Predicate Logic: Lucia Moura

Predicates and Quantifiers Nested Quantifiers Using Predicate Calculus
Predicate Logic
Lucia Moura
Winter 2010
CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers
Predicates
A Predicate is a declarative sentence whose true/false value depends on
one or more variables.
The statement “x is greater than 3” has two parts:
the subject: x is the subject of the statement
the predicate: “is greater than 3” (a property that the subject can
have).
We denote the statement “x is greater than 3” by P (x), where P is the
predicate “is greater than 3” and x is the variable.
The statement P (x) is also called the value of propositional function P
at x.
Assign a value to x, so P (x) becomes a proposition and has a truth value:
P (5) is the statement “5 is greater than 3”, so P (5) is true.
P (2) is the statement “2 is greater than 3”, so P (2) is false.

Predicates: Examples
Given each propositional function determine its true/false value when

variables are set as below.
Prime(x) = “x is a prime number.”
Prime(2) is true, since the only numbers that divide 2 are 1 and itself.
Prime(9) is false, since 3 divides 9.
C(x, y)=“x is the capital of y”.
C(Ottawa,Canada) is true.
C(Buenos Aires,Brazil) is false.
E(x, y, z) = “x + y = z”.
E(2, 3, 5) is ...
E(4, 4, 17) is ...

Quantifiers
Assign a value to x in P (x) =“x is an odd number”, so the resulting
statement becomes a proposition: P (7) is true, P (2) is false.
Quantification is another way to create propositions from a propositional

functions:
universal quantification: ∀xP (x) says
“the predicate P is true for every element under consideration.”
Under the domain of natural numbers, ∀xP (x) is false.
existencial quantification: ∃xP (x) says
“there is one or more element under consideration for which the
predicate P is true.”
Under the domain of natural numbers, ∃xP (x) is true, since for
instance P (7) is true.
Predicate calculus: area of logic dealing with predicates and quantifiers.
Domain, domain of discourse, universe of discourse
Before deciding on the truth value of a quantified predicate, it is

mandatory to specify the domain (also called domain of discourse or
universe of discourse).
P (x) =“x is an odd number”
∀xP (x) is false for the domain of integer numbers; but

∀xP (x) is true for the domain of prime numbers greater than 2.

Universal Quantifier
The universal quantification of P (x) is the statement:
“P (x) for all values of x in the domain” denoted ∀xP (x).
∀xP (x) is true when P (x) is true for every x in the domain.
∀xP (x) is false when there is an x for which P (x) is false.

An element for which P (x) is false is called a counterexample of ∀xP (x).
If the domain is empty, ∀xP (x) is true for any propositional function
P (x), since there are no counterexamples in the domain.
If the domain is finite {x1 , x2 , . . . , xn }, ∀xP (x) is the same as
P (x1 ) ∧ P (x2 ) ∧ · · · ∧ P (xn ).

Universal quantifiers: example
Let P (x) be “x2 > 1000 .

What is the truth value of ∀xP (x) for each of the following domains:

Let P (x) be “x2 > 1000 .

I the set of real numbers: R

Let P (x) be “x2 > 1000 .

I False. 3 is a counterexample.

Let P (x) be “x2 > 1000 .

I the set of positive integers not exceeding 4: {1, 2, 3, 4}

Let P (x) be “x2 > 1000 .


Let P (x) be “x2 > 1000 .

I Also note that here ∀P (x) is P (1) ∧ P (2) ∧ P (3) ∧ P (4), so its enough
to observe that P (3) is false.

Let P (x) be “x2 > 1000 .

I the set of real numbers in the interval [10, 39.5]

Let P (x) be “x2 > 1000 .

I True. It takes a bit longer to verify than in false statements.
Let x ∈ [10, 39.5]. Then x ≥ 10 which implies x2 ≥ 102 = 100 > 10,
and so x2 > 10.

Existencial Quantifier
The existential quantification of P (x) is the statement:
“There exists an element x in the domain such that P (x)” denoted
∃xP (x).
∃xP (x) is true when P (x) is true for one or more x in the domain.
An element for which P (x) is true is called a witness of ∃xP (x).
∃xP (x) is false when P (x) is false for every x in the domain
(if domain nonempty).
If the domain is empty, ∃xP (x) is false for any propositional function
P (x), since there are no witnesses in the domain.
If the domain is finite {x1 , x2 , . . . , xn }, ∃xP (x) is the same as
P (x1 ) ∨ P (x2 ) ∨ · · · ∨ P (xn ).

Existencial quantifiers: example
Let P (x) be “x2 > 1000 .

What is the truth value of ∃xP (x) for each of the following domains:

Let P (x) be “x2 > 1000 .


Let P (x) be “x2 > 1000 .

I True. 10 is a witness.

Let P (x) be “x2 > 1000 .


Let P (x) be “x2 > 1000 .


Let P (x) be “x2 > 1000 .

I Also note that here ∃P (x) is P (1) ∨ P (2) ∨ P (3) ∨ P (4), so its enough
to observe that P (4) is true.

Let P (x) be “x2 > 1000 .

to observe that P (4) is true. √

Let P (x) be “x2 > 1000 .

to observe that P (4) is true. √
I False. It takes a bit longer to conclude
√ than in true statements.
Let x ∈√[0, 9.8]. Then 0 ≤ x ≤ 9.8 which implies
x2 ≤ ( 9.8)2 = 9.8 < 10, and so x2 < 10.
What we have shown is that ∀x¬P (x), which (we will see) is
equivalent to ¬∃xP (x)

Other forms of quantification

Other Quantifiers
The most important quantifiers are ∀ and ∃, but we could define
many different quantifiers: “there is a unique”, “there are exactly
two”, “there are no more than three”, “there are at least 100”, etc.
A common one is the uniqueness quantifier, denoted by ∃!.
∃!xP (x) states “There exists a unique x such that P (x) is true.”
Advice: stick to the basic quantifiers. We can write ∃!xP (x) as
∃x(P (x) ∧ ∀y(P (y) → y = x)) or more compactly
∃x∀y(P (y) ↔ y = x)
Restricting the domain of a quantifier
Abbreviated notation is allowed, in order to restrict the domain of
certain quantifiers.
I ∀x > 0(x2 > 0) is the same as ∀x(x > 0→x2 > 0).
I ∀y 6= 0(y 3 6= 0) is the same as ∀y(y 6= 0→y 3 6= 0).
I ∃z > 0(z 2 = 2) is the same as ∃x(z > 0∧z 2 = 2)
Precedence and scope of quantifiers

∀ and ∃ have higher precedence than logical operators.
Example: ∀xP (x) ∨ Q(x) means (∀xP (x)) ∨ Q(x), it doesn’t mean
∀x(P (x) ∨ Q(x)).
(Note: This statement is not a proposition since there is a free variable!)
Binding variables and scope
When a quantifier is used on the variable x we say that this
occurrence of x is bound. When the occurrence of a variable is not
bound by a quantifier or set to a particular value, the variable is said
to be free.
The part of a logical expression to which a quantifier is applied is the
scope of the quantifier. A variable is free if it is outside the scope of
all quantifiers.
In the example above, (∀xP (x)) ∨ Q(x), the x in P (x) is bound by
the existencial quantifier, while the x in Q(x) is free. The scope of
the universal quantifier is underlined.
Logical Equivalences Involving Quantifiers
Definition
Two statements S and T involving predicates and quantifiers are logically
equivalent if and only if they have the same truth value regardless of the
interpretation, i.e. regardless of
the meaning that is attributed to each propositional function,
the domain of discourse.
