
• (a1 , a2 , . . . , ak ) ∈ X
• (h(a1 ), h(a2 ), . . . , h(ak )) ∈ X
then X is not definiable in M.
Example. Let L = {R} where R is a binary relation symbol, and let M be the L-structure where M = Z
and RM = {(a, b) ∈ Z2 : a < b}. We show that a set X ⊆ M is definable in M if and only if either X = ∅ or
X = Z. First notice that ∅ is definable as witnessed by ¬(x = x) and Z as witnessed by x = x. Suppose now
that X ⊆ Z is such that X 6= ∅ and X 6= Z. Fix a, b ∈ Z such that a ∈ X and b ∈ / X. Define h : M → M
by letting h(c) = c + (b − a) for all c ∈ M . Notice that h is automorphism of M because it is bijective (the
map g(c) = c − (b − a) is clearly an inverse) and a homomorphism because if c1 , c2 ∈ Z, then have have
(c1 , c2 ) ∈ RM ⇔ c1 < c2
⇔ c1 + (b − a) < c2 + (b − a)
⇔ h(c1 ) < h(c2 )
⇔ (h(c1 ), h(c2 )) ∈ RM
Notice also that h(a) = a + (b − a) = b, so a ∈ X but h(a) ∈

/ X. It follows from the proposition that X is
not definable in M.
4.3.3 Substructures
Definition 4.3.9. Let L be a language and let M and A be L-structures. We say that A is a substructure
of M, and we write A ⊆ M if
1. A ⊆ M .
2. cA = cM for all c ∈ C.
3. RA = RM ∩ Ak for all R ∈ Rk .
4. f A = f M Ak for all f ∈ Fk .
Remark 4.3.10. Let L be a language and let M and A be L-structures with A ⊆ M . We then have that
A ⊆ M if and only if the identity map ι : A → M is a homomorphism.
Remark 4.3.11. Suppose that M is an L-structure and that A ⊆ M . A is the universe of a substructure
of M if and only if {cM : c ∈ C} ⊆ A and f M (a1 , a2 , . . . , ak ) ∈ A for all f ∈ Fk and all a1 , a2 , . . . , ak ∈ A.
Proposition 4.3.12. Let M be an L-structure and let B ⊆ M . Suppose either that B 6= ∅ or C 6= ∅. If we

let A = G(M, B ∪ {cM : c ∈ C}, {f M : f ∈ F }), then A is the universe of a substructure of M. Moreover, if
N ⊆ M with B ⊆ N , then A ⊆ N .
Proposition 4.3.13. Let L be a language.
1. A Σ1 -formula is an element of G(Sym∗L , QuantF reeF ormL , {h∃,x : x ∈ V ar}).
2. A Π1 -formula is an element of G(Sym∗L , QuantF reeF ormL , {h∀,x : x ∈ V ar}).
Proposition 4.3.14. Suppose that A ⊆ M.
1. For any ϕ ∈ QuantF reeF ormL and any s : V ar → A, we have
(A, s) ϕ if and only if (M, s) ϕ
2. For any Σ1 -formula ϕ ∈ F ormL and any s : V ar → A, we have
If (A, s) ϕ, then (M, s) ϕ
3. For any Π1 -formula ϕ ∈ F ormL and any s : V ar → A, we have
If (M, s) ϕ, then (A, s) ϕ

Proof.
1. This follows from Remark 4.3.10 and Theorem 4.3.4.
2. We prove this by induction. If ϕ is quantifier-free, this follows from part 1. Suppose that we know the
result for ϕ, and suppose that (A, s) ∃xϕ. Fix a ∈ A such that (A, s[x ⇒ a]) ϕ. By induction, we
know that (M, s[x ⇒ a]) ϕ, hence (M, s) ∃xϕ.
3. We prove this by induction. If ϕ is quantifier-free, this follows from part 1. Suppose that we know the
result for ϕ, and suppose that (M, s) ∀xϕ. For every a ∈ A, we then have (M, s[x ⇒ a]) ϕ, and
hence (A, s[x ⇒ a]) ϕ by induction. It follows that (A, s) ∀xϕ.
4.3.4 Elementary Substructures

Definition 4.3.15. Let L be a language and let M and A be L-structures. We say that A is an elementary
substructure of M if A ⊆ M and for all ϕ ∈ F ormL and all s : V ar → A, we have
(A, s) ϕ if and only if (M, s) ϕ
We write A M to mean that A is an elementary substructure of M.

Example. Let L = {f} where f is a unary function symbol. Let M be the L-structure with M = N and
f M (n) = n + 1. Let A be L-structure with A = N+ and f A (n) = n + 1. We then have that A ⊆ M.
Furthermore, we have M ∼= A, hence for all σ ∈ SentL we have
A σ if and only if M σ
However, notice that A 6 M because if ϕ(x) is the formula ¬∃y(fy = x), we then have that (A, 1) ϕ but
(M, 1) 6 ϕ.
Theorem 4.3.16 (Tarski-Vaught Test). Suppose that A ⊆ M. The following are equivalent.
1. A M.
2. Whenever ϕ ∈ F ormL , x ∈ V ar, and s : V ar → A satisfy (M, s) ∃xϕ, there exists a ∈ A such that
(M, s[x ⇒ a]) ϕ
Proof. We first prove that 1 implies 2. Suppose then that A M. Let ϕ ∈ F ormL and s : V ar → A be
such that (M, s) ∃xϕ. Using the fact that A M, it follows that (A, s) ∃xϕ. Fix a ∈ A such that
(A, s[x ⇒ a]) ϕ. Using again the fact that A M, we have (M, s[x ⇒ a]) ϕ.
We now prove that 2 implies 1. We prove by induction on ϕ ∈ F ormL that for all s : V ar → A, we have
(A, s) ϕ if and only if (M, s) ϕ. That is, we let
X = {ϕ ∈ F ormL : For all s : V ar → A we have (A, s) ϕ if and only if (M, s) ϕ}
and prove that X = F ormL by induction. First notice that ϕ ∈ X for all quantifier-free ϕ because A ⊆ M.
Suppose now that ϕ ∈ X. For any s : V ar → A, we have
(A, s) ¬ϕ ⇔ (A, s) 6 ϕ
⇔ (M, s) 6 ϕ (since ϕ ∈ X)
⇔ (M, s) ¬ϕ
Therefore, ¬ϕ ∈ X.
Suppose now that ϕ, ψ ∈ X. For any s : V ar → A, we have
(A, s) ϕ ∧ ψ ⇔ (A, s) ϕ and (A, s) ψ
⇔ (M, s) ϕ and (M, s) ψ (since ϕ, ψ ∈ X)
⇔ (M, s) ϕ ∧ ψ
Therefore, ϕ ∧ ψ ∈ X. Similarly, we have ϕ ∨ ψ ∈ X and ϕ → ψ ∈ X.
Suppose now that ϕ ∈ X and x ∈ V ar. For any s : V ar → A, we have
(A, s) ∃xϕ ⇔ There exists a ∈ A such that (A, s[x ⇒ a]) ϕ
⇔ There exists a ∈ A such that (M, s[x ⇒ a]) ϕ (since ϕ ∈ X)
⇔ (M, s) ∃xϕ (by our assumption 2)
Therefore, ∃xϕ ∈ X.
Suppose now that ϕ ∈ X and x ∈ V ar. We then have that ¬ϕ ∈ X from above, hence ∃x¬ϕ ∈ X from
above, hence ¬∃x¬ϕ ∈ X again from above. Thus, for any s : V ar → A, we have
(A, s) ∀xϕ ⇔ (A, s) ¬∃x¬ϕ
⇔ (M, s) ¬∃x¬ϕ (since ¬∃x¬ϕ ∈ X)
⇔ (M, s) ∀xϕ
Therefore, ∀xϕ ∈ X.
Theorem 4.3.17 (Countable Lowenheim-Skolem-Tarski Theorem). Suppose that L is countable, that M is
an L-structure, and that X ⊆ M is countable. There exists a countable A M such that X ⊆ A.
Proof. Fix an element d ∈ M (this will be used as a “dummy” element of M to ensure that we always have
something to go to when all else fails).
For each ϕ ∈ F ormL and x ∈ V ar such that F reeV ar(ϕ) = {x}, we define an element nϕ,x ∈ M as
follows. If M ∃xϕ, fix an arbitrary m ∈ M such that (M, m) ϕ, and let nϕ,x = m. Otherwise, let
nϕ,x = d.
Now for each ϕ ∈ F ormL and x ∈ V ar such that {x} ( F reeV ar(ϕ), we define a function. Suppose that
F reeV ar(ϕ) = {y1 , y2 , . . . , yk , x}. We define a function hϕ,x : M k → M as follows. Let b1 , b2 , . . . , bk ∈
M . If (M, b1 , b2 , . . . , bk ) ∃xϕ, fix an arbitrary a ∈ M such that (M, b1 , b2 , . . . , bk , a) ϕ, and let
hϕ,x (b1 , b2 , . . . , bk ) = a. Otherwise, let hϕ,x (b1 , b2 , . . . , bk ) = d.
We now let
B = X ∪ {d} ∪ {cM : c ∈ C} ∪ {nϕ,x : x ∈ V ar, ϕ ∈ F ormL , and F reeV ar(ϕ) = {x}}
and we let
A = G(M, B, {f M : f ∈ Fk } ∪ {hϕ,x : ϕ ∈ F ormP , x ∈ V ar})
We then have that A is the universe of a substructure A of M. Notice that X ⊆ A and that by a problem
on Homework 1, we have that A is countable. Thus, we need only show that A M, which we do by the
Tarski-Vaught test. Suppose that ϕ ∈ F ormL , x ∈ V ar, and s : V ar → A are such that (M, s) ∃xϕ.
Suppose first that x ∈ / F reeV ar(ϕ). Since (M, s) ∃xϕ, we may fix m ∈ M such that (M, s[x ⇒ m]) ϕ.
Now using the fact that x ∈ / F reeV ar(ϕ), it follows that (M, s[x ⇒ d]) ϕ.
Suppose now that F reeV ar(ϕ) = {x}, and let a = nϕ,x ∈ A. Since M ∃xϕ, hence (M, a) ϕ by
definition of nϕ,x . It follows that there exists a ∈ A such that (M, s[x ⇒ a]) ϕ.
Finally, suppose that F reeV ar(ϕ) = {y1 , y2 , . . . , yk , x}. For each i with 1 ≤ i ≤ k, let bi = s(yi ), and let
a = hϕ,x (b1 , b2 , . . . , bk ) ∈ A. Since (M, b1 , b2 , . . . , bk ) ∃xϕ, hence (M, b1 , b2 , . . . , bk , a) ϕ by definition of
hϕ,x . It follows that there exists a ∈ A such that (M, s[x ⇒ a]) ϕ.
4.4. CHANGING THE LANGUAGE 73
Corollary 4.3.18. Suppose that L is countable and that M is an L-structure. There exists a countable
L-structure N such that N ≡ M.
Proof. Let N be a countable elementary substructure of M. For any σ ∈ SentL , we then have that N σ
if and only if M σ, so N ≡ M.
This is our first indication that first-order logic is not powerful enough to distinguish certain aspects of
cardinality, and we’ll see more examples of this phenomenon after the Compactness Theorem (for first-order
logic) and once we talk about infinite cardinalities and extend the Lowenheim-Skolem-Tarski result.
This restriction already has some interesting consequences. For example, you may be familiar with the
result that (R, 0, 1, <, +, ·) is the unique (up to isomorphism) Dedekind-complete ordered field.
Corollary 4.3.19. The Dedekind-complete ordered fields are not a weak elementary class in the language
L = {0, 1, <, +, ·}.
Proof. Let K be the class of all Dedekind-complete ordered fields. Suppose that Σ ⊆ SentL is such that
K = M od(Σ). By the Countable Lowenheim-Skolem-Tarski Theorem, there exists a countable N such that
N ≡ (R, 0, 1, <, +, ·). Since (R, 0, 1, <, +, ·) ∈ K, we have (R, 0, 1, <, +, ·) σ for all σ ∈ Σ, so N σ for all
σ ∈ Σ, and hence N ∈ K. However, this is a contradiction because all Dedekind-complete ordered fields are
isomorphic to (R, 0, 1, <, +, ·), hence are uncountable.
Definition 4.3.20. Let L be a language and let M and N be L-structures. Suppose that h : N → M . We
say that h is an elementary embedding if h is an embedding and for all ϕ ∈ F ormL and all s : V ar → N ,
we have
(N , s) ϕ if and only if (M, h ◦ s) ϕ
4.4 Changing the Language

4.4.1 Expansions and Restrictions
Definition 4.4.1. Let L ⊆ L0 be languages, let M be an L-structure, and let M0 be an L0 -structure. We
say that M is the restriction of M0 to L, and that M0 is an expansion of M to L0 , if
• M = M 0.
0
• cM = cM for all c ∈ C.
0
• RM = RM for all R ∈ R.
0
• f M = f M for all f ∈ F.
Proposition 4.4.2. Let L ⊆ L0 be languages, let M0 be an L0 -structure, and let M be the restriction of M0
to L. For all ϕ ∈ F ormL and all s : V ar → M , we have (M, s) ϕ if and only if (M0 , s) ϕ.
Proof. By induction.
4.4.2 Adding Constants to Name Elements

Definition 4.4.3. Let L be a language and let M be an L-structure. For each a ∈ M , introduce a new
constant ca (not appearing in the original language L and all distinct). Let LM = L ∪ {ca : a ∈ M } and let
M
Mexp be the LM -structure which is the expansion of M in which ca exp = a for all a ∈ M . We call Mexp
the expansion of M obtained by adding names for elements of M .
Definition 4.4.4. Let M be an L-structure, and let s : V ar → M be a variable assignment. Define a

function N ames : T ermL → T ermLM by plugging in names for free variables according to s. Define a
function N ames : F ormL → SentLM again by plugging in names for free variables according to s.
Proposition 4.4.5. Let M be an L-structure, and let s : V ar → M be a variable assignment. For every
ϕ ∈ F ormL , we have
(M, s) ϕ if and only if Mexp N ames (ϕ)
Definition 4.4.6. Let M be an L-structure.
• We let AtomicDiag(M) = {σ ∈ SentLM ∩ AtomicF ormLM : Mexp σ}.
• We let Diag(M) = {σ ∈ SentLM : Mexp σ}.
Proposition 4.4.7. Let L be a language and let M and N be the L-structures. The following are equivalent:
• There exists an embedding h from M to N .
• There exists an expansion of N to an LM -structure which is a model of AtomicDiag(M).
Proposition 4.4.8. Let L be a language and let M and N be the L-structures. The following are equivalent:
• There exists an elementary embedding h from M to N .
• There exists an expansion of N to an LM -structure which is a model of Diag(M).
Chapter 5
Semantic and Syntactic Implication
5.1 Semantic Implication and Theories

5.1.1 Definitions
Definition 5.1.1. Let L be a language and let Γ ⊆ F ormL . A model of Γ is a pair (M, s) where
• M is an L-structure.
• s : V ar → M is a variable assignment.
• (M, s) γ for all γ ∈ Γ.
Definition 5.1.2. Let L be a language. Let Γ ⊆ F ormL and let ϕ ∈ F ormL . We write Γ ϕ to mean that
whenever (M, s) is a model of Γ, we have that (M, s) ϕ. We pronounce Γ ϕ as Γ semantically implies
ϕ.
Definition 5.1.3. Let L be a language and let Γ ⊆ F ormL . We say that Γ is satisfiable if there exists a
model of Γ.
Definition 5.1.4. Let L be a language. A set Σ ⊆ SentL is an L-theory if
• Σ is a satisfiable.
• For every τ ∈ SentL with Σ τ , we have τ ∈ Σ.
There are two standard ways to get theories. One is to take a stucture, and consider all of the sentences
that are true in that structure.
Definition 5.1.5. Let M be an L-structure. We let T h(M) = {σ ∈ SentL : M σ}. We call T h(M) the
theory of M.
Proposition 5.1.6. Let L be a language and let M be an L-structure. T h(M) is an L-theory.
Proof. First notice that T h(M) is satisfiable because M is a model of T h(M) (since M σ for all σ ∈
T h(M) by definition. Suppose now that τ ∈ SentL is such that T h(M) τ . Since M is a model of T h(M),
it follows that M τ , and hence τ ∈ T h(M).
Another standard way to get a theory is to take an arbitrary satisfiable set of sentences, and close it off
under semantic implication.
75
76 CHAPTER 5. SEMANTIC AND SYNTACTIC IMPLICATION
Definition 5.1.7. Let L be a language and let Σ ⊆ SentL . We let Cn(Σ) = {τ ∈ SentL : Σ τ }. We call
Cn(Σ) the set of consequences of Σ.
Proposition 5.1.8. Let L be a language and let Σ ⊆ SentL be satisfiable. We then have that Cn(Σ) is an
L-theory.
Proof. We first show that Cn(Σ) is satisfiable. Since Σ is satsfiable, we may fix a model M of Σ. Let
τ ∈ Cn(Σ). We then have that Σ τ , so using the fact that M is a model of Σ we conclude that M τ .
Therefore, M is a model of Cn(Σ), hence Cn(Σ) is satisfiable.
Suppose now that τ ∈ SentL and that Cn(Σ) τ . We need to show that τ ∈ Cn(Σ), i.e. that Σ τ .
Let M be a model of Σ. Since Σ σ for all σ ∈ Cn(Σ), it follows that M σ for all σ ∈ Cn(Σ). Thus, M
is a model of Cn(Σ). Since Cn(Σ) τ , it follows that M τ . Thus, Σ τ , and so τ ∈ Cn(Σ).
Definition 5.1.9. An L-theory Σ is complete if for all τ ∈ SentL , either τ ∈ Σ or ¬τ ∈ Σ.
Proposition 5.1.10. Let L be a language and let M be an L-theory. T h(M) is a complete L-theory.
Proof. We’ve already seen that T h(M) is a theory. Suppose now that σ ∈ SentL . If M σ, we then have
that σ ∈ T h(M). Otherwise, we have M 6 σ, so by definition M ¬σ, and hence ¬σ ∈ T h(M).
Example. Let L = {f, e} where f is a binary function symbol and e is a constant symbol. Consider the
following sentences.
ϕ1 = ∀x∀y∀z(f(f(x, y), z) = f(x, f(x, y)))
ϕ2 = ∀x(f(x, e) = x ∧ f(e, x) = x)
ϕ3 = ∀x∃y(f(x, y) = e ∧ f(y, x) = e)
The theory T = Cn({ϕ1 , ϕ2 , ϕ3 }) is the theory of groups. T is not complete because it neither contains
∀x∀y(f(x, y) = f(y, x)) nor its negation because there are both abelian groups and nonabelian groups.
Definition 5.1.11. Let L = {R} where R is a binary relation symbol. Consider the following sentences
ϕ1 = ∀x¬Rxx
ϕ2 = ∀x∀y∀z((Rxy ∧ Ryz) → Rxz)
ϕ3 = ∀x∀y¬(Rxy ∧ Ryx)
ϕ4 = ∀x∀y(x = y ∨ Rxy ∨ Ryx)
and let LO = Cn({ϕ1 , ϕ2 , ϕ3 , ϕ4 }). LO is called the theory of (strict) linear orderings. LO is not complete
because it neither contains ∃y∀x(x = y ∨ x < y) nor its negation because there are linear ordering with greatest
elements and linear orderings without greatest elements.
Definition 5.1.12. Let L = {R} where R is a binary relation symbol. Consider the following sentences
ϕ1 = ∀x¬Rxx
ϕ2 = ∀x∀y∀z((Rxy ∧ Ryz) → Rxz)
ϕ3 = ∀x∀y¬(Rxy ∧ Ryx)
ϕ4 = ∀x∀y(x = y ∨ Rxy ∨ Ryx)
ϕ5 = ∀x∀y(Rxy → ∃z(Rxz ∧ Rzy))
ϕ6 = ∀x∃yRxy
ϕ7 = ∀x∃yRyx
and let DLO = Cn({ϕ1 , ϕ2 , ϕ3 , ϕ4 , ϕ5 , ϕ6 , ϕ7 }). DLO is called the theory of dense (strict) linear orderings
without endpoints. DLO is complete as we’ll see below.
5.1. SEMANTIC IMPLICATION AND THEORIES 77
Theorem 5.1.13 (Countable Lowenheim-Skolem Theorem). Suppose that L is countable and that Γ ⊆
F ormL is satisfiable. There exists a countable model (M, s) of Γ.
Proof. Since Γ is satisfiable, we may fix a model (N , s) of Γ. Let X = ran(s) ⊆ N and notice that X
is countable. By the Countable Lowenheim-Skolem-Tarski Theorem, there exists a countable elementary
substructure M N such that X ⊆ M . Notice that s is also a variable assigment on M . Now for any
γ ∈ Γ, we have that (N , s) γ because (N , s) is a model of Γ, hence (M, s) γ because M N . It follows
that (M, s) is a model of Γ.
5.1.2 Finite Models of Theories

Given a theory T and an n ∈ N+ , we want to count the number of models of T of cardinality n up to
isomorphism. There are some technical set-theoretic difficulties here which will be elaborated upon later,
but the key fact that limits the number of isomorphism classes is the following result.
Proposition 5.1.14. Let L be a language and let n ∈ N+ . For every L-structure M with |M | = n, there
exists an L-stucture N with N = [n] such that M ∼
= N.
Proof. Let M be an L-structure with |M | = n. Fix a bijection h : M → [n]. Let N be the L-structure where
• N = [n].
• cN = h(cM ) for all c ∈ C.
• RN = {(b1 , b2 , . . . , bk ) ∈ N k : (h−1 (b1 ), h−1 (b2 ), . . . , h−1 (bk )) ∈ RN } for all R ∈ Rk .
• f N is the function from N k to N defined by f N (b1 , b2 , . . . , bk ) = h(f M (h−1 (b1 ), h−1 (b2 ), . . . , h−1 (bk )))
for all f ∈ F k .
We then have that h is an isomorphism from M to N .
Proposition 5.1.15. If L is finite and n ∈ N+ , then there are only finitely many L-structures with universe
[n].
Definition 5.1.16. Let L be a finite language and let T be an L-theory. For each n ∈ N+ , let I(T, n) be the
number of models of T of cardinality n up to isomorphism. Formally, we consider the set of all L-structures
with universe [n], and count the number of equivalence classes under the equivalence relation of isomorphism.
Example 5.1.17. If T is the theory of groups, then I(T, n) is a very interesting function that you study
in algebra courses. For example, you show that I(T, p) = 1 for all primes p, that I(T, 6) = 2, and that
I(T, 8) = 5.
Example 5.1.18. Let L = ∅ and let T = Cn(∅). We have I(T, n) = 1 for all n ∈ N+ .
Proof. First notice that for every n ∈ N+ , the L-structure M with universe [n] is a model of T of cardinality
n, so I(T, n) ≥ 1. Now notice that if M and N are models of T of cardinality n, then any bijection
h : M → N is an isomorphism (because L = ∅), so I(T, n) ≤ 1. It follows that I(T, n) = 1 for all n ∈ N.
Example 5.1.19. I(LO, n) = 1 for all n ∈ N+ .
Proof. First notice that for every n ∈ N, the L-strucutre M where M = [n] and RM = {(k, `) ∈ [n]2 : k < `}
is a model of LO of cardinality n, so I(LO, n) ≥ 1. Next notice that any two linear orderings of cardinality
n are isomorphic. Intuitively, this works as follows. Notice (by induction on the number of elements) that
every finite linear ordering has a least element. Let M and N be two linear orderings of cardinality n.
Each must have a least element, so map the least element of M to that of N . Remove these elements, then
map the least element remaining in M to the least element remaining in N , and continue. This gives an
isomorphism. Formally, you can turn this into a proof by induction on n.
Example 5.1.20. Let L = {f} where f is a unary function symbol, and let T = Cn({∀x(ffx = x)}). We have
I(T, n) = b n2 c + 1 for all n ∈ N+ .
Proof. Let’s first analyze the finite models of T . Suppose that M is a model of T of cardinality n. For every
a ∈ M , we then have f M (f M (a)) = a. There are now two cases. Either f M (a) = a, or f M (a) = b 6= a in
which case f M (b) = a. Let
• F ixM = {a ∈ M : f M (a) = a}.
• M oveM = {a ∈ M : f M (a) 6= a}.
From above, we then have that |M oveM | is even and that |F ixM | + |M oveM | = n. Now the idea is that
two models M and N of T of cardinality n are isomorphic if and only if they have the same number of fixed
points, because then we can match up the fixed points and then match up the “pairings” left over to get an
isomorphism. Here’s a more formal argument.
We know show that if M and N are models of T of cardinality n, then M ∼ = N if and only if |F ixM | =
|F ixN |. Clearly, if M ∼= N , then |F ixM | = |F ixN |. Suppose conversely that |F ixM | = |F ixN |. We then
must have |M oveM | = |M oveN |. Let XM ⊆ M oveM be a set of cardinality |M ove 2
M|
such that f M (x) 6= y
M
for all x, y ∈ X (that is, we pick out one member from each pairing given by f ), and let XN be such
a set for N . Define a function h : M → N . Fix a bijection from α : F ixM → F ixN and a bijection
β : XM → XN . Define h by letting h(a) = α(a) for all a ∈ F ixM , letting h(x) = β(x) for all x ∈ XM , and
letting h(y) = f N (β(f M (y))) for all y ∈ M oveM \X. We then have that h is an isomophism from M to N .
Now we need only count how many possible values there are for |F ixM |. Let n ∈ N+ . Suppose first that n
is even. Since |M oveM | must be even, it follows that |F ixM | must be even. Thus, |F ixM | ∈ {0, 2, 4, . . . , n},
so there are n2 + 1 many possibilities, and it’s easy to construct models in which each of these possibilities
occurs. Suppose now that n is odd. Since |M oveM | must be even, it follows that |F ixM | must be odd.
Thus, |F ixM | ∈ {1, 3, 5, . . . , n}, so there are n−1
2 + 1 many possibilities, and it’s easy to construct models in
which each of these possibilities occurs. Thus, in either case, we have I(T, n) = b n2 c + 1.
Example 5.1.21. I(DLO, n) = 0 for all n ∈ N+ .
Proof. As mentioned in the LO example, every finite linear ordering has a least element.
Definition 5.1.22. Suppose that L is a finite language and σ ∈ SentL . Let
Spec(σ) = {n ∈ N+ : I(Cn(σ), n) > 0}
Proposition 5.1.23. There exists a finite language L and a σ ∈ SentL such that Spec(σ) = {2n : n ∈ N+ }.
Proof. We’ll give two separate arguments. First, let L = {e, f} be the language of group theory. Let σ be
the conjunction of the group axioms with the sentence ∃x(¬(x = e) ∧ fxx = e) expressing that there is an
element of order 2. Now for every n ∈ N+ , the group Z/(2n)Z is a model of σ of cardinality 2n because n is
an element of order 2. Thus, {2n : n ∈ N+ } ⊆ Spec(σ). Suppose now that k ∈ Spec(σ), and fix a model M
of σ of order k. We then have that M is a group with an element of order 2, so by Lagrange’s Theorem it
follows that 2 | k, so k ∈ {2n : n ∈ N+ }. It follows that Spec(σ) = {2n : n ∈ N+ }.
For a second example, let L = {R} where R is a binary relation symbol. Let σ be the conjunction of the
following sentences:
• ∀xRxx.
• ∀x∀y(Rxy → Ryx).
• ∀x∀y∀z((Rxy ∧ Ryz) → Rxz).

• ∀x∃y(¬(y = x) ∧ Rxy ∧ ∀z(Rxz → (z = x ∨ z = y))).

Notice that a model of σ is simply an equivalence relation in which every equivalence class has exactly 2
elements. It is now straightforward to show that Spec(σ) = {2n : n ∈ N+ }.
Proposition 5.1.24. There exists a finite language L and a σ ∈ SentL such that Spec(σ) = {2n : n ∈ N+ }.
Proof. Again, let’s give two separate arguments. First, let L = {e, f} be the language of group theory. Let
σ be the conjunction of the group axioms with the sentences ∃x¬(x = e) and ∀x(fxx = e) expressing that
the group is nontrivial and that there every nonidentity element has order 2. Now for every n ∈ N+ , the
group (Z/2Z)n is a model of σ of cardinality 2n . Thus, {2n : n ∈ N+ } ⊆ Spec(σ). Suppose now that
k ∈ Spec(σ), and fix a model M of σ of order k. We then have that k > 1 and that M is a group such that
every nonidentity element has order 2. Now for any prime p 6= 2, it is not the case that p divides k because
otherwise M would have to have an element of order p by Cauchy’s Theorem. Thus, the only prime that
divides k is 2, and so k ∈ {2n : n ∈ N+ }. It follows that Spec(σ) = {2n : n ∈ N+ }.
For a second example, let L = {0, 1, +, ·} be the language where 0, 1 are constant symbols and +, ·
are binary function symbols. Let σ be the conjunction of the field axioms together with 1 + 1 = 0. Thus,
the models of σ are exactly the fields of characteristic 2. By results in algebra, there is a finite field of
characteristic 2 of order k if and only if k is a power of 2.
5.1.3 Countable Models of Theories

Theorem 5.1.25. Suppose that M and N are two countably infinite models of DLO. We then have that
M∼= N.
Proof. Back-and-forth construction. See Damir’s carefully written proof.
Corollary 5.1.26 (Countable Los-Vaught Test). Let L be a countable language. Suppose that T is an L-
theory such that all models of T are infinite, and suppose also that every two countably infinite models of T
are isomorphic. We then have that T is complete.
Proof. Suppose that T is not complete and fix σ ∈ SentL such that σ ∈
/ T and ¬σ ∈ / T . We then have that
T ∪ {σ} and T ∪ {¬σ} are both satisfiable by infinite models (because all models of T are infinite), so by
the Countable Lowenheim-Skolem Theorem we may fix countably infinite models M1 of T ∪ {σ} and M2
of T ∪ {¬σ}. We then have that M1 and M2 are countably infinite models of T which are not isomorphic
(because they are not elementarily equivalent), a contradiction.
Corollary 5.1.27. DLO is complete.
Proposition 5.1.28. Suppose that T is a complete L-theory. If M and N are models of T , then M ≡ N .
Proof. Let σ ∈ SentL . If σ ∈ T , we then have that both M σ and N σ. Suppose that σ ∈
/ T . Since T is
complete, we then have that ¬σ ∈ T , hence M ¬σ and N ¬σ. It follows that both M 6 σ and N 6 σ.
Therefore, for all σ ∈ SentL , we have that M σ if and only if N σ, so M ≡ N .
Corollary 5.1.29. In the language L = {R} where R is a binary relation symbol, we have (Q, <) ≡ (R, <).
5.2 Syntactic Implication

5.2.1 Definitions
Basic Proofs:
• Γ ` ϕ if ϕ ∈ Γ (AssumeL )
• Γ ` t = t for all t ∈ T ermL (EqRef l)
Proof Rules:
Γ`ϕ∧ψ Γ`ϕ∧ψ Γ`ϕ Γ`ψ
(∧EL) (∧ER) (∧I)
Γ`ϕ Γ`ψ Γ`ϕ∧ψ
Γ`ϕ Γ`ψ
(∨IL) (∨IR)
Γ`ϕ∨ψ Γ`ϕ∨ψ
Γ`ϕ→ψ Γ ∪ {ϕ} ` ψ
(→ E) (→ I)
Γ ∪ {ϕ} ` ψ Γ`ϕ→ψ
Γ ∪ {ϕ} ` θ Γ ∪ {ψ} ` θ Γ ∪ {ψ} ` ϕ Γ ∪ {¬ψ} ` ϕ
(∨P C) (¬P C)
Γ ∪ {ϕ ∨ ψ} ` θ Γ`ϕ
Γ ∪ {¬ϕ} ` ψ Γ ∪ {¬ϕ} ` ¬ψ
(Contr)
Γ`ϕ
Equality Rules:
Γ ` ϕtx Γ ` t = u
if V alidSubsttx (ϕ) = 1 = V alidSubstux (ϕ) (= Sub)
Γ ` ϕux
Existential Rules:
Γ ` ϕtx
if V alidSubsttx (ϕ) = 1 (∃I)
Γ ` ∃xϕ
Γ ∪ {ϕyx } ` ψ
/ F reeV ar(Γ ∪ {∃xϕ, ψ}) and V alidSubstyx (ϕ) = 1 (∃P )
if y ∈
Γ ∪ {∃xϕ} ` ψ
Universal Rules:
Γ ` ∀xϕ
if V alidSubsttx (ϕ) = 1 (∀E)
Γ ` ϕtx
Γ ` ϕyx
/ F reeV ar(Γ ∪ {∀xϕ}) and V alidSubstyx (ϕ) = 1 (∀I)
if y ∈
Γ ` ∀xϕ
Superset Rule:
Γ`ϕ
if Γ0 ⊇ Γ (Super)
Γ0 ` ϕ
Definition 5.2.1. A deduction is a witnessing sequence in (P(F ormL ) × F ormL , AssumeL ∪ EqRef l, H).
Definition 5.2.2. Let Γ ⊆ F ormP and let ϕ ∈ F ormP . We write Γ ` ϕ to mean that
(Γ, ϕ) ∈ (P(F ormL ) × F ormL , AssumeL ∪ EqRef l, H)
We pronounce Γ ` ϕ as “Γ syntactically implies ϕ”.
Notation 5.2.3.
1. If Γ = ∅, we write ` ϕ instead of ∅ ` ϕ.
2. If Γ = {γ}, we write γ ` ϕ instead of {γ} ` ϕ.
Definition 5.2.4. Γ is inconsistent if there exists θ ∈ F ormP such that Γ ` θ and Γ ` ¬θ. Otherwise, we
say that Γ is consistent.
5.2.2 Some Fundamental Deductions

Proposition 5.2.5. For any t, u ∈ T ermL , we have t = u ` u = t.
Proof. Fix t, u ∈ T ermL .
{t = u} ` t = t (EqRef l) (1)
{t = u} ` t = u (AssumeL ) (2)
{t = u} ` u = t (= Sub on 1 and 2 with x = t) (3)
Proposition 5.2.6. For any t, u, w ∈ T ermL , we have {t = u, u = w} ` t = w.
Proof. Fix t, u, w ∈ T ermL .
{t = u, u = w} ` t = u (AssumeL ) (1)
{t = u, u = w} ` u = w (AssumeL ) (2)
{t = u, u = w} ` t = w (= Sub on 1 and 2 with t = x) (3)
Proposition 5.2.7. For any R ∈ Rk and any t1 , t2 , . . . , tk ∈ T ermL , we have
{Rt1 t2 · · · tk , t1 = u1 , t2 = u2 , . . . , tk = uk } ` Ru1 u2 · · · uk
Sk
Proof. Fix R ∈ Rk and t1 , t2 , . . . , tk ∈ T ermL . Fix x ∈ / i=1 (OccurV ar(ti ) ∪ OccurV ar(ui )). Let Γ =
{Rt1 t2 · · · tk , t1 = u1 , t2 = u2 , . . . , tk = uk }. We have
Γ ` Rt1 t2 · · · tk (AssumeL ) (1)

Γ ` t1 = u1 (AssumeL ) (2)
Γ ` Ru1 t2 t3 · · · tk (= Sub on 1 and 2 with Rxt2 t3 · · · tk ) (3)
Γ ` t2 = u2 (AssumeL ) (4)
Γ ` Ru1 u2 t3 · · · tk (= Sub on 3 and 4 with Ru1 xt3 · · · tk ) (5)
..
.
Γ ` tk = uk (AssumeL ) (2k)
Γ ` Ru1 u2 · · · uk (= Sub on 2k − 1 and 2k with Ru1 u2 · · · x) (2k + 1)
Proposition 5.2.8. For any f ∈ Fk and any t1 , t2 , . . . , tk ∈ T ermL , we have
{t1 = u2 , t2 = u2 , . . . , tk = uk } ` ft1 t2 · · · tk = fu1 u2 · · · uk

Sk
Proof. Fix f ∈ Fk and t1 , t2 , . . . , tk ∈ T ermL . Fix x ∈
/ i=1 (OccurV ar(ti ) ∪ OccurV ar(ui )). Let Γ = {t1 =
u1 , t2 = u2 , . . . , tk = uk }. We have
Γ ` ft1 t2 · · · tk = f t1 t2 · · · tk (EqRef l) (1)

Γ ` t1 = u1 (AssumeL ) (2)
Γ ` ft1 t2 · · · tk = fu1 t2 · · · tk (= Sub on 1 and 2 with ft1 t2 · · · tk = fxt2 · · · tk ) (3)
Γ ` t2 = u2 (AssumeL ) (4)
Γ ` ft1 t2 · · · tk = fu1 u2 · · · tk (= Sub on 1 and 2 with ft1 t2 · · · tk = fu1 x · · · tk ) (3)
..
.
Γ ` tk = uk (AssumeL ) (2k)
Γ ` ft1 t2 · · · tk = fu1 u2 · · · uk (= Sub on 2k − 1 and 2k with ft1 t2 · · · tk = fu1 u2 · · · x) (2k + 1)
Similar to the previous proposition, but start with the line ∅ ` ft1 t2 · · · tk = ft1 t2 · · · tk using the EqRef l
rule.
Proposition 5.2.9. ∃xϕ ` ¬∀x¬ϕ.
Proof. Fix y 6= x with y ∈
/ OccurV ar(ϕ).
{ϕyx , ¬¬∀x¬ϕ, ¬∀x¬ϕ} ` ¬∀x¬ϕ (AssumeL ) (1)

{ϕyx , ¬¬∀x¬ϕ, ¬∀x¬ϕ} ` ¬¬∀x¬ϕ (AssumeL ) (2)
{ϕyx , ¬¬∀x¬ϕ} ` ∀x¬ϕ (Contr on 1 and 2) (3)
{ϕyx , ¬¬∀x¬ϕ} ` ¬(ϕyx ) (∀E on 3) (4)
{ϕyx , ¬¬∀x¬ϕ} ` ϕyx (AssumeL ) (5)
{ϕyx } ` ¬∀x¬ϕ (Contr on 4 and 5) (6)
{∃xϕ} ` ¬∀x¬ϕ (∃P on 6) (7)
Proposition 5.2.10. ¬∃x¬ϕ ` ∀xϕ.

Proof. Fix y 6= x with y ∈
/ OccurV ar(ϕ).
{¬∃x¬ϕ, ¬ϕyx } ` ¬∃x¬ϕ (AssumeL ) (1)

{¬∃x¬ϕ, ¬ϕyx } ` (¬ϕ)yx (AssumeL ) (2)
{¬∃x¬ϕ, ¬ϕyx } ` ∃x¬ϕ (∃I on 2) (3)
{¬∃x¬ϕ} ` ϕyx (Contr on 1 and 3) (4)
{¬∃x¬ϕ} ` ∀xϕ (∀I on 4) (5)
5.2.3 Theorems About `

Proposition 5.2.11. If Γ is inconsistent, then Γ ` ϕ for all ϕ ∈ F ormP .
Proof. Fix θ such that Γ ` θ and Γ ` ¬θ, and fix ϕ ∈ F ormL . We have that Γ ∪ {¬ϕ} ` θ and Γ ∪ {¬ϕ} ` ¬θ
by the Super rule. Therefore, Γ ` ϕ by using the Contr rule.
Proposition 5.2.12.
1. If Γ ∪ {ϕ} is inconsistent, then Γ ` ¬ϕ.

2. If Γ ∪ {¬ϕ} is inconsistent, then Γ ` ϕ.
Proof.
1. Since Γ ∪ {ϕ} is inconsistent, we know that Γ ∪ {ϕ} ` ¬ϕ by Proposition 5.2.11. Since we also have
that Γ ∪ {¬ϕ} ` ¬ϕ by Assume, it follows that Γ ` ¬ϕ by the ¬P C rule.
2. Since Γ ∪ {¬ϕ} is inconsistent, we know that Γ ∪ {¬ϕ} ` ϕ by Proposition 5.2.11. Since we also have
that Γ ∪ {ϕ} ` ϕ by Assume, it follows that Γ ` ϕ by the ¬P C rule.
Corollary 5.2.13. If Γ ⊆ F ormL is consistent and ϕ ∈ F ormL , then either Γ ∪ {ϕ} is consistent or
Γ ∪ {¬ϕ} is consistent.
Proof. If both Γ ∪ {ϕ} and Γ ∪ {¬ϕ} are inconsistent, then both Γ ` ¬ϕ and Γ ` ϕ by Proposition 5.2.12,
so Γ is inconsistent.
Proposition 5.2.14.
1. If Γ ` ϕ and Γ ∪ {ϕ} ` ψ, then Γ ` ψ.
2. If Γ ` ϕ and Γ ` ϕ → ψ, then Γ ` ψ.
Proof.
1. Since Γ ` ϕ, it follows from the Super rule that Γ ∪ {¬ϕ} ` ϕ. Since we also have Γ ∪ {¬ϕ} ` ¬ϕ by
Assume, we may conclude that Γ ∪ {¬ϕ} is inconsistent. Therefore, by Proposition 5.2.11, we have
that Γ ∪ {¬ϕ} ` ψ. Now we also have Γ ∪ {ϕ} ` ψ by assumption, so the ¬P C rule gives that Γ ` ψ.
2. Since Γ ` ϕ → ψ, we can conclude that Γ ∪ {ϕ} ` ψ by rule → E. The result follows from part 1.
Proposition 5.2.15. Let Gf in = G(Pf in (F ormL ) × F ormL , AssumeL ∪ EqRef l, H), i.e. we insist that the
set Γ is finite but otherwise have exactly the same proof rules. Let Γ `f in ϕ denote that (Γ, ϕ) ∈ Gf in
1. If Γ `f in ϕ, then Γ ` ϕ.
2. If Γ ` ϕ, then there exists a finite Γ0 ⊆ Γ such that Γ0 `f in ϕ

In particular, if Γ ` ϕ, then there exists a finite Γ0 ⊆ Γ such that Γ0 ` ϕ.
Proof. 1 is a completely straightforward induction because the starting points are the same and we have the
exact same rules. The proof of 2 goes in much the same way as the corresponding result for propositional
logic.
Corollary 5.2.16. If every finite subset of Γ is consistent, then Γ is consistent.
Proof. Suppose that Γ is inconsistent, and fix θ ∈ F ormL such that Γ ` θ and Γ ` ¬θ. By Proposition
5.2.15, there exists finite sets Γ0 ⊆ Γ and Γ1 ⊆ Γ such that Γ0 ` θ and Γ1 ` ¬θ. Using the Super rule, it
follows that Γ0 ∪ Γ1 ` θ and Γ0 ∪ Γ1 ` ¬θ, so Γ0 ∪ Γ1 is a finite inconsistent subset of Γ.
Chapter 6
Soundness, Completeness, and

Compactness
6.1 Soundness
Theorem 6.1.1 (Soundness Theorem).
1. If Γ ` ϕ, then Γ ϕ.
2. Every satisfiable set of formulas is consistent.
Proof.
1. The proof is by induction. We let X = {(Γ, ϕ) ∈ G : Γ ϕ} and we show by induction on G that
X = G. We begin by noting that if ϕ ∈ Γ, then Γ ϕ because if (M, s) is a model of Γ, then (M, s)
is a model of γ simply because ϕ ∈ Γ. Therefore, (Γ, ϕ) ∈ X for all (Γ, ϕ) ∈ AssumeL . Also, for any
Γ ⊆ F ormL and any t ∈ T ermL , we have Γ t = t because for any any model (M, s) of Γ we have
s(t) = s(t), hence (M, s) t = t.
We now handle the inductive steps. All of the old rules go through in a similar manner as before, and
the Super rule is trivial.
• We first handle the = Sub rule. Suppose that Γ ϕtx , that Γ t = u, and that V alidSubsttx (ϕ) =
1 = V alidSubstux (ϕ). We need to show that Γ ϕux . Fix a model (M, s) of Γ. Since Γ ϕtx , we
have that (M, s) ϕtx . Also, since Γ t = u, we htave that (M, s) t = u, and hence s(t) = s(u).
Now using the fact that V alidSubsttx (ϕ) = 1 = V alidSubstux (ϕ) = 1, we have
(M, s) ϕtx ⇒ (M, s[x ⇒ s(t)]) ϕ (by the Substitution Theorem)
⇒ (M, s[x ⇒ s(u)]) ϕ
⇒ (M, s) ϕux (by the Substitution Theorem)
• We now handle the ∃I rule. Suppose that Γ ϕtx where V alidSubstt,x (ϕ) = 1. We need to
show that Γ ∃xϕ. Fix a model (M, s) of Γ. Since Γ ϕtx , it follows that (M, s) ϕtx . Since
V alidSubsttx (ϕ) = 1, we have
(M, s) ϕtx ⇒ (M, s[x ⇒ s(t)]) ϕ (by the Substitution Theorem)
⇒ There exists a ∈ M such that (M, s[x ⇒ a]) ϕ
⇒ (M, s) ∃xϕ
85
86 CHAPTER 6. SOUNDNESS, COMPLETENESS, AND COMPACTNESS
• Let’s next attack the ∃P rule. Suppose that Γ ∪ {ϕyx } ψ, that y ∈

/ F reeV ar(Γ ∪ {∃xϕ, ψ}), and
that V alidSubstyx (ϕ) = 1. We need to show that Γ ∪ {∃xϕ} ψ. Fix a model (M, s) of Γ ∪ {∃xϕ}.
Since (M, s) ∃xϕ, we may fix a ∈ M such that (M, s[x ⇒ a]) ϕ. We first divide into two
cases to show that (M, s[y ⇒ a]) ϕyx .
Case 1: Suppose that y = x. We then have ϕyx = ϕxx = ϕ and s[x ⇒ a] = s[y ⇒ a], hence
(M, s[y ⇒ a]) ϕyx because (M, s[x ⇒ a]) ϕ.
Case 2: Suppose that y 6= x. We then have
(M, s[x ⇒ a]) ϕ ⇒ (M, (s[y ⇒ a])[x ⇒ a]) ϕ (since y ∈

/ F reeV ar(ϕ) and y 6= x)
⇒ (M, (s[y ⇒ a])[x ⇒ s[y ⇒ a](y)]) ϕ
⇒ (M, s[y ⇒ a]) ϕyx (by the Substitution Theorem)
Thus, (M, s[y ⇒ a]) ϕyx in either case. Now since (M, s) γ for all γ ∈ Γ and y ∈
/ F reeV ar(Γ),
we have (M, s[y ⇒ a]) γ for all γ ∈ Γ. Thus, (M, s[y ⇒ a]) ψ because Γ ∪ {ϕyx } ψ. Finally,
since y ∈
/ F reeV ar(ψ), it follows that (M, s) ψ.
• We next do the ∀E rule. Suppose that Γ ∀xϕ and that t ∈ T ermL is such that V alidSubsttx (ϕ) =
1. We need to show that Γ ϕtx . Fix a model (M, s) of Γ. Since Γ ∀xϕ, it follows that that
(M, s) ∀xϕ. Since V alidSubsttx (ϕ) = 1, we have
(M, s) ∀xϕ ⇒ For all a ∈ M, we have (M, s[x ⇒ a]) ϕ

⇒ (M, s[x ⇒ s(t)]) ϕ
⇒ (M, s) ϕtx (by the Substitution Theorem)
• We finally end with the ∀I rule. Suppose that Γ ϕyx , that y ∈

/ F reeV ar(Γ ∪ {∀xϕ}), and that
V alidSubstyx (ϕ) = 1. We need to show that Γ ∀xϕ. Fix a model (M, s) of Γ.
Case 1: Suppose that y = x. Since y = x, we have ϕyx = ϕxx = ϕ. Fix a ∈ M . Since (M, s) γ
for all γ ∈ Γ and x = y ∈/ F reeV ar(Γ), we may conclude that (M, s[x ⇒ a]) γ for all γ ∈ Γ.
Therefore, (M, s[x ⇒ a]) ϕyx , i.e. (M, s[x ⇒ a]) ϕ, because Γ ϕyx . Now a ∈ M was arbitrary,
so (M, s[x ⇒ a]) ϕ for every a ∈ M , hence (M, s) ∀xϕ.
Case 2: Suppose that y 6= x. Fix M . Since (M, s) γ for all γ ∈ Γ and y ∈ / F reeV ar(Γ), we
may conclude that (M, s[y ⇒ a] γ for all γ ∈ Γ. Therefore, (M, s[y ⇒ a]) ϕyx because Γ ϕyx .
Since V alidSubstyx (ϕ) = 1, we have
(M, s[y ⇒ a]) ϕyx ⇒ (M, (s[y ⇒ a])[x ⇒ s[y ⇒ a](y)]) ϕ (by the Substitution Theorem)
⇒ (M, (s[y ⇒ a])[x ⇒ a]) ϕ
⇒ (M, s[x ⇒ a]) ϕ (since y ∈
/ F reeV ar(ϕ) and y 6= x)
Now a ∈ M was arbitrary, so (M, s[x ⇒ a]) ϕ for every a ∈ M , hence (M, s) ∀xϕ.
2. Let Γ be a satisfiable set of formulas. Fix a model (M, s) of Γ. Suppose that Γ is inconsistent, and fix
θ ∈ F ormL such that Γ ` θ and Γ ` ¬θ. We then have Γ θ and Γ ¬θ by part 1, hence (M, s) θ
and (M, s) (¬θ), a contradiction. It follows that Γ is consistent.
6.2. PRIME FORMULAS 87
6.2 Prime Formulas

Definition 6.2.1. A formula ϕ ∈ F ormL is prime if one the following holds.
1. ϕ ∈ AtomicF ormL .
2. ϕ = ∃xψ for some x ∈ V ar and some ψ ∈ F ormL .
3. ϕ = ∀xψ for some x ∈ V ar and some ψ ∈ F ormL .
We denote the set of prime formulas by P rimeF ormL .
Definition 6.2.2. Let L be a language. Let P (L) = {Aϕ : ϕ ∈ P rimeF ormL }.
• We define a function h : F ormL → F ormP (L) recursively as follows.
1. h(ϕ) = Aϕ for all ϕ ∈ AtomicF ormL .
2. h(¬ϕ) = ¬h(ϕ) for all ϕ ∈ F ormL .
3. h(3ϕψ) = 3h(ϕ)h(ψ) for all ϕ, ψ ∈ F ormL and all 3 ∈ {∧, ∨, →}.
4. h(Qxϕ) = AQxϕ for all ϕ ∈ F ormL , x ∈ V ar, and Q ∈ {∀, ∃}.
For ϕ ∈ F ormL , we write ϕ# for h(ϕ). For Γ ⊆ F ormL , we write Γ# for {γ # : γ ∈ Γ}.
• We also define a function g : F ormP (L) → F ormL recursively as follows.
1. g(Aϕ ) = ϕ for all ϕ ∈ P rimeF ormL .
2. g(¬α) = ¬g(α) for all α ∈ F ormPL .
3. g(3αβ) = 3g(α)g(β) for all α, β ∈ F ormPL and all 3 ∈ {∧, ∨, →}.
For α ∈ F ormP (L) , we write α? for g(α). For Γ ⊆ F ormP (L) , we write Γ? for {γ ? : γ ∈ Γ}.
Proposition 6.2.3.
1. For all ϕ ∈ F ormL , we have (ϕ# )? = ϕ.
2. For all α ∈ F ormP (L) , we have (α? )# = α.
Proof. A trivial induction.
Proposition 6.2.4. Let L be a language, let Γ ⊆ F ormP (L) , and let ϕ ∈ F ormP (L) .
1. If Γ `P (L) ϕ (in the propositional language P (L)), then Γ? `L ϕ? (in the first-order language L).
2. If Γ P (L) ϕ (in the propositional language P (L)), then Γ? `L ϕ? (in the first-order language L).
Proof.
1. This follows by induction because all propositional rules are included as first-order logic rules.
2. If Γ P (L) ϕ, then Γ `P (L) ϕ by the Completeness Theorem for propositional logic, hence Γ? `L ϕ? by
part 1.
Corollary 6.2.5. Let L be a language, let Γ ⊆ F ormL , and let ϕ ∈ F ormL . If Γ# P (L) ϕ# (in the
propositional language P (L)), then Γ `L ϕ (in the first-order language L).
Proof. Suppose that Γ# P (L) ϕ# . By Proposition 6.2.4, it follows that (Γ# )? P (L) (ϕ# )? , hence Γ `L ϕ
by Proposition 6.2.3
Example 6.2.6. Let L be a languague and let ϕ, ψ ∈ F ormL We have ¬(ϕ → ψ) `L ϕ ∧ (¬ψ)
Proof. We show that (¬(ϕ → ψ))# P (L) (ϕ ∧ (¬ψ))# . Notice that
1. (¬(ϕ → ψ))# = ¬(ϕ# → ψ # )
2. (ϕ ∧ (¬ψ))# = ϕ# ∧ ¬(ψ # ).
Suppose that v : P (L) → {0, 1} is a truth assignment such that v(¬(ϕ# → ψ # )) = 1. We then have v(ϕ# →
ψ # ) = 0, hence v(ϕ# ) = 1 and v(ψ # ) = 0. We therefore have v(¬(ψ # )) = 1 and hence v(ϕ# ∧ ¬(ψ # )) = 1.
It follows that (¬(ϕ → ψ))# P (L) (ϕ ∧ (¬ψ))# .
Corollary 6.2.7. Let L be a language, let Γ ⊆ F ormL , and let ϕ, ψ ∈ F ormL . If Γ `L ϕ and ϕ# P (L) ψ # ,
then Γ `L ψ.
Proof. Since ϕ# P (L) ψ # , we have that ϕ `L ψ by Corollary 6.2.5. It follows from the Super rule that
Γ ∪ {ϕ} `L ψ. Using Proposition 5.2.14 (since Γ `L ϕ and Γ ∪ {ϕ} `L ψ), we may conclude that Γ `L ψ.
6.3 Completeness
6.3.1 Motivating the Proof
We first give an overview of the key ideas in our proof of completeness. Let L be a language, and suppose
that Γ ⊆ L is consistent.
Definition 6.3.1. Suppose that L is a language and that ∆ ⊆ F ormL . We say that ∆ is complete if for all
ϕ ∈ F ormL , either ϕ ∈ ∆ or ¬ϕ ∈ ∆.
As we saw in propositional logic, it will aid use greatly to extend Γ to a set ∆ ⊇ Γ which is both consistent
and complete, so let’s assume that we can do that (we will prove it exactly the same way below). We need
to construct an L-structure M and a variable assignment s : V ar → M such that (M, s) δ for all δ ∈ ∆.
Now all that we have is the syntactic information that ∆ provides, so it seems that the only way to proceed
is to define our M from these syntactic objects. Since terms intuitively name elements, it is natural to try
to define the universe M to simply be T ermL . We would then define the structure as follows
1. cM = c for all c ∈ C.
2. RM = {(t1 , t2 , . . . , tk ) ∈ M k : Rt1 t2 . . . tk ∈ ∆} for all R ∈ Rk .
3. f M (t1 , t2 , . . . , tk ) = ft1 t2 · · · tk for all f ∈ Fk and all t1 , t2 , . . . , tk ∈ M .
and let s : V ar → M be the variable assignment defined by s(x) = x for all x ∈ V ar.
However, there are two problems with this approach, one of which is minor and the other is quite serious.
First, let’s think about the minor problem. Suppose that L = {f, e} where f is a binary function symbol
and e is a constant symbol, and that Γ is the set of group axioms. Suppose that ∆ ⊇ Γ is consistent and
complete. We then have fee = e ∈ ∆ because Γ ` fee = e. However, the two terms fee and e are syntactically
different objects, so if we were to let M be T ermL this would cause a problem because fee and e are distinct
despite the fact that ∆ says they must be equal. Of course, when you have distinct objects which you want
to consider equivalent, you should define an equivalence relation. Thus, we should define ∼ on T ermL by
letting t ∼ u if t = u ∈ ∆. We would then need to check that ∼ is an equivalence relation and that the
6.3. COMPLETENESS 89
definition of the structure above is independent of our choice of representatives for the classes. This is all
fairly straightfoward, and will be carried out below.
On to the more serious obstactle. Suppose that L = {P} where P is a unary relation symbol. Let
Γ = {¬Px : x ∈ V ar} ∪ {¬(x = y) : x, y ∈ V ar with x 6= y} ∪ {∃xPx} and notice that Γ is consistent because
it is satisfiable. Suppose that ∆ ⊇ Γ is consistent and complete. In the model M constructed, we have
M = T ermL = V ar (notice that the equivalence relation defined above will be trivial in this case). Thus,
since (M, s) ¬Px for all x ∈ V ar, it follows that (M, s) 6 ∃xPx. Hence, M is not a model of ∆.
The problem in the above example is that there was an existential statement in ∆, but whenever you
plugged a term in for the quantied variable, the resulting formula was not in ∆. Since we are building our
model from the terms, this is a serious problem. However, if ∆ had the following, then this problem would
not arise.
Definition 6.3.2. Let L be a language and let Γ ⊆ F ormL . We say that Γ contains witnesses if for all
ϕ ∈ F ormL and all x ∈ V ar, there exists c ∈ C such that (∃xϕ) → ϕcx ∈ Γ.
Our goal then is to show that if Γ is consistent, then there exists a ∆ ⊇ Γ which is consistent, complete,
and contains witnesses. On the face of it, this is not true, as the above example shows (because there are
no constant symbols). However, if we allow ourselves to expand our language with new constant symbols,
we can repeatedly add witnessing statements by using these fresh constant symbols as our witnesses.
6.3.2 The Proof

We can also define substitution of variables for constants in the obvious recursive fashion. Ignore the following
lemma until you see why we need it later.
Lemma 6.3.3. Let ϕ ∈ F ormL , let t ∈ T ermL , let c ∈ C, and let x, z ∈ V ar. Suppose that z ∈
/ OccurV ar(ϕ).
tz
• (ϕtx )zc equals (ϕzc )xc .
tz
• If V alidSubsttx (ϕ) = 1, then V alidSubstxc (ϕzc ) = 1.
Lemma 6.3.4. Let L be a language, and let L0 be L together with a new constant symbol c. Suppose that
Γ0 `L0 ϕ0
Γ1 `L0 ϕ1
Γ2 `L0 ϕ2
..
.
Γn `L0 ϕn
Sn
is an L0 -deduction. For any z ∈ V ar with z ∈
/ OccurV ar( i=0 (Γi ∪ {ϕi })), we have that
(Γ0 )zc `L (ϕ0 )zc

(Γ1 )zc `L (ϕ1 )zc
(Γ2 )zc `L (ϕ2 )zc
..
.
(Γn )zc `L (ϕn )zc
is an L-deduction.
Proof. We prove by induction on i that
(Γ0 )zc `L (ϕ0 )zc

(Γ1 )zc `L (ϕ1 )zc
(Γ2 )zc `L (ϕ2 )zc
..
.
(Γi )zc `L (ϕi )zc
is an L-deduction.
If ϕ ∈ Γ, then ϕzc ∈ Γzc .
Suppose that line i is Γ ` t = t where t ∈ T ermL0 . Since (t = t)zc equals tzc = tzc , we can place Γzc ` tzc = tzc
on line i by the EqRef l rule.
Suppose that Γ `L0 ϕ∧ψ was a previous line and we inferred Γ `L0 ϕ. Inductively, we have Γzc `L (ϕ ∧ ψ)zc
on the corresponding line. Since (ϕ ∧ ψ)zc = ϕzc ∧ ψcz , we may use the ∧EL rule to put Γzc `L ϕzc on the
corresponding line. The other propositional rules are similarly uninteresting.
Suppose that Γ `L0 ϕtx and Γ `L0 t = u were previous lines, that V alidSubsttx (ϕ) = 1 = V alidSubstux (ϕ),
and we inferred Γ `L0 ϕux . Inductively, we have Γzc `L (ϕtx )zc and Γzc `L (t = u)zc on the corresponding lines.
tz tz
Now (ϕtx )zc equals (ϕzc )xc by the previous lemma, and (t = u)zc equals tzc = uzc . Thus, we have Γzc `L (ϕzc )xc
and Γzc `L tzc = uzc on the corresponding lines. Using the fact that V alidSubsttx (ϕ) = 1 = V alidSubstux (ϕ),
tz uz
we can use the previous lemma to conclude that that V alidSubstxc (ϕzc ) = 1 = V alidSubstx c (ϕzc ). Hence, we
z
u uz
may use that = Sub rule to put Γzc `L (ϕzc )x c on the corresponding line. We now need only note that (ϕzc )x c
equals (ϕux )zc by the previous lemma.
Suppose that Γ `L0 ϕtx where V alidSubsttx (ϕ) = 1 was a previous line and we inferred Γ `L0 ∃xϕ.
tz tz
Inductively, we have Γzc `L (ϕtx )zc on the corresponding line. Now (ϕtx )zc equals (ϕzc )xc and V alidSubstxc (ϕzc ) = 1
by the previous lemma. Hence, we may use the ∃I rule to put Γzc `L ∃x(ϕzc ) on the corresponding line. We
now need only note that ∃x(ϕzc ) equals (∃xϕ)zc .
The other rules are similarly awful.
Corollary 6.3.5 (Generalization on Constants). Let L be a language, and let L0 be L together with a new
constant symbol c. Suppose that Γ ⊆ F ormL and ϕ ∈ F ormL . If Γ `L0 ϕcx , then Γ `L ∀xϕ.
Proof. Since Γ `L0 ϕ, we may use Proposition 5.2.15 to fix an L0 -deduction
Γ0 `L0 ϕ0
Γ1 `L0 ϕ1
Γ2 `L0 ϕ2
..
.
Γn `L0 ϕn
such that each Γi ⊆ F ormL0 is finite and Γn ⊆ Γ. Fix y ∈ V ar such that
n
[
y∈
/ OccurV ar( (Γi ∪ {ϕi }))
i=0
From Lemma 6.3.4, we have that
(Γ0 )yc `L (ϕ0 )yc

(Γ1 )yc `L (ϕ1 )yc
(Γ2 )yc `L (ϕ2 )yc
..
.
(Γn )yc `L (ϕn )yc
is an L-deduction. Since Γn ⊆ Γ ⊆ F ormL , we have (Γn )yc = Γn . Now (ϕn )yc = (ϕcx )yc = ϕyx . We therefore
have Γn `L ϕyx . We may then use the ∀I rule to conclude that Γn `L ∀xϕ. Finally, Γ `L ∀xϕ by the Super
rule.
Corollary 6.3.6. Let L be a language, let Γ ⊆ F ormL , and let ϕ ∈ F ormL .
1. Let L0 be L together with a new constant symbol. If Γ `L0 ϕ, then Γ `L ϕ.
2. Let L0 be L together with finitely many new constant symbols. If Γ `L0 ϕ, then Γ `L ϕ.
3. Let L0 be L together with (perhaps infinitely many) new constant symbols. If Γ `L0 ϕ, then Γ `L ϕ.
Proof.
1. Since Γ `L0 ϕ, we may use Proposition 5.2.15 to fix an L0 -deduction
Γ0 `L0 ϕ0
Γ1 `L0 ϕ1
Γ2 `L0 ϕ2
..
.
Γn `L0 ϕn
such that each Γi ⊆ F ormL0 is finite, Γn ⊆ Γ, and ϕn = ϕ. Fix y ∈ V ar such that

n
[
y∈
/ OccurV ar( (Γi ) ∪ {ϕi }))
i=0
From Lemma 6.3.4, we have that
(Γ0 )yc `L (ϕ0 )yc

(Γ1 )yc `L (ϕ1 )yc
(Γ2 )yc `L (ϕ2 )yc
..
.
(Γn )yc `L (ϕn )yc
is an L-deduction. Since Γn ⊆ Γ ⊆ F ormL and ϕ ∈ F ormL , it follows that (Γn )yc = Γn and (ϕn )yc = ϕ.
Thus, Γn `L ϕ and so Γ `L ϕ by the Super rule.
2. This is proved by induction on the number of new constant symbols, using part 1 for the base case
and the inductive step.
3. Suppose that Γ `L0 ϕ, and use Proposition 5.2.15 to fix an L0 -deduction
Γ0 `L0 ϕ0
Γ1 `L0 ϕ1
Γ2 `L0 ϕ2
..
.
Γn `L0 ϕn
SnΓi ⊆ F ormL0 is finite, Γn ⊆ Γ, and ϕn = ϕ. Let {c0 , c1 , . . . , cm } be all constants

such that each
appearing in i=0 (Γi ∪ {ϕi }) and let L0 = L ∪ {c0 , c1 , . . . , cm }. We then have that
Γ0 `L0 ϕ0
Γ1 `L0 ϕ1
Γ2 `L0 ϕ2
..
.
Γn `L0 ϕn
is an L0 -deduction, so Γn `L0 ϕ. Using the Super rule, we conclude that Γ `L0 ϕ. Therefore, Γ `L ϕ
by part 2.
Corollary 6.3.7. Let L be a language and let L0 be L together with (perhaps infinitely many) new constant
symbols. Let Γ ⊆ F ormL . Γ is L-consistent if and only if Γ is L0 -consistent.
Proof. Since any L-deduction is also a L0 -deduction, if Γ is L-inconsistent then it is L0 -inconsistent . Suppose
that Γ is L0 -inconsistent. We then have that Γ `L0 ϕ for all ϕ ∈ F ormL by Proposition 5.2.11, hence Γ `L ϕ
for all ϕ ∈ F ormL by Corollary 6.3.6. Therefore, Γ is L-inconsistent.
Lemma 6.3.8. Let L be a language, and let L0 be L together with a new constant symbol c. Suppose that
Γ ⊆ F ormL is L-consistent and that ϕ ∈ F ormL . We then have that Γ ∪ {(∃xϕ) → ϕcx } is L0 -consistent.
Proof. Suppose that Γ ∪ {(∃xϕ) → ϕcx } is L0 -inconsistent. We then have that Γ `L0 ¬((∃xϕ) → ϕcx ), hence
# #
Γ `L0 (∃xϕ) ∧ ¬(ϕcx ) by Corollary 6.2.7 (because (¬((∃xϕ) → ϕcx )) P (L0 ) ((∃xϕ) ∧ ¬(ϕcx )) ). Thus, Γ `L0
∃xϕ by the ∧EL rule, so Γ `L0 ¬∀x¬ϕ (by Proposition 5.2.9 and Proposition 5.2.14), and hence Γ `L ¬∀x¬ϕ
by Corollary 6.3.6. We also have Γ `L0 (¬ϕ)cx by the ∧ER rule, so Γ `L ∀x¬ϕ by Generalization on
Constants. This contradicts the fact that Γ is L-consistent.
Lemma 6.3.9. Let L be a language and let Γ ⊆ F ormL be L-consistent. There exists a language L0 ⊇ L
and Γ0 ⊆ F ormL0 such that
1. Γ ⊆ Γ0 .
2. Γ0 is L0 -consistent.
3. For all ϕ ∈ F ormL and all x ∈ V ar, there exists c ∈ C such that (∃xϕ) → ϕcx ∈ Γ0 .
Proof. For each ϕ ∈ F ormL and each x ∈ V ar, let cϕ,x be a new constant symbol (distinct from all symbols
in L). Let L0 = L ∪ {cϕ,x : ϕ ∈ F ormL and x ∈ V ar}. Let
c
Γ0 = Γ ∪ {(∃xϕ) → ϕxϕ,x : ϕ ∈ F ormL and x ∈ V ar}}
Conditions 1 and 3 are clear, so we need only check that Γ0 is L0 -consistent. By Corollary 5.2.16, it suffices
to check that all finite subsets of Γ0 are L0 -consistent, and for this it suffices to show that
cϕ ,x cϕ ,x c
Γ ∪ {(∃x1 ϕ1 ) → (ϕ1 )x1 1 1 , (∃x2 ϕ2 ) → (ϕ2 )x2 2 2 , . . . , (∃xn ϕn ) → (ϕn )xϕn n ,xn }
is L0 -consistent whenever ϕ1 , ϕ2 , . . . , ϕn ∈ F ormL and x1 , x2 , . . . , xn ∈ V ar. Formally, one can prove this by
induction on n. A slightly informal argument is as as follows. Fix ϕ1 , ϕ2 , . . . , ϕn ∈ F ormL and x1 , x2 , . . . , xn ∈
V ar. We have
• Γ is L-consistent, so
cϕ ,x
• Γ ∪ {(∃x1 ϕ1 ) → (ϕ1 )x1 1 1 } is (L ∪ {cϕ1 ,x1 })-consistent by Lemma 6.3.8, so
cϕ ,x cϕ ,x
• Γ ∪ {(∃x1 ϕ1 ) → (ϕ1 )x1 1 1 , (∃x2 ϕ2 ) → (ϕ2 )x2 2 2 } is (L ∪ {cϕ1 ,x1 , cϕ2 ,x2 })-consistent by Lemma 6.3.8, so
• ...
cϕ ,x cϕ ,x c
• Γ ∪ {(∃x1 ϕ1 ) → (ϕ1 )x1 1 1 , (∃x2 ϕ2 ) → (ϕ2 )x2 2 2 , . . . , (∃xn ϕn ) → (ϕn )xϕn n ,xn } is
(L ∪ {cϕ1 ,x1 , cϕ2 ,x2 , . . . , cϕn ,xn })-consistent
Therefore,
cϕ ,x cϕ ,x c
Γ ∪ {(∃x1 ϕ1 ) → (ϕ1 )x1 1 1 , (∃x2 ϕ2 ) → (ϕ2 )x2 2 2 , . . . , (∃xn ϕn ) → (ϕn )xϕn n ,xn }
is L0 -consistent by Corollary 6.3.7.
Proposition 6.3.10. Let L be a language and let Γ ⊆ F ormL be consistent. There exists a language L0 ⊇ L
and Γ0 ⊆ F ormL0 such that
1. Γ ⊆ Γ0 .
2. Γ0 is L0 -consistent.
3. Γ0 contains witnesses.
Proof. Let L0S= L and Γ0 = Γ. ForSeach n ∈ N, use the previous lemma to get Ln+1 and Γn+1 from Ln and
Γn . Set L0 = n∈N L and set Γ0 = n∈N Γn .
Proposition 6.3.11. (Suppose that L is countable.) If Γ is consistent, then there exists a set ∆ ⊇ Γ which
is consistent and complete.
Proof. Exactly the same proof as the propositional logic case, using Zorn’s Lemma in the uncountable
case.
Proposition 6.3.12. Let L be a language. If Γ ⊆ L is consistent, then there a language L0 ⊇ L (which is
L together with new constant symbols) and ∆ ⊆ F ormL0 such that
• Γ ⊆ ∆.
• ∆ is consistent.
• ∆ is complete.
• ∆ contains witnesses.
Proof.
Lemma 6.3.13. Suppose that ∆ is consistent and complete. If ∆ ` ϕ, then ϕ ∈ ∆.
Proof. Suppose that ∆ ` ϕ. Since ∆ is complete, we have that either ϕ ∈ ∆ or ¬ϕ ∈ ∆. Now if ¬ϕ ∈ ∆,

then ∆ ` ¬ϕ, hence ∆ is inconsistent contradicting our assumption. It follows that ϕ ∈ ∆.
Lemma 6.3.14. Suppose that ∆ is consistent, complete, and contains witnesses. For every t ∈ T ermL ,
there exists c ∈ C such that t = c ∈ ∆.
Proof. Let t ∈ T ermL . Fix x ∈ V ar such that x ∈ / OccurV ar(t). Since ∆ contains witnesses, we may fix
c ∈ C such that (∃x(t = x)) → (t = c) ∈ ∆ (using the formula t = x). Now ∅ ` (t = x)tx , so we may use the
∃I rule (because V alidSubsttx (t = x) = 1) to conclude that ∅ ` ∃x(t = x). From here we can use the Super
rule to conclude that ∆ ` ∃x(t = x). We therefore have ∆ ` ∃x(t = x) and ∆ ` (∃x(t = x)) → (t = c), hence
∆ ` t = c by Proposition 5.2.14. Using Lemma 6.3.13, we conclude that t = c ∈ ∆.
Lemma 6.3.15. Suppose that ∆ is consistent, complete, and contains witnesses. We have
1. ¬ϕ ∈ ∆ if and only if ϕ ∈
/ ∆.
2. ϕ ∧ ψ ∈ ∆ if and only if ϕ ∈ ∆ and ψ ∈ ∆.
3. ϕ ∨ ψ ∈ ∆ if and only if ϕ ∈ ∆ or ψ ∈ ∆.
4. ϕ → ψ ∈ ∆ if and only if ϕ ∈
/ ∆ or ψ ∈ ∆.
5. ∃xϕ ∈ ∆ if and only if there exists c ∈ C such that ϕcx ∈ ∆.
6. ∀xϕ ∈ ∆ if and only if ϕcx ∈ ∆ for all c ∈ C.
Proof.
1. If ¬ϕ ∈ ∆, then ϕ ∈
/ ∆ because otherwise ∆ ` ϕ and so ∆ would be inconsistent.
Conversely, if ϕ ∈
/ ∆, then ¬ϕ ∈ ∆ because ∆ is complete.
2. Suppose first that ϕ ∧ ψ ∈ ∆. We then have that ∆ ` ϕ ∧ ψ, hence ∆ ` ϕ by the ∧EL rule and ∆ ` ψ
by the ∧ER rule. Therefore, ϕ ∈ ∆ and ψ ∈ ∆ by Lemma 6.3.13.
Conversely, suppose that ϕ ∈ ∆ and ψ ∈ ∆. We then have ∆ ` ϕ and ∆ ` ψ, hence ∆ ` ϕ ∧ ψ by the
∧I rule. Therefore, ϕ ∧ ψ ∈ ∆ by Lemma 6.3.13.
3. Suppose first that ϕ ∨ ψ ∈ ∆. Suppose that ϕ ∈/ ∆. Since ∆ is complete, we have that ¬ϕ ∈ ∆. From
Proposition 3.4.10, we know that {¬ϕ, ϕ ∨ ψ} ` ψ, hence ∆ ` ψ by the Super rule. Therefore, ψ ∈ ∆
by Lemma 6.3.13. It follows that either ϕ ∈ ∆ or ψ ∈ ∆.
Conversely, suppose that either ϕ ∈ ∆ or ψ ∈ ∆.
Case 1: Suppose that ϕ ∈ ∆. We have ∆ ` ϕ, hence ∆ ` ϕ ∨ ψ by the ∨IL rule. Therefore, ϕ ∨ ψ ∈ ∆
by Lemma 6.3.13.
Case 2: Suppose that ψ ∈ ∆. We have ∆ ` ψ, hence ∆ ` ϕ ∨ ψ by the ∨IR rule. Therefore, ϕ ∨ ψ ∈ ∆
by Lemma 6.3.13.
4. Suppose first that ϕ → ψ ∈ ∆. Suppose that ϕ ∈ ∆. We then have that ∆ ` ϕ and ∆ ` ϕ → ψ, hence
∆ ` ψ by Proposition 5.2.14. Therefore, ψ ∈ ∆ by Lemma 6.3.13. It follows that either ϕ ∈
/ ∆ or
ψ ∈ ∆.
Conversely, suppose that either ϕ ∈
/ ∆ or ψ ∈ ∆.
Case 1: Suppose that ϕ ∈
/ ∆. We have ¬ϕ ∈ ∆ because ∆ is complete, hence ∆ ∪ {ϕ} is inconsistent
(as ∆ ∪ {ϕ} ` ϕ and ∆ ∪ {¬ϕ} ` ¬ϕ). It follows that ∆ ∪ {ϕ} ` ψ by Proposition 5.2.11, hence
∆ ` ϕ → ψ by the → I rule. Therefore, ϕ → ψ ∈ ∆ by Lemma 6.3.13.
Case 2: Suppose that ψ ∈ ∆. We have ψ ∈ ∆ ∪ {ϕ}, hence ∆ ∪ {ϕ} ` ψ, and so ∆ ` ϕ → ψ by the
→ I rule. Therefore, ϕ → ψ ∈ ∆ by Lemma 6.3.13.
5. Suppose first that ∃xϕ ∈ ∆. Since ∆ contains witnesses, we may fix c ∈ C such that (∃xϕ) → ϕcx ∈ ∆.
We therefore have ∆ ` ∃xϕ and ∆ ` (∃xϕ) → ϕcx , hence ∆ ` ϕcx by Proposition 5.2.14. Using Lemma
6.3.13, we conclude that ϕcx ∈ ∆.
Conversely, suppose that there exists c ∈ C such that ϕcx ∈ ∆. We then have ∆ ` ϕcx , hence ∆ ` ∃xϕ
using the ∃I rule (notice that V alidSubstcx (ϕ) = 1). Using Lemma 6.3.13, we conclude that ∃xϕ ∈ ∆.
6. Suppose first that ∀xϕ ∈ ∆. We then have ∆ ` ∀xϕ, hence ∆ ` ϕcx for all c ∈ C using the ∀E rule
(notice that V alidSubstcx (ϕ) = 1 for all c ∈ C). Using Lemma 6.3.13, we conclude that ϕcx ∈ ∆ for all
c ∈ C.
Conversely, suppose that ϕcx ∈ ∆ for all c ∈ C. Since ∆ is consistent, this implies that there does not
exist c ∈ C with ¬(ϕcx ) = (¬ϕ)cx ∈ ∆. Therefore, ∃x¬ϕ ∈ / ∆ by part 5, so ¬∃x¬ϕ ∈ ∆ by part 1. It
follows from Proposition 5.2.10 that ∆ ` ∀xϕ. Using Lemma 6.3.13, we conclude that ∀xϕ ∈ ∆.
Proposition 6.3.16. If ∆ is consistent, complete, and contains witnesses, then ∆ is satisfiable.
Proof. Suppose that ∆ is consistent, complete, and contains witnesses.

Define a relation ∼ on T ermL by letting t ∼ u if t = u ∈ ∆. We first check that ∼ is an equivalence
relation. Reflexivity follows from the EqRef l rule and Lemma 6.3.13. Symmetry and transitivity follow
from Proposition 5.2.5 and Proposition 5.2.6, together with the Super rule and Lemma 6.3.13).
We now define our L-structure M. We first let M = T ermL / ∼. For each t ∈ T ermL , we let [t] denote
the equivalence class of t. Notice that M = {[c] : c ∈ C} be Lemma 6.3.14. We now finish our description of
the L-structure M by saying how to interpret the constant, relation, and function symbols. We let
1. cM = [c] for all c ∈ C.
2. RM = {([t1 ], [t2 ], . . . , [tk ]) ∈ M k : Rt1 t2 · · · tk ∈ ∆} for all R ∈ Rk .
3. f M ([t1 ], [t2 ], . . . , [tk ]) = [ft1 t2 · · · tk ] for all f ∈ Fk .
Notice that our definitions of RM do not depend on our choice of representatives for the equivalence classes
by Proposition 5.2.7. Similarly, our definitions of f M do not depend on our choice of representatives for the
equivalences classes by Proposition 5.2.8. Finally, define s : V ar → M by letting s(x) = [x] for all x ∈ V ar.
We first show that s(t) = [t] for all t ∈ T ermL by induction. We have s(c) = cM = [c] for all c ∈ C and
s(x) = s(x) = [x] for all x ∈ V ar. Suppose that f ∈ Fk and t1 , t2 , . . . , tk ∈ T ermL are such that s(ti ) = [ti ]
for all i. We then have
s(ft1 t2 · · · tk ) = f M (s(t1 ), s(t2 ), . . . , s(tk ))

= f M ([t1 ], [t2 ], . . . , [tk ]) (by induction)
= [ft1 t2 · · · tk ]
Therefore, s(t) = [t] for all t ∈ T ermL .

We now show that ϕ ∈ ∆ if and only if (M, s) ϕ for all ϕ ∈ F ormL by induction. We first prove the
result for ϕ ∈ AtomicF ormL . Suppose that R ∈ Rk and t1 , t2 , . . . , tk ∈ T ermL . We have
Rt1 t2 · · · tk ∈ ∆ ⇔ ([t1 ], [t2 ], . . . , [tk ]) ∈ RM

⇔ (s(t1 ), s(t2 ), . . . , s(tk )) ∈ RM
⇔ (M, s) Rt1 t2 · · · tk
Suppose now that t1 , t2 ∈ T ermL . We have

t1 = t2 ∈ ∆ ⇔ [t1 ] = [t2 ]
⇔ s(t1 ) = s(t2 )
⇔ (M, s) t1 = t2
Suppose that the result holds for ϕ. We have
¬ϕ ∈ ∆ ⇔ ϕ ∈
/∆ (by Lemma 6.3.15)
⇔ (M, s) 6 ϕ (by induction)
⇔ (M, s) ϕ
ϕ ∧ ψ ∈ ∆ ⇔ ϕ ∈ ∆ and ψ ∈ ∆ (by Lemma 6.3.15)
⇔ (M, s) ϕ and (M, s) ψ (by induction)
⇔ (M, s) ϕ ∧ ψ
and
ϕ ∨ ψ ∈ ∆ ⇔ ϕ ∈ ∆ or ψ ∈ ∆ (by Lemma 6.3.15)
⇔ (M, s) ϕ or (M, s) ψ (by induction)
⇔ (M, s) ϕ ∨ ψ
and finally
ϕ→ψ∈∆⇔ϕ∈
/ ∆ or ψ ∈ ∆ (by Lemma 6.3.15)
⇔ (M, s) 6 ϕ or (M, s) ψ (by induction)
⇔ (M, s) ϕ → ψ
Suppose that the result holds for ϕ and that x ∈ V ar. We have
∃xϕ ∈ ∆ ⇔ There exists c ∈ C such that ϕcx ∈ ∆ (by Lemma 6.3.15)
⇔ There exists c ∈ C such that (M, s) ϕcx (by induction)
⇔ There exists c ∈ C such that (M, s[x ⇒ s(c)]) ϕ (by the Substitution Theorem)
⇔ There exists c ∈ C such that (M, s[x ⇒ [c]]) ϕ
⇔ There exists a ∈ M such that (M, s[x ⇒ a]) ϕ
⇔ (M, s) ∃xϕ
and also
∀xϕ ∈ ∆ ⇔ For all c ∈ C, we have ϕcx ∈ ∆ (by Lemma 6.3.15)
⇔ For all c ∈ C, we have (M, s) ϕcx (by induction)
⇔ For all c ∈ C, we have (M, s[x ⇒ s(c)]) ϕ (by the Substitution Theorem)
⇔ For all c ∈ C, we have (M, s[x ⇒ [c]])
⇔ For all a ∈ M, we have (M, s[x ⇒ a]) ϕ
⇔ (M, s) ∀xϕ
Therefore, by induction, we have ϕ ∈ ∆ if and only if (M, s) ϕ. In particular, we have (M, s) ϕ for all
ϕ ∈ ∆, hence ∆ is satisfiable.
6.4. COMPACTNESS 97
Theorem 6.3.17 (Completeness Theorem). (Suppose that L is countable.)
1. Every consistent set of formulas is satisfiable.
2. If Γ ϕ, then Γ ` ϕ.
Proof.
1. Suppose that Γ is consistent. By Proposition 6.3.12, we may fix a language L0 ⊇ L and ∆ ⊆ F ormL0
such that ∆ ⊇ Γ is consistent, complete, and contains witnesses. Now ∆ is satisfiable by Proposition
6.3.16, so we may fix an L0 -structure M0 together with s : V ar → M 0 such that (M0 , s) ϕ for all
ϕ ∈ ∆. We then have (M0 , s) γ for all γ ∈ Γ. Letting M be the restiction of M0 to L, we then have
(M, s) γ for all γ ∈ Γ. Therefore, Γ is satisfiable.
2. Suppose that Γ ϕ. We then have that Γ ∪ {¬ϕ} is unsatisfiable, hence Γ ∪ {¬ϕ} is inconsistent by
part 1. It follows from Proposition 5.2.12 that Γ ` ϕ.
We now give another proof of the Countable Lowenheim-Skolem Theorem which does not go through the
concept of elementary substructures.
Corollary 6.3.18 (Countable Lowenheim-Skolem Theorem). Suppose that L is countable and Γ ⊆ F ormL
is consistent. There exists a countable model of Γ.
Proof. Notice that if L is consistent, then the L0 formed in Lemma 6.3.9 is countable because F ormL × V ar
is countable. Thus, each Ln in the proof of Proposition 6.3.10 is countable, so the L0 formed in Proposition
6.3.10 is countable. It follows that T ermL0 is countable, and since the L0 -structure M we construct in the
proof of Proposition 6.3.16 is formed by taking the quotient from an equivalence relation on the countable
T ermL0 , we can conclude that M is countable. Therefore, the L-structure which is the restriction of M to
L from the proof of the Completeness Theorem is countable.
6.4 Compactness
Corollary 6.4.1 (Compactness Theorem).
1. If Γ ϕ, then there exists a finite Γ0 ⊆ Γ such that Γ0 ϕ.
2. If every finite subset of Γ is satisfiable, then Γ is satisfiable.
Proof.
1. Suppose that Γ ϕ. By the Completeness Theorem, we have Γ ` ϕ. Using Proposition 5.2.15, we may
fix a finite Γ0 ⊆ Γ such that Γ0 ` ϕ. By the Soundness Theorem, we have Γ0 ϕ.
2. If every finite subset of Γ is satisfiable, then every finite subset of Γ is consistent by the Soundness
Theorem, hence Γ is consistent by Corollary 5.2.16, and so Γ is satisfiable by the Soundness Theorem.
6.5 Applications of Compactness

The next proposition is another result which expresses that first-order logic is not powerful enough to
distinguish certain aspects of cardinality. Here the distinction is between large finite numbers and the
infinite.
Proposition 6.5.1. Let L be a language. Suppose that Γ ⊆ F ormL is such that for all n ∈ N, there exists
a model (M, s) of Γ such that |M | > n. We then have that there exists a model (M, s) of Γ such that M is
infinite.
Proof. Let L0 = L ∪ {ck : k ∈ N} where the ck are new distinct constant symbols. Let
Γ0 = Γ ∪ {ck 6= c` : k, ` ∈ N and k 6= `}
We claim that every finite subset of Γ0 is satisfiable. Fix a finite Γ00 ⊆ Γ0 . Fix N ∈ N such that
Γ00 ⊆ Γ ∪ {ck 6= c` : k, ` ≤ N and k 6= `}
By assumption, we may fix a model (M, s) of Γ such that |M | > N . Let M0 be the L0 structure M together
with interpreting the constants c0 , c1 , . . . , cN as distinct elements of M and interpreting each ci for i > N
arbitrarily. We then have (M0 , s) is a model of Γ0 . Hence, every finite subset of Γ0 is satisfiable.
By the Compactness Theorem we may conclude that Γ0 is satisfiable. Fix a model (M0 , s) of Γ0 . If we
let M be the restriction of M0 to L, then (M, s) is a model of Γ which is infinite.
Corollary 6.5.2. The class of all finite groups is not a weak elementary class in the language L = {F, e}.
Proof. If Σ ⊆ SentL is such that M od(Σ) includes all finite groups, then we may use the trivial fact that there
are arbitrarily large finite groups and Proposition 6.5.1 to conclude that it contains an infinite structure.
Proposition 6.5.3. Let L be a language. Suppose that Γ ⊆ F ormL is such there exists a model (M, s) of
Γ with M infinite. We then have that there exists a model (M, s) of Γ such that M is uncountable.
Proof. Let L0 = L ∪ {cr : r ∈ R} where the cr are new distinct constant symbols. Let
Γ0 = Γ ∪ {cr 6= ct : r, t ∈ R and r 6= t}
We claim that every finite subset of Γ0 is satisfiable. Fix a finite Γ00 ⊆ Γ0 . Fix a finite Z ⊆ R such that
Γ00 ⊆ Γ ∪ {cr 6= ct : r, t ∈ Z}
By assumption, we may fix a model (M, s) of Γ such that M is infinite. Let M0 be the L0 structure M
together with interpreting the constants cr for r ∈ Z as distinct elements of M and interpreting each ct for
t∈/ Z arbitrarily. We then have (M0 , s) is a model of Γ0 . Hence, every finite subset of Γ0 is satisfiable.
By the Compactness Theorem we may conclude that Γ0 is satisfiable. Fix a model (M0 , s) of Γ0 . If we
let M be the restriction of M0 to L, then (M, s) is a model of Γ which is uncountable.
Proposition 6.5.4. The class K of all torsion groups is not a weak elementary class in the language
L = {f, e}.
Proof. Suppose that Σ ⊆ SentL is such that K ⊆ M od(Σ). Let L0 = L ∪ {c} where c is new constant symbol.
For each n ∈ N+ , let τn ∈ SentL0 be cn 6= e (more formally, fcfc · · · fcc where there are n − 1 f’s). Let
Σ0 = Σ ∪ {τn : n ∈ N}
6.5. APPLICATIONS OF COMPACTNESS 99
We claim that every finite subset of Σ0 has a model. Suppose that Σ0 ⊆ Σ0 is finite. Fix N ∈ N such that
Σ0 ⊆ Σ ∪ {τn : n < N }
0
Notice that if we let M0 be the group Z/N Z and let cM = 1, then M0 is a model of Σ0 . Thus, every finite
subset of Σ0 has a model, so Σ0 has a model by Compactness. If we restrict this model to L, we get an
element of M od(Σ) which is not int K because it has an element of infinite order.
Proposition 6.5.5. The class K of all equivalence relations in which all equivalence classes are finite is not
a weak elementary class in the language L = {R}.
Proof. Suppose that Σ ⊆ SentL is such that K ⊆ M od(Σ). Let L0 = L ∪ {c} where c is new constant symbol.
For each n ∈ N+ , let τn ∈ SentL0 be
^ n
^
∃x1 ∃x2 · · · ∃xn ( (xi 6= xj ) ∧ Rcxi )
1≤i<j≤n i=1
and let
Σ0 = Σ ∪ {τn : n ∈ N}
We claim that every finite subset of Σ0 has a model. Suppose that Σ0 ⊆ Σ0 is finite. Fix N ∈ N such that
Σ0 ⊆ Σ ∪ {τn : n ≤ N }
0 0
Notice that if we let M 0 = {0, 1, 2, . . . , N }, RM = (M 0 )2 , and cM = 0, then M0 is a model of Σ0 . Thus,
every finite subset of Σ0 has a model, so Σ0 has a model by Compactness. If we restrict this model to L, we
get an element of M od(Σ) which is not int K because it has an infinite equivalence class.
Proposition 6.5.6. Suppose that K is an elementary class, that Σ ⊆ SentL , and that K = M od(Σ). There
exists a finite Σ0 ⊆ Σ such that K = M od(Σ0 ).
Proof. Since K is an elementary class, we may fix τ ∈ SentL with K = M od(τ ). We then have Σ τ , so by
the Compactness Theorem we may fix a finite Σ0 ⊆ Σ such that Σ0 τ . Notice that K = M od(Σ0 ).
Corollary 6.5.7. The class K of all fields of characteristic 0 is a weak elementary class, but not an ele-
mentary class, in the language L = {0, 1, +, ·}.
Proof. We already know that K is a weak elementary class because if we let σ be the conjunction of the
fields axioms and let τn be 1 + 1 + · · · + 1 6= 0 (where there are n 1’s) for each n ∈ N+ , then K = M od(Σ)
where
Σ = {σ} ∪ {τn : n ∈ N+ }
Suppose that K is an elementary class. By the previous proposition, we may fix a finite Σ0 ⊆ Σ such that
K = M od(Σ0 ). Fix N ∈ N such that
Σ0 ⊆ {σ} ∪ {τn : n ≤ N }
Now if fix a prime p > N we see that (Fp , 0, 1, +, ·) (the field with p elements) is a model of Σ0 which is not
an element of K. This is a contradiction, so K is not an elementary class.
6.6 An Example: The Random Graph

Throughout this section, we work in the language L = {R} where R is binary relation symbol. We think of
graphs as L-structures which are model of {∀x¬Rxx, ∀x∀y(Rxy → Ryx)}.
Definition 6.6.1. For each n ∈ N+ , let Gn be the set of of all models of {∀x¬Rxx, ∀x∀y(Rxy → Ryx)} with
universe [n].
Definition 6.6.2. For each A ⊆ Gn , we let
|A|
P rn (A) =
|Gn |
For each σ ∈ SentL , we let
|{M ∈ Gn : M σ}|
P rn (σ) =
|Gn |
We use the suggestive P r because we think of constucting a graph randomly by flipping a fair coin for
each pair i, j to determine whether or not there is an edge linking them. In this context, P rn (A) is the
probability the graph so constucted on {1, 2, . . . , n} is in A. Notice that given i, j and i0 , j 0 distinct two
element subsets of {1, 2, . . . , n}, the question of whether there is an edge linking i, j and the question of
whether there is an edge linking i0 , j 0 are independent. We aim to prove the following.
Theorem 6.6.3. For all σ ∈ SentL , either lim P rn (σ) = 1 or lim P rn (σ) = 0.
n→∞ n→∞
Definition 6.6.4. For each r, s ∈ N with max{r, s} > 0, let σr,s be the sentence
^ ^ r ^
^ s
∀x1 ∀x2 · · · ∀xr ∀y1 ∀y2 · · · ∀ys ( (xi 6= xj ) ∧ (yi 6= yj ) ∧ (xi 6= yj )
1≤i<j≤r 1≤i<j≤s i=1 j=1
r
^ s
^ r
^ s
^
→ ∃z( (z 6= xi ) ∧ (z 6= yj ) ∧ Rxi z ∧ ¬Ryj z))
i=1 j=1 i=1 j=1
Proposition 6.6.5. For all r, s ∈ N with max{r, s} > 0, we have lim P rn (σr,s ) = 1.
n→∞
Proof. Fix r, s ∈ N. Suppose that n ∈ N with n > r, s. Fix distinct a1 , a2 , . . . , ar , b1 , b2 , . . . , bs ∈ {1, 2, . . . , n}.
For each c distinct from the ai and bj , let
Ac = {M ∈ Gn : c is linked to each ai and to no bj }
1 1
For each such c, we have P rn (Ac ) = 2r+s , so P rn (Ac ) = 1 − 2r+s and hence the probability that no c works
is
1
(1 − )n−r−s
2r+s
Therefore,

n n−r 1
P rn (¬σr,s ) ≤ (1 − r+s )n−r−s
r s 2
1
≤ nr+s (1 − r+s )n−r−s
2
1 −r−s r+s 1
= (1 − r+s ) · n (1 − r+s )n
2 2
1 2r+s − 1
= (1 − r+s )−r−s · nr+s ( r+s )n
2 2
1 −r−s nr+s
= (1 − r+s ) · 2r+s
2 ( 2r+s −1 )n
6.6. AN EXAMPLE: THE RANDOM GRAPH 101
Thus, lim P rn (¬σr,s ) = 0, and hence lim P rn (σr,s ) = 1.

n→∞ n→∞
Proposition 6.6.6. Let Σ = {∀x¬Rxx, ∀x∀y(Rxy → Ryx)} ∪ {σr,s : r, s ∈ N+ and max{r, s} > 0} and let
RG = Cn(Σ).
Proposition 6.6.7. RG is satisfiable.
Proof. We build a countable model M of RG with M = N. Notice first that since Pf in (N) (the set of all
finite subsets of N) is countable, so is the set Pf in (N)2 , Hence the set
{(A, B) ∈ Pf in (N)2 : A ∩ B = ∅ and A ∪ B 6= ∅}
is countable. Therefore, we may list it as
(A1 , B1 ), (A2 , B2 ), (A3 , B3 ), . . .
and furthermore we may assume that max(An ∪ Bn ) < n for all n ∈ N. Let M be the L-structure where
M = N and RM = {(k, n) : k ∈ An } ∪ {(n, k) : k ∈ An }. Suppose now that A, B ⊆ N are finite with
A ∩ B = ∅ and A ∪ B 6= ∅. Fix n ∈ N with A = An . We then have that (k, n) ∈ RM for all k ∈ A and
/ RM for all ` ∈ B. Fix n ∈ N with A = An and B = Bn . We then have that (k, n) ∈ RM for all
(`, n) ∈
k ∈ A (because k ∈ An ) and (`, n) ∈/ RM for all ` ∈ B (because ` ∈
/ An and n ∈
/ A` since ` < n). Therefore,
M σr,s for all r, s ∈ N with max r, s > 0. Thus, M is a model of RG.
Theorem 6.6.8. All models of RG are infinite, and any two countable models of RG are isomorphic.
Proof. Suppose that M is model of RG which is finite. Let n = |M |. Since M σn,0 , there exists b ∈ M
such that (b, a) ∈ RM for all a ∈ M . However, this is a contradiction because (a, a) ∈ / RM for all a ∈ M . It
follows that all models of RG are infinite.
Suppose now that M and N are two countable models of RG. From above, we know that M and N are
both countably infinite. List M as m0 , m1 , m2 , . . . and list N as n0 , n1 , n2 , . . . . We build an isomorphism
via a back-and-forth construction as in the proof of the corresponding result for DLO. That is, we define
σk ∈ Pf in (M × N ) for k ∈ N recursively such that
1. σk ⊆ σk+1 .
2. If (m, n) ∈ σk and (m0 , n) ∈ σk , then m = m0 .
3. If (m, n) ∈ σk and (m, n0 ) ∈ σk , then n = n0 .
4. mi ∈ dom(σ2i ).
5. nj ∈ ran(σ2j+1 ).
6. If (m, n) ∈ σ and (m0 , n0 ) ∈ σ, then (m, m0 ) ∈ RM if and only if (n, n0 ) ∈ RN
Suppose
S that we are successful. Define h : M → N be letting h(m) be the unique n such that (m, n) ∈
k∈N σk , and notice that h is isomorphism.
We now define the σk . Let σ0 = (m0 , n0 ). Suppose that k ∈ N and we’ve defined σk . Suppose first
that k is odd, say k = 2i + 1. If mi ∈ dom(σk ), let σk+1 = σk . Suppose then that mi ∈ / dom(σk ). Let
A = {m ∈ dom(σk ) : (m, mi ) ∈ RM } and let B = {m ∈ dom(σk ) : (m, mi ) ∈ / RM }. Since N is a model of RG
and A ∩ B = ∅, we may fix n ∈ N \ran(σk ) such that (σk (m), n) ∈ RM for all m ∈ A and (σk (m), n) ∈ / RM
for all m ∈ B. Let σk+1 = σk ∪ {(mi , n)}.
Suppose now that k is even, say k = 2j. If nj ∈ ran(σk ), let σk+1 = σk . Suppose then that nj ∈/ ran(σk ).
Let A = {n ∈ ran(σk ) : (n, nj ) ∈ RN } and let B = {n ∈ ran(σk ) : (n, nj ) ∈
/ RN }. Since M is a model of RG
and A∩B = ∅, we may fix m ∈ M \dom(σk ) such that (σk−1 (n), m) ∈ RM for all n ∈ A and (σk−1 (n), m) ∈ / RM
for all n ∈ B. Let σk+1 = σk ∪ {(m, nj )}.
Corollary 6.6.9. RG is a complete theory.

Proof. Apply the Countable Los-Vaught Test.
Theorem 6.6.10. Let τ ∈ SentL .
1. If τ ∈ RG, then lim P rn (τ ) = 1.
n→∞
2. If τ ∈
/ RG, then lim P rn (τ ) = 0.
n→∞
Proof.
1. Suppose that τ ∈ RG. We then have Σ τ , so by Compactness we may fix N ∈ N such that
{∀x¬Rxx, ∀x∀y(Rxy → Ryx)} ∪ {σr,s : r, s ≤ N } τ
We then have that if M ∈ Gn is such that M ¬τ , then

_
M ¬σr,s
0≤r,s≤N,max{r,s}>0
Hence for every n ∈ N we have

X
P rn (¬τ ) ≤ P rn (¬σr,s )
0≤r,s≤N,max{r,s}>0
Therefore, lim P rn (¬τ ) = 0, and hence lim P rn (τ ) = 1.

n→∞ n→∞
2. Suppose that τ ∈
/ RG. Since RG is complete, it follows that ¬τ ∈ RG. Thus, lim P rn (¬τ ) = 1 by
n→∞
part 1, and hence lim P rn (τ ) = 0.
n→∞
Chapter 7
Quantifier Elimination
7.1 Motivation and Definition

Quantifiers make life hard, so it’s always nice when we can find a way to express a statement involving
quantifiers using an equivalent statement without quantifiers.
Examples.
1. Let L = {0, 1, +, ·} and let ϕ(a, b, c) (where a, b, c ∈ V ar) be the formula
∃x(ax2 + bx + c = 0)
or more formally
∃x(a · x · x + b · x + c = 0)
Let M be the L-structure (C, 0, 1, +, ·). Since C is algebraically closed, we have that
(M, s) ϕ ↔ (a 6= 0 ∨ b 6= 0 ∨ c = 0)
for all s : V ar → C.
2. Let L = {0, 1, +, ·, <} and let ϕ(a, b, c) (where a, b, c ∈ V ar) be the formula
∃x(ax2 + bx + c = 0)
Let M be the L-structure (R, 0, 1, +, ·). Using the quadratic formula, we have
(M, s) ϕ ↔ ((a 6= 0 ∧ b2 − 4ac ≥ 0) ∨ (a = 0 ∧ b 6= 0) ∨ (a = 0 ∧ b = 0 ∧ c = 0))
for all s : V ar → R.
The above examples focused on one structure rather than a theory (which could have many models).
The next example uses a theory.
Example. Let L = {0, 1, +, ·} and let T be the theory of fields, i.e. T = Cn(Σ) where Σ is the set of field
axioms. Let ϕ(a, b, c, d) (where a, b, c, d ∈ V ar) be the formula
∃w∃x∃y∃z(wa + xc = 1 ∧ wb + xd = 0 ∧ ya + zc = 0 ∧ yb + zd = 1)
Using determinants, we have
T ϕ ↔ ad 6= bc
103
104 CHAPTER 7. QUANTIFIER ELIMINATION
Definition 7.1.1. Let T be a theory. We say that T has quantifier elimination, or has QE, if for every
k ≥ 1 and every ϕ(x1 , x2 , . . . , xk ) ∈ F ormL , there exists a quantifier-free ψ(x1 , x2 , . . . , xk ) such that
T ϕ↔ψ
7.2 What Quantifier Elimination Provides

The first application of a using QE is to show that certain theories are complete. QE itself is not sufficient,
but a very mild additional assumption gives us what we want.
Proposition 7.2.1. Let T be a theory that has QE. If there exists an L-structure N such that for every
model M of T there is an embedding h : N → M from N to M, then T is complete. (Notice, there is no
assumption that N is a model of T .)
Proof. Fix an L-structure N such that for every model M of T there is an embedding h : N → M from N to
M, and fix n ∈ N . Let M1 and M2 be two models of T . For each i ∈ {1, 2}, fix an embedding hi : N → Mi
from N to Mi . For each i, let Ai = ran(hi ), and notice that Ai is the universe of a substructure Ai of Mi .
Furthermore, notice that hi is an isomorphism from N to Ai .
Let σ ∈ SentL and let ϕ(x) ∈ F ormL be the formula σ ∧ (x = x). Since T has QE, we may fix a
quantifier-free ψ(x) ∈ F ormL such that T ϕ ↔ ψ. We then have
M1 σ ⇔ (M1 , h1 (n)) ϕ
⇔ (M1 , h1 (n)) ψ
⇔ (A1 , h1 (n)) ψ (since ψ is quantifier-free)
⇔ (N , n) ψ (since h1 is an isomorphism from N to A1 )
⇔ (A2 , h2 (n)) ψ (since h2 is an isomorphism from N to A2 )
⇔ (M2 , h2 (n)) ψ (since ψ is quantifier-free)
⇔ (M2 , h2 (n)) ϕ
⇔ M2 σ
Proposition 7.2.2. Let T be a theory that has QE. Suppose that A and M are models of T and that
A ⊆ M. We then have that A M.
Proof. Let ϕ ∈ F ormL and let s : V ar → A be a variable assignment. Suppose first that ϕ ∈
/ SentL . Since
T has QE, we may fix a quantifier-free ψ(x) ∈ F ormL such that T ϕ ↔ ψ. We then have
(M, s) ϕ ⇔ (M, s) ψ
⇔ (A, s) ψ (since ψ is quantifier-free)
⇔ (A, s) ϕ
If ϕ is a sentence, we may tack on a dummy x = x as in the previous proof.

Proposition 7.2.3. Let T be a theory that has QE, let M be a model of T , and let k ∈ N+ . Let Z be the
set of all subsets of M k which are definable by atomic formuals. The set of definable subsets of M k equals
G(P(M k ), Z, {h1 , h2 }) where h1 : P(M k ) → P(M k ) is the complement function and h2 : P(M k )2 → P(M k )
is the union function.
Proof.
7.3. QUANTIFIER MANIPULATION RULES 105
7.3 Quantifier Manipulation Rules

Definition 7.3.1. Let L be a language, and let ϕ, ψ ∈ F ormL . We say that ϕ and ψ are semantically
equivalent if ϕ ψ and ψ ϕ.
We now list a bunch of simple rules for manipulating formulas while maintaing.
1. ¬(∃xϕ) and ∀x(¬ϕ) are s.e.
2. ¬(∀xϕ) and ∃x(¬ϕ) are s.e.
3. (∃xϕ) ∧ ψ and ∃x(ϕ ∧ ψ) are s.e. if x ∈

/ F reeV ar(ψ).
4. (∀xϕ) ∧ ψ and ∀x(ϕ ∧ ψ) are s.e. if x ∈

/ F reeV ar(ψ).
5. (∃xϕ) ∨ ψ and ∃x(ϕ ∨ ψ) are s.e. if x ∈

/ F reeV ar(ψ).
6. (∀xϕ) ∨ ψ and ∀x(ϕ ∨ ψ) are s.e. if x ∈

/ F reeV ar(ψ).
7. (∃xϕ) → ψ and ∀x(ϕ → ψ) are s.e. if x ∈

/ F reeV ar(ψ).
8. (∀xϕ) → ψ and ∃x(ϕ → ψ) are s.e. if x ∈

/ F reeV ar(ψ).
We’ll need the following to change annoying variables.
1. ∃xϕ and ∃y(ϕyx ) are s.e. if y ∈

/ OccurV ar(ϕ).
2. ∀xϕ and ∀y(ϕyx ) are s.e. if y ∈

/ OccurV ar(ϕ).
We’ll also need to know that if ϕ and ψ are s.e., then
1. ¬ϕ and ¬ψ are s.e.
2. ∃xϕ and ∃xψ are s.e.
3. ∀xϕ and ∀xψ are s.e.
and also that if ϕ1 are ϕ2 s.e., and ψ1 and ψ2 are s.e., then
1. ϕ1 ∧ ψ1 and ϕ2 ∧ ψ2 are s.e.
2. ϕ1 ∨ ψ1 and ϕ2 ∨ ψ2 are s.e.
3. ϕ1 → ψ1 and ϕ2 → ψ2 are s.e.
Definition 7.3.2. A quantifier-free formula ϕ is in disjunctive normal form if there exists ψi for 1 ≤ i ≤ n
such that
ϕ = ψ1 ∨ ψ2 ∨ · · · ∨ ψn
where for each i, there exists αi,j which is either an atomic formula or the negation of an atomic formula
for 1 ≤ j ≤ mi such that
ψi = αi,1 ∧ αi,2 ∧ · · · ∧ αi,mi
Proposition 7.3.3. Suppose that ϕ(x1 , x2 , . . . , xk ) ∈ F ormL is quantifier-free. There exists a quantifier-free
formula θ(x1 , x2 , . . . , xk ) in disjunctive normal form such that ϕ and θ are s.e.
Proof.
Definition 7.3.4. A formula ϕ is called a prenex formula if it is an element of
G(Sym∗L , QuantF reeF ormL , {h∀,x , h∃,x : x ∈ V ar})
Proposition 7.3.5. For every ϕ ∈ F ormL , there exists a prenex formula ψ such that ϕ and ψ are seman-
tically equivalent.
Proposition 7.3.6. Let T be a theory. The following are equivalent
1. T has QE.
2. For all k ≥ 1 and all α1 , . . . , αm , β1 , . . . , βn ∈ F ormL with
(a) F reeV ar(α1 , . . . , αm , β1 , . . . , βn ) ⊆ {y, x1 , x2 , . . . , xk }.

(b) y ∈ F reeV ar(αi ) for all i and y ∈ F reeV ar(βj ) for all j.
(c) Each αi and βj is an atomic formula.
there exists a quantifier-free ψ(x1 , x2 , . . . , xk ) ∈ F ormL such that

m
^ n
^
T ∃y( αi ∧ ¬βj ) ↔ ψ
i=1 j=1
Proof.
7.4 Examples of Theories With QE

Theorem 7.4.1. Let L = ∅. For each n ∈ N+ , let σn be the sentence
^
∃x1 ∃x2 · · · ∃xn ¬(xi = xj )
1≤i<j≤n
Let T = Cn({σn : n ∈ N+ }). T has QE and is complete.
Proof. Suppose that α1 , . . . , αm , β1 , . . . , βn ∈ F ormL with
1. F reeV ar(α1 , . . . , αm , β1 , . . . , βn ) ⊆ {y, x1 , x2 , . . . , xk }.
2. y ∈ F reeV ar(αi ) for all i and y ∈ F reeV ar(βj ) for all j.
3. Each αi and βj is an atomic formula.
We need to show that there exists a quantifier-free ψ(x1 , x2 , . . . , xk ) ∈ F ormL such that
m
^ n
^
T ∃y( αi ∧ ¬βj ) ↔ ψ
i=1 j=1
Now each αi and βj is s.e. with, and hence we may assume is, one of the following:
1. x` = y
2. y = y
7.4. EXAMPLES OF THEORIES WITH QE 107
If some αi is x` = y, then
m
^ n
^ m
^ n
^
T ∃y( αi ∧ ¬βj ) ↔ ( αi ∧ ¬βj )xy`
i=1 j=1 i=1 j=1
If some βj is y = y, then
m
^ n
^
T ∃y( αi ∧ ¬βj ) ↔ ¬(x1 = x1 )
i=1 j=1
Suppose then that each αi is y = y and each βj is x` = y. We then have

m
^ n
^
T ∃y( αi ∧ ¬βj ) ↔ x1 = x1
i=1 j=1
because all models of T are infinite. Therefore, T has QE.

Notice that T is complete because the structure M given by M = {0} trivially embeds into all models
of T .
Theorem 7.4.2. RG has QE and is complete.
Proof. Suppose that α1 , . . . , αm , β1 , . . . , βn ∈ F ormL with
1. F reeV ar(α1 , . . . , αm , β1 , . . . , βn ) ⊆ {y, x1 , x2 , . . . , xk }.
2. y ∈ F reeV ar(αi ) for all i and y ∈ F reeV ar(βj ) for all j.
3. Each αi and βj is an atomic formula.
We need to show that there exists a quantifier-free ψ(x1 , x2 , . . . , xk ) ∈ F ormL such that
m
^ n
^
RG ∃y( αi ∧ ¬βj ) ↔ ψ
i=1 j=1
Now each αi and βj is RG-equivalent with, and hence we may assume is, one of the following:
1. x` = y
2. Rx` y
3. y = y
4. Ryy
If some αi is x` = y, then
m
^ n
^ m
^ n
^
RG ∃y( αi ∧ ¬βj ) ↔ ( αi ∧ ¬βj )xy`
i=1 j=1 i=1 j=1
If some αi is Ryy or some βj is y = y, then

m
^ n
^
RG ∃y( αi ∧ ¬βj ) ↔ ¬(x1 = x1 )
i=1 j=1
Suppose then that no αi is x` = y, no αi is Ryy, and no βj is y = y. Let
A = {` ∈ {1, 2, . . . , k} : there exists i such that αi is Rx` y}
and let
B = {` ∈ {1, 2, . . . , k} : there exists j such that βj is Rx` y}
We then have that
m
^ n
^ ^ ^
RG ∃y( αi ∧ ¬βj ) ↔ ¬(xa = xb )
i=1 j=1 a∈A b∈B
because in models of RG, given disjoint finite sets A and B of vertices, there are infinitely many vertices
linked to everything in A and not linked to everything in B. Therefore, RG has QE.
Notice that RG is complete because the structure M given by M = {0} and RM = ∅ trivially embeds
into all models of RG.
7.5 Algebraically Closed Fields

Definition 7.5.1. Let L = {0, 1, +, ·}. Let Σ ⊆ SentL be the field axioms together with the sentences
∀a0 ∀a1 · · · ∀an (an 6= 0 → ∃x(an xn + · · · + a1 x + a0 = 0))
for each n ∈ N+ . Let ACF = Cn(Σ).

Theorem 7.5.2. ACF has QE.
Proof Sketch. The fundamental observation is that we can think of atomic formulas with free variables in
{y, x1 , x2 , . . . , xk } as equations p(~x, y) = 0 where p(~x, y) ∈ Z[~x, y] is a polynomial.
Thus, we have to find quantifier-free equivalents to formulas of the form
m
^ n
^
∃y[ (pi (~x, y) = 0) ∧ (qj (~x, y) 6= 0)]
i=1 j=1
Qn
which, letting q(~x, y) = j=1 qj (~
x, y), is equivalent in ACF to
m
^
∃y[ (pi (~x, y) = 0) ∧ q(~x, y) 6= 0]
i=1
Suppose now that R is a ring and p1 , p2 , . . . , pm , q ∈ R[y] listed in decreasing order of degrees. Let the
leading term of p1 be ay n and let the leading term of pm be by k . We then have that there is a simlutaneous
root of polynomials p1 , p2 , . . . , pm which is not a root of q if and only if one of the following happens:
1. b = 0 and there is simlutaneous root of polynomials p1 , p2 , . . . , pm−1 , pm − by k which is not a root of q.
2. b 6= 0 and there is a simultaneous root of the polynomials bp1 − ay n−k pm , p2 , . . . , pm which is not a
root of q.
Repeating this, we may assume that we have a forumla of the form
∃y[p(~x, y) = 0 ∧ q(~x, y) 6= 0]
If q is not present, then we may use the fact that in an algebraically closed field, the polynomial an y n +
· · · + a1 y + a0 has a root if and only if some ai 6= 0 for i > 0, or a0 = 0. If p is not present, then we use
7.5. ALGEBRAICALLY CLOSED FIELDS 109
the fact that every algebraically closed field is infinite and also that every nonzero has finitely many roots to
conclude that the there is an element which is not a root of the polynomial an y n + · · · a1 y + a0 if and only
if some ai 6= 0.
Suppose then that both p and q are present. The key fact is to use here is that if p and q are polynomials
over an algebraically closed field and the degree of p is at most n, then every root of p is a root of q if and
only if p | q n .
Thus, we have two polynomials p and q, and we want to find a quantifier formula equivlent to p | q. We
may now use the Euclidean algorithm to repeatedly reduce the problem.
Corollary 7.5.3. ACF0 is complete and ACFp is complete for all primes p.
Corollary 7.5.4. If F and K are algebraically closed fields such that F is a subfield of K, then (F, 0, 1, +, ·)
(K, 0, 1, +, ·).
Corollary 7.5.5. (Q, 0, 1, +, ·) (C, 0, 1, +, ·).

Corollary 7.5.6. Suppose that F is an algebraically closed field. A subset X ⊆ F is definable in (F, 0, 1, +, ·)
if and only if X is either finite or cofinite.
Proposition 7.5.7. Let σ ∈ SentL . The following are equivalent.

1. ACF0 σ.
2. There exists m such that ACFp σ for all primes p > m.
3. ACFp σ for infinitely many primes p.
Proof. 1 imples 2 is Compactness. 2 implies 3 is trivial. 3 implies 1 using completeness of ACF0 and the
ACFp for each prime p, together with 1 implies 2.
Proposition 7.5.8. Let p be prime. Every finitely generated subfield of Fp is finite.
n
Proof. Let p be prime. For every n, let Kn be the set of roots of xp − x in Fp . By standard results
in algebra, we have that Kn is a field of order pn , and furthermore is the unique subfield of Fp of order
d 2·d d d d
pn . If d | n, we then have that Kd ⊆ Kn because if ap = a, then ap = (ap )p = ap = a, so
3·d 2·d d d
ap = (ap )p = ap = a, etc. Let K = n∈N Kn . Notice that K is a subfield of Fp because if a ∈ Kn
S
and b ∈ Km , then a + b, a · b ∈ Km·n . Furthermore, notice that K is algebraically closed because a finite
extension of a finite field is finite. Therefore, K = Fp .
Theorem 7.5.9. Every injective polynomial map from Cn to Cn is surjective.
Proof. Let σn,d ∈ SentL be the sentence expressing that every injective polynomial map from F n to F n
where each polynomial has degree at most d is surjective. We want to show that C σn,d for all n, d. To do
this, it suffices to show that Fp σn,d for all primes p and all n, d ∈ N. Thus, it suffices to show that for all
n n
primes p, every injective polynomial map from Fp to Fp is surjective.
n n
Fix a prime p and an n ∈ N. Suppose that f : Fp → Fp is an injective polynomial map. Let
n n
(b1 , b2 , . . . , bn ) ∈ Fp . We need to show that there exists (a1 , a2 , . . . , an ) ∈ Fp with f (a1 , a2 , . . . , an ) =
(b1 , b2 , . . . , bn ). Let f1 , f2 , . . . , fn ∈ Fp [x1 , x2 , . . . , xn ] be such that f = (f1 , f2 , . . . , fn ), and let C be the fi-
nite set of coefficients appearing in f1 , f2 , . . . , fn . Let K be the subfield of Fp generated by C ∪{b1 , b2 , . . . , bn }
and notice that K is a finite field. Now f K n maps K n into K n and is injective, so it’s surjective because
n
K n is finite. Thus, there exists (a1 , a2 , . . . , an ) ∈ K n ⊆ Fp such that f (a1 , a2 , . . . , an ) = (b1 , b2 , . . . , bn ).
Chapter 8
Nonstandard Models of Arithmetic

and Analysis
8.1 Nonstandard Models of Arithmetic

Throughout this section, we work in the language L = {0, 1, <, +, ·} where 0, 1 are constant symbols, < is
a binary relation symbol, and +, · are binary function symbols. We also let N = (N, 0, 1, <, +, ·) where the
symbol 0 is interpreted as the “real” 0, the symbol + is interpreted as “real” addition, etc. Make sure that
you understand when + means the symbol in the language L and when it mean the addition function on N.
A basic question is whether T h(N) compeletely determines the model N. More precisely, we have the
following question.
Question 8.1.1. Are all models of T h(N) isomorphic to N?
Using Proposition 6.5.3, we can immediately give a negative answer to this question because there is an
uncountable model of T h(N), and an uncountable model can’t be isomorphic to N. What would such a
model look like? In order to answer this, let’s think a little about the kinds of sentences that are in T h(N).
Definition 8.1.2. For each n ∈ N, we define a term n ∈ T ermL as follows. Let 0 = 0 and let 1 = 1. Now
define the n recursively by letting n + 1 = n + 1 for each n ≥ 1. Notice here that the 1 and the + in n + 1
mean the actual number 1 and the actual addition function, whereas the 1 and + in n + 1 mean the symbols
1 and + in our language L. Thus, for example, 2 is the term 1 + 1 and 3 is the term (1 + 1) + 1.
Definition 8.1.3. Let M be an L-structure. We know that given any t ∈ T ermL containing no variables, t
corresponds to an element of M given by s(t) for some (any) variable assignment s : V ar → M . We denote
this value by tM .
Notice that nN = n for all n ∈ N be a simple induction. Here are some important examples of the kinds
of things in T h(N).
Examples of Sentences in T h(N).
1. 2 + 2 = 4 and in general m + n = m + n and m · n = m · n.
2. ∀x∀y(x + y = y + x)
3. ∀x(x 6= 0 → ∃y(y + 1 = x))
111
112 CHAPTER 8. NONSTANDARD MODELS OF ARITHMETIC AND ANALYSIS
4. For each ϕ(x) ∈ F ormL , the sentence
(ϕ0x ∧ ∀x(ϕ → ϕx+1

x )) → ∀xϕ
Now any model M of T h(N) must satisfy all of these sentences. The basic sentences in 1 above roughly
tell us that M has a piece which looks just like N. We make this precise as follows.
Proposition 8.1.4. For any model M of T h(N), the function h : N → M given by h(n) = nM is an
embedding of N into M.
Proof. Notice that
h(0N ) = h(0) = 0M = 0M
and
h(1N ) = h(1) = 1M = 1M
Now let m, n ∈ N. We have
m<n⇔Nm<n
⇔ m < n ∈ T h(N)
⇔Mm<n
⇔ mM <M nM
⇔ h(m) <M h(n)
Also, since m + n = m + n ∈ T h(N) we have
h(m + n) = (m + n)M
= m M +M n M
= h(m) +M h(n)
and since m · n = m · n ∈ T h(N) we have
h(m · n) = (m · n)M
= mM ·M nM
= h(m) ·M h(n)
Finally, for any m, n ∈ N with m 6= n, we have m 6= n ∈ T h(N), so M m 6= n, and hence h(m) 6= h(n).
Proposition 8.1.5. Let M be a model of T h(N). The following are equivalent.
1. M ∼
= N.
2. M = {nM : n ∈ N}.
Proof. If 2 holds, then the h of the Proposition 8.1.4 is surjective and hence an isomorphism. Suppose then
that 1 holds and fix an isomorphism h : N → M from N to M. We show that h(n) = nM for all n ∈ N by
induction. We have
h(0) = h(0N ) = 0M
and
h(1) = h(1N ) = 1M
8.2. THE STRUCTURE OF NONSTANDARD MODELS OF ARITHMETIC 113
Suppose that n ∈ N and h(n) = nM . We then have
h(n + 1) = h(n) +M h(1)

= n M +M 1 M
= (n + 1)M
Therefore, h(n) = nM for all n ∈ N, so M = {nM : n ∈ N} because h is surjective.
Definition 8.1.6. A nonstandard model of arithmetic is a model M of T h(N) such that M ∼

6 N.
=
We’ve already seen that there are nonstandard models of arithmetic by cardinality considerations, but
we can also build countable nonstandard models of arithmetic using the Compactness Theorem and the
Countable Lowenheim-Skolem Theorem.
Theorem 8.1.7. There exists a countable nonstandard model of arithmetic.
Proof. Let L0 = L ∪ {c} where c is a new constant symbol. Consider the following set of L0 -sentences.
Σ0 = T h(N) ∪ {c 6= n : n ∈ N}
Notice that every finite subset of Σ0 has a model (by taking N and interpreting c large enough), so Σ0
has a countable model M (notice that L0 is countable) by the Compactness Theorem and the Countable
Lowenheim-Skolem Theorem. Restricting this model to the original language L, we may use the Proposition
8.1.5 to conclude that M is a countable nonstandard model of arithmetic.
8.2 The Structure of Nonstandard Models of Arithmetic

Throughout this section, let M be a nonstandard model of arithmetic. Anything we can express in the
first-order language of L which is true of N is in T h(N), and hence is true in M. For example, we have the
following.
Proposition 8.2.1.
• +M is associative on M .
• +M is commutative on M .
• <M is a linear ordering on M .
• For all a ∈ M with a 6= 0M , there exists b ∈ M with a + 1 = b.
Proof. The sentences
• ∀x∀y∀z(x + (y + z) = (x + y) + x)
• ∀x∀y(x + y = y + x)
• ∀x∀y(x < y ∨ y < x ∨ x = y)
• ∀x(x 6= 0 → ∃y(y + 1 = x))
are in T h(N).
Since we already know that N is naturally embedded in M, and it gets tiresome to write +M , ·M , and
<M , we’ll abuse notation by using just +, ·, and < to denote these. Thus, these symbols now have three
different meanings. They are used as formal symbols in our language, as the normal functions and relations
in N, and as their interpretations in M. Make sure you know how each appearance of these symbols is being
used.
Definition 8.2.2. We let Mf in = {nM : n ∈ N} and we call Mf in the set of finite elements of M. We also
let Minf = M \Mf in and we call Minf the set of infinite elements of M.
The following definition justifies our choice of name.
Proposition 8.2.3. Let a ∈ Minf . For any n ∈ N, we have nM < a.
Proof. For each n ∈ N, the sentence
n−1
_
∀x(x < n → (x = i))
i=0
is in T h(N). Since a 6= nM for all n ∈ N, it follows that it’s not the case that a < nM for all n ∈ N. Since
< is a linear ordering on M , we may conclude that nM < a for all n ∈ N.
Definition 8.2.4. Define a relation ∼ on M by letting a ∼ b if either
• a = b.
• a < b and there exists n ∈ N such that a + nM = b.
• b < a and there exists n ∈ N such that b + nM = a.
In other words, a ∼ b if a and b are “finitely” far apart.
Proposition 8.2.5. ∼ is an equivalence relation on M .
Proof. ∼ is clearly relexive and symmetric. Suppose that a, b, c ∈ M , that a ∼ b, and that b ∼ c. We handle
one case. Suppose that a < b and b < c. Fix m, n ∈ N with a + mM = b and b + nM = c. We then have
a + (m + n)M = a + (mM + nM )
= (a + mM ) + nM
= b + nM
=c
so a ∼ c. The other cases are similar.
Definition 8.2.6. Let a, b ∈ M . We write a b to mean that a < b and a 6∼ b.
We’d like to know that that relation is well-defined on the equivalence classes of ∼. The following
lemma is useful.
Lemma 8.2.7. Let a, b, c ∈ M be such that a ≤ b ≤ c and suppose that a ∼ c. We then have a ∼ b and
b ∼ c.
Proof. If either a = b or b = c, this is trivial, so assume that a < b < c. Since a < c and a ∼ c, there exists
n ∈ N+ with a + nM = c. Now the sentence
∀x∀z∀w(x + w = z → ∀y((x < y ∧ y < z) → ∃u(u < w ∧ x + u = y)))
is in T h(N), so there exists d ∈ M such that d < nM and a + d = b. Since d < nM , there exists i ∈ N with
d = iM . We then have a + iM = b, hence a ∼ b. The proof that b ∼ c is similar.
8.2. THE STRUCTURE OF NONSTANDARD MODELS OF ARITHMETIC 115
Proposition 8.2.8. Suppose that a0 , b0 ∈ M are such that a0 b0 . For any a, b ∈ M with a ∼ a0 and
b ∼ b0 , we have a b.
Proof. We first show that a 6∼ b. If a ∼ b, then using a0 ∼ a and b0 ∼ b, together with the fact that ∼ is an
equivalence relation, we can conclude that a0 ∼ b0 , a contradiction. Therefore, a 6∼ b.
Thus, we need only show that a < b. Notice that a0 < b because otherwise a0 ∼ b0 by Lemma 8.2.7.
Similarly, a < b0 because otherwise a0 ∼ b0 by Lemma 8.2.7. Thus, if b ≤ a, we have
a0 < b ≤ a < b0 .
so b ∼ a0 by Lemma 8.2.7, hence a0 ∼ b0 , a contradiction. It follows that a < b.
This allows us to define an ordering on the equivalence classes.
Definition 8.2.9. Given a, b ∈ M , we write [a] ≺ [b] to mean that a b.
The next proposition implies that there is no largest equivalence class under the ordering ≺.
Proposition 8.2.10. For any a ∈ Minf , we have a a + a.
Proof. Let a ∈ Minf . For each n ∈ N, the sentence
∀x(n < x → x + n < x + x)
is in T h(N). Using this when n = 0, we see that a = a + 0M < a + a. Since a ∈ Minf , we have nM < a and
hence a + nM < a + a for all n ∈ N. Therefore, a + nM 6= a + a for all n ∈ N, and so a 6∼ a + a.
Lemma 8.2.11. For all a ∈ M , one of the following holds
1. There exists b ∈ M such that a = 2M · b.
2. There exists b ∈ M such that a = 2M · b + 1M .
Proof. The sentence

∀x∃y(x = 2 · y ∨ x = 2 · y + 1)
is in T h(N).
Proposition 8.2.12. For any a ∈ Minf , there exists b ∈ Minf with b a.
Proof. Suppose first that we have a b ∈ M such that a = 2M · b. We then have a = b + b (because
∀x(2 · x = x + x) is in T h(N)). Notice that b ∈
/ Mf in because otherwise we would have a ∈ Mf in . Therefore,
b b + b = a using Proposition 8.2.10. Suppose instead that we have a b ∈ M such that a = 2M · b + 1M .
We then have a = (b + b) + 1M because ∀x(2 · x + 1 = (x + x) + 1) is in T h(N). Notice that b ∈
/ Mf in because
otherwise we would have a ∈ Mf in . Therefore, b b + b using Proposition 8.2.10, so b (b + b) + 1 = a
since b + b ∼ (b + b) + 1.
Proposition 8.2.13. For any a, b ∈ Minf with a b, there exists c ∈ Minf with a c b.
Proof. Suppose first that we have a c ∈ M such that a + b = 2M · c. We then have a + b = c + c. Since
∀x∀y∀z((x < y ∧ x + y = z + z) → (x < z ∧ z < y))
is in T h(N) it follows that a < c < b.

Suppose that a ∼ c and fix n ∈ N with a + nM = c. We then have that a + b = c + c = a + a + (2n)M ,
so b = a + (2n)M contradicting the fact that a b. Therefore a 6∼ c.
Suppose that c ∼ b and fix n ∈ N with c + nM = b. We then have that
a + (2n)M + b = c + c + (2n)M
= (c + nM ) + (c + nM )
=b+b
so a + (2n)M = b contradicting the fact that a 6∼ b. Therefore, b 6∼ c.

A similar argument handles the case when we have a c ∈ M such that a + b = 2M · c + 1M .
This last proposition shows how nonstandard models can simplify quantifiers. It says that asking whether
a first-order statement holds for infinitely many n ∈ N is equivalent to asking whether it holds for at least
one infinite element of a nonstandard model.
Proposition 8.2.14. Let ϕ(x) ∈ F ormL . The following are equivalent.
1. There are infinitely many n ∈ N such that (N, n) ϕ.
2. There exists a ∈ Minf such that (M, a) ϕ.
Proof. Suppose first that there are infinitely many n ∈ N such that (N, n) ϕ. In this case, the sentence
∀y∃x(y < x ∧ ϕ)
is in T h(N), so it holds in M. Fixing any b ∈ Minf , we may conclude that there exists a ∈ M with b < a
such that (M, a) ϕ. Since b < a and b ∈ Minf , we may conclude that a ∈ Minf .
Conversely, suppose that there are only finitely many n ∈ N such that (N, a) ϕ. Fix N ∈ N such that
n < N for all n with (N, n) ϕ. We then have that the sentence
∀x(ϕ → x < N )
is in T h(N), so it holds in M. Since there is no a ∈ Minf with a < N M , it follows that there is no a ∈ Minf
such that M ϕ(a).
8.3 Nonstandard Models of Analysis

With a basic understanding of nonstandard models of arithmetic, let’s think about nonstandard models of
other theories. One of the more amazing and useful such theories is the theory of the real numbers. The
idea is that we will have nonstandard models of the theory of the reals which contain in “infinite” and
“infinitesimal” elements. We can then transfer first-order statements back-and-forth, and do “calculus” in
this expanded stucture where the basic definitions (of say continuity) are simpler and more intuitive.
The first thing we need to decide on is what our language will be. Since we want to do calculus, we want
to have analogs of all of our favorite functions (such as sin) in the nonstandard models. Once we through
these in, it’s hard to know where to draw the line. In fact, there is no reason to draw a line at all. Simply
throw in relation symbols for every possible subset of Rk , and throw in function symbols for every possible
function f : Rk → R. Thus, throughout this section, we work in the language L = {r : r ∈ R} ∪ {P : P ⊆
Rk } ∪ {f : f : Rk → R} where the P and f have the corresponding arities. We also let R be the structure
with universe R and where we interpret all symbols in the natural way.
Proposition 8.3.1. For any model M of T h(R), the function h : R → M given by h(n) = rM is an
embedding of R into M.
8.3. NONSTANDARD MODELS OF ANALYSIS 117
Proof. Notice that

h(rR ) = h(r) = rM
for every r ∈ R. Now let P ⊆ Rk and let r1 , r2 , . . . , rk ∈ R. We have
(r1 , r2 , . . . , rn ) ∈ P R ⇔ R P r1 r2 · · · rk
⇔ P r1 r2 · · · rk ∈ T h(R)
⇔ M P r1 r2 · · · rk
⇔ (r1 M , r2 M , . . . , rk M ) ∈ P M
⇔ (h(r1 ), h(r2 ), . . . , h(rk )) ∈ P M
Now let f : Rk → R and let r1 , r2 , . . . , rk ∈ R. Since f r1 r2 · · · rk = f (r1 , r2 , . . . , rk ) ∈ T h(R) we have
h(f R (r1 , r2 , . . . , rk )) = h(f (r1 , r2 , . . . , rk ))

M
= f (r1 , r2 , . . . , rk )
= f M (r1 M , r2 M , . . . , rk M )
= f M (h(r1 ), h(r2 ), . . . , h(rk ))
Finally, for any r1 , r2 ∈ R with r1 6= r2 , we have r1 6= r2 ∈ T h(R), so M r1 6= r2 , and hence h(r1 ) 6=

h(r2 ).
Proposition 8.3.2. Let M be a model of T h(R). The following are equivalent.
1. M ∼
= R.
2. M = {rM : r ∈ R}.
Proof. If 2 holds, then the h of the Proposition 8.3.1 is surjective and hence an isomorphism. Suppose
then that 1 holds and fix an isomorphism h : R → M from R to M. For any r ∈ R, we must have
h(r) = h(rR ) = rM . Therefore, M = {rM : r ∈ R} because h is surjective.
Definition 8.3.3. A nonstandard model of analysis is a model M of T h(R) such that M ∼

6 R.
=
Theorem 8.3.4. There exists a nonstandard model of analysis.
Proof. Let L0 = L ∪ {c} where c is a new constant symbol. Consider the following set of L0 -sentences.
Σ0 = T h(R) ∪ {c 6= r : r ∈ R}
Notice that every finite subset of Σ0 has a model (by taking R and interpreting c distinct from each r
such that r appears in Σ0 ), so Σ0 has a model M by the Compactness Theorem. Restricting this model to
the original language L, we may use the Proposition 8.3.2 to conclude that M is a nonstandard model of
analysis.
Definition 8.3.5. ∗For the rest of this section, fix a nonstandard model of analysis and denote it by ∗ R.
Instead of wrting f R for each f : Rk → R, we simply write ∗ f . We use similar notation for each P ⊆ Rk .
Also, since there is a natural embedding (the h above) from R into ∗ R, we will identify R with it’s image
and hence think of R as a subset of ∗ R. Finally, for operations like + and ·, we will abuse notation and omit
the ∗’s.
Proposition 8.3.6. There exists z ∈ ∗ R such that z > 0 and z < ε for all ε ∈ R with ε > 0.
Proof. Fix b ∈ ∗ R such that b 6= r for all r ∈ R.

Case 1: Suppose that b > r for all r ∈ R. Let f : R → R be the function
(
1
if r 6= 0
f (r) = r
0 otherwise
Let z = ∗ f (b). We then have have that z > 0 using the sentence
∀x(0 < x → 0 < f x)
Also, for any ε ∈ R with ε > 0, we have that b > 1ε , hence z < ε using the sentence
∀x(f ε < x → f x < ε)
Case 2: Suppose that b < r for all r ∈ R. We then have that b < −r for all r ∈ R and hence r < −b for
all r ∈ R. Thus, we may tak z = ∗ f (−b) by the argument in Case 1.
Case 3: Suppose then that there exists r ∈ R with r < b and there exists r ∈ R with b < r. Let
X = {r ∈ R : r < b}
Notice that X is downward closed (if r1 , r2 ∈ R with r2 ∈ X and r1 < r2 , then r1 ∈ X), nonempty, bounded
above, Let s = sup X ∈ R. Now b = s is impossible, so either s < b or b < s.
Subcase 1: Suppose that s < b. We claim that we may take z = b − s. Since s < b, we have z = b − s > 0.
Suppose that ε ∈ R and ε > 0. We then have that s + ε > s = sup X, so s + ε ∈ / X and hence s + ε ≥ b.
Now s + ε 6= b because s + ε ∈ R, so s + ε > b. It follows that z = b − s < ε.
Subcase 2: Suppose that b < s. We claim that we may take z = s − b. Since b < s, we have z = s − b > 0.
Suppose that ε ∈ R and ε > 0. We then that s − ε < s = sup X, so we may fix r ∈ X with s − ε < r. Since
X is downward closed, we have that s − ε ∈ X, so s − ε < b. It follows that z = s − b < ε.
From now on, we’ll use the more natural notation 1

b for ∗ f (b) whenever b 6= 0.
Definition 8.3.7.
1. Z = {a ∈ ∗ R : |a| < ε for all ε ∈ R with ε > 0}. We call Z the set of infinitesimals.
2. F = {a ∈ ∗ R : |a| < r for some r ∈ R with r > 0}. We call F the set of finite or limited elements.
3. I = ∗ R\F. We call I the set of infinite or unlimited elements.
Proposition 8.3.8.
1. Z is a subring of ∗ R.
2. F is a subring of ∗ R.
3. Z is a prime ideal of F.
Proof.
1. First notice that Z =

6 ∅ because 0 ∈ Z (or we can use Proposition 8.3.6). Suppose that a, b ∈ Z. Let
ε ∈ R with ε > 0.
ε ε ε
We have that 2 ∈ R and 2 > 0, hence |a| < 2 and |b| < 2ε . It follows that
|a − b| ≤ |a + (−b)|
≤ |a| + | − b|
≤ |a| + |b|
ε ε
< +
2 2
=ε
Therefore, a − b ∈ Z. We also have that |a| < 1 and |b| < ε, hence
|a · b| = |a| · |b|
<1·ε
=ε
Therefore, a · b ∈ Z.
2. Clearly, F 6= ∅. Suppose that a, b ∈ F, and fix r1 , r2 ∈ R with r1 , r2 > 0 such that |a| < r1 and |b| < r2 .
We have
|a − b| ≤ |a + (−b)|
≤ |a| + | − b|
≤ |a| + |b|
< r1 + r2
so a − b ∈ F. We also have
|a · b| = |a| · |b|
< r1 · r2
so a · b ∈ F.
3. We first show that Z is an ideal of F. Suppose that a ∈ F and b ∈ Z. Fix r ∈ R with r > 0 and
|a| < r. Let ε ∈ R with ε > 0. We then have that rε ∈ R and rε > 0, hence |a| < rε . It follows that
|a · b| = |a| · |b|
ε
< ·r
r
=ε
Therefore, a · b ∈ Z.
We now show that Z is a prime ideal of F. Suppose that a, b ∈ F\Z. We have a · b ∈ F by part 2.
Fix ε, δ ∈ R with ε, δ > 0 such that |a| > ε and |b| > δ. We then have |a · b| = |a| · |b| > ε · δ, hence
a·b∈ / Z.
Definition 8.3.9. Let a, b ∈ ∗ R.
1. We write a ≈ b to mean that a − b ∈ Z.

2. We write a ∼ b to mean that a − b ∈ F.

Proposition 8.3.10. ≈ and ∼ are equivalence relations on ∗ R.
Definition 8.3.11. Let a ∈ ∗ R. The ≈-equivalence class of a is called the halo of a. The ∼-equivalence
class of a is called the galaxy of a.
Proposition 8.3.12. Let a1 , b1 , a2 , b2 ∈ ∗ R with a1 ≈ b1 and a2 ≈ b2 .
1. a1 + a2 ≈ b1 + b2 .
2. a1 − a2 ≈ b1 − b2 .
3. If a1 , b1 , a2 , b2 ∈ F, then a1 · a2 ≈ b1 · b2 .
a1 b1
4. If a1 , b1 , a2 , b2 ∈ F\Z, then a2 ≈ b2 .
Proof.
1. We have a1 − b1 ∈ Z and a2 − b2 ∈ Z, hence
(a1 + a2 ) − (b1 + b2 ) = (a1 − b1 ) + (a2 − b2 )
is in Z by Proposition 8.3.8.
2. We have a1 − b1 ∈ Z and a2 − b2 ∈ Z, hence
(a1 − a2 ) − (b1 − b2 ) = (a1 − b1 ) − (a2 − b2 )
is in Z by Proposition 8.3.8.
3. We have a1 − b1 ∈ Z and a2 − b2 ∈ Z. Now
a1 · a2 − b1 · b2 = a1 · a2 − a1 · b2 + a1 · b2 − b1 · b2 = a1 · (a2 − b2 ) + b2 · (a1 − b1 )
so a1 · a2 − b1 · b2 ∈ Z by Proposition 8.3.8.
4. We have a1 − b1 ∈ Z and a2 − b2 ∈ Z. Now
a1 b1 a1 · b2 − a2 · b1 1
− = = · (a1 · b2 − a2 · b1 )
a2 b2 a2 · b2 a2 · b2
and we know by part 3 that a1 · b2 − a2 · b1 ∈ Z. Since a2 , b2 ∈ F\Z, it follows that a2 · b2 ∈ F\Z
by Proposition 8.3.8. Therefore, a21·b2 ∈ F (if ε > 0 is such that |a2 · b2 | > ε, then | a21·b2 | < 1ε ), so
a1 b1
a2 − b2 ∈ Z by Proposition 8.3.8.
Proposition 8.3.13. For every a ∈ F, there exists a unique r ∈ R such that a ≈ r.

Proof. Fix a ∈ F. We first prove existence. Let
X = {r ∈ R : r < a}
and notice that X is downward closed, nonempty, and bounded above because a ∈ F. Now let s = sup X
and argue as in Case 3 of Proposition 8.3.6 that a ≈ s.
Suppose now that r1 , r2 ∈ R are such that a ≈ r1 and a ≈ r2 . We then have that r1 ≈ r2 because ≈ is
an equivalence relation. However, this is a contradiction because |r1 − r2 | > |r1 −r
2
2|
∈ R.
Definition 8.3.14. We define a map st : F → R by letting st(a) be the unique r ∈ R such that a ≈ r. We
call st(a) the standard part or shadow of a.
Corollary 8.3.15. The function st : F → R is a ring homomorphism and ker(st) = Z.
Proposition 8.3.16. Suppose that A ⊆ R, that f : A → R, and that r, ` ∈ R. Suppose also that there exists
δ > 0 such that (r − δ, r + δ)\{r} ⊆ A. The following are equivalent.
1. lim f (x) = `.
x→r
2. For all a ≈ r with a 6= r, we have ∗ f (a) ≈ `.

Proof. Suppose first that lim f (x) = `. Fix a ∈ ∗ A\{r} with a ≈ r. Let ε ∈ R with ε > 0. Since
x→r
lim f (x) = `, we may fix δ ∈ R with δ > 0 such that |f (x) − `| < ε whenever x ∈ A and 0 < |x − r| < δ.
x→r
Now the sentence
∀x((x ∈ A ∧ 0 < |x − r| < δ) → |f (x) − `| < ε)
is in T h(R) = T h(∗ R). Now we have a ∈ ∗ A and 0 < |a − r| < δ, hence |∗ f (a) − `| < ε. Since ε was arbitrary,
it follows that ∗ f (a) ≈ `.
Suppose now that for all a ≈ r with a 6= r, we have ∗ f (a) ≈ `. Fix z ∈ Z with z > 0. Let ε ∈ R with
ε > 0. By assumption, whenever a ∈ ∗ A and 0 < |a − r| < z, we have that ∗ f (a) ≈ `. Thus, the sentence
∃δ(δ > 0 ∧ ∀x((x ∈ A ∧ 0 < |x − r| < δ) → |f (x) − `| < ε))
is in T h(∗ R) = T h(R). By fixing a witnessing δ, we see that the limit condition holds for ε.
Proposition 8.3.17. Suppose that A ⊆ R, that f, g : A → R, and that r, `, m ∈ R. Suppose also that there
exists δ > 0 such that (r − δ, r + δ)\{r} ⊆ A, that lim f (x) = ` and lim g(x) = m. We then have
x→r x→r
1. lim (f + g)(x) = ` + m.
x→r
2. lim (f − g)(x) = ` + m.
x→r
3. lim (f · g)(x) = ` · m.
x→r
4. If m 6= 0, then lim ( fg )(x) = `

m.
x→r
Proof. Fix a ≈ r with a 6= r. We then have ∗ f (a) ≈ ` and ∗ g(a) ≈ m. We have

1. ∗ (f + g)(a) = ∗ f (a) + ∗ g(a) ≈ ` + m.
2. ∗ (f − g)(a) = ∗ f (a) − ∗ g(a) ≈ ` − m.
3. ∗ (f · g)(a) = ∗ f (a) · ∗ g(a) ≈ ` · m.
∗
f (a)
4. ∗ ( fg )(a) = ∗ g(a) ≈ `
m (notice ∗ g(a) ∈
/ Z because m 6= 0).
Corollary 8.3.18. Suppose that A ⊆ R, that f : A → R, and that r ∈ R. Suppose also that there exists
δ > 0 such that (r − δ, r + δ) ⊆ A. The following are equivalent.
1. f is continuous at r.
2. For all a ≈ r, we have ∗ f (a) ≈ f (r).
Taking the standard definition of the derivative, we immediately get the following.
Corollary 8.3.19. Suppose that A ⊆ R, that f : A → R, and that r, ` ∈ R. Suppose also that there exists
δ > 0 such that (r − δ, r + δ) ⊆ A. The following are equivalent.
1. f is differentiable at r with f 0 (r) = `.
∗
f (a)−f (r)
2. For all a ≈ r with a 6= r, we have a−r ≈ `.
Proposition 8.3.20. If f is differentiable at r, then f is continuous at r.

Proof. Fix a ≈ r with a 6= r. Since f is differentiable at r, we have
∗
f (a) − f (r)
≈ f 0 (r)
a−r
∗
f (a)−f (r)
Now f 0 (r) ∈ F, so a−r ∈ F, and hence ∗ f (a) − f (r) ∈ Z because a − r ∈ Z. It follows that
∗
f (a) ≈ f (r).
Proposition 8.3.21. Suppose that f, g : R → R and r ∈ R. Suppose also that g is differentiable at r and f
is differentiable at g(r). We then have that f ◦ g is differentiable at r and (f ◦ g)0 (r) = f 0 (g(r)) · g 0 (r).
Proof. We know that for all a ≈ r with a 6= r, we have
∗
g(a) − g(r)
≈ g 0 (r)
a−r
Also, for all b ≈ g(r) with b 6= g(r), we have
∗
f (b) − f (g(r))
≈ f 0 (g(r))
b − g(r)
Now fix a ≈ r with a 6= r. Since g is continuous at r, we have ∗ g(a) ≈ g(r). If ∗ g(a) 6= g(r), then
∗ ∗ ∗
(f ◦ g)(a) − (f ◦ g)(r) f ( g(a)) − f (g(r))
=
a−r a−r
∗ ∗
f ( g(a)) − f (g(r)) ∗ g(a) − g(r)
= ∗ g(a) − g(r)
·
a−r
0 0
≈ f (g(r)) · g (r)
Suppose then that ∗ g(a) = g(r). Since the first line above holds for every a ≈ r with a 6= r, we must have
g 0 (r) ≈ 0 and hence g 0 (r) = 0 because g 0 (r) ∈ R. Therefore,
∗ ∗ ∗
(f ◦ g)(a) − (f ◦ g)(r) f ( g(a)) − f (g(r))
=
a−r a−r
=0
= f 0 (g(r)) · g 0 (r)

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Mathematical Logic for Mathematicians, Part I

September 20, 2007

2 Induction and Recursion 15

3.6.1 The Compactness Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4 First-Order Logic : Syntax and Semantics 51

5 Semantic and Syntactic Implication 75

6 Soundness, Completeness, and Compactness 85

7 Quantifier Elimination 103

8 Nonstandard Models of Arithmetic and Analysis 111

1.1 The Nature of Mathematical Logic

1. The language of mathematics.

2. The basic assumptions of mathematics.

3. The permissible rules of proof.

We gave a mathematical definition of a deduction, so what’s wrong with using mathematics to

1.2 The Language of Mathematics

1. ∧ will denote and.

2. ∨ will denote or.

3. ¬ will denote not.

4. → will denote implies.

1. If ϕ and ψ are in F ormP , then (ϕ ∧ ψ) is in F ormP ..

2. If ϕ and ψ are in F ormP , then (ϕ ∨ ψ) is in F ormP .

3. If ϕ is in F ormP , then (¬ϕ) is in F ormP .

4. If ϕ and ψ are in F ormP , then (ϕ → ψ) is in F ormP .

Thus, the following is an element of F ormP :

((¬(B ∨ ((¬A) → C))) ∨ A)

1. Associativity: For all x, y, z ∈ G, we have (x · y) · z = x · (y · z).

2. Identity: For all x ∈ G, we have x · e = x = e · x.

3. Inverses: For all x ∈ G, there exists y ∈ G such that x · y = e = y · x.

∀x∀y(f(x, y) = f(y, x))

or that the center is nontrivial:

1. Reflexive: For all x ∈ P , we have x ≤ x.

2. Antisymmetric: If x, y ∈ P are such that x ≤ y and y ≤ x, then x = y.

3. Transitivity: If x, y, z ∈ P are such that x ≤ y and y ≤ z, then x ≤ z.

2. ∀x∀y((R(x, y) ∧ R(y, x)) → (x = y))

3. ∀x∀y∀z((R(x, y) ∧ R(y, z)) → R(x, z))

∀x∀y(R(x, y) ∨ R(y, x))

or that there exists a maximal element:

1.3 Syntax and Semantics

∀x∀y(f(x, y) = f(y, x))

is either true or false, according to whether G is abelian or not.

1.4 The Point of It All

1.5 Some Basic Terminology and Notation

Definition 1.5.2. For each n ∈ N, we let [n] = {m ∈ N : m < n}.

Induction and Recursion

2.1 Induction and Recursion on N

2. f (S(n)) = g(n, f (n)) for all n ∈ N.

2. f (S(n)) = g(n, f (n)) = S(n) · f (n) for all n ∈ N.

(a) h1 : G2 → G is given by h1 (x, y) = x · y for all x, y ∈ G.

(G, B, H) is a simple generating system.

(a) g : V 2 → V is given by g(v, w) = v + w for all v, w ∈ V .

(V, B, H) is a simple generating system.

(a) h1 : K 2 → P(K) is given by h1 (a, b) = {a + b} for all a, b ∈ K.

(d) h4 : K → P(K) is given by (

(K, B, H) is a generating system.

2.2.1 From Above

2.2.2 From Below: Building by Levels

The following remarks are immediate from our definition.

2.2.3 From Below: Witnessing Sequences

2.2.4 Equivalence of the Definitions

2. h Gk is injective for every h ∈ Hk .

3. ran(h1 Gk ) ∩ ran(h2 G` ) = ∅ whenever h1 ∈ Hk and h2 ∈ H` with h1 6= h2 .

Since c1 , c2 , . . . , c` ∈ G, it follows that h = ĥ because ran(h Gk ) ∩ ran(ĥ G` ) = ∅ if h 6= ĥ, and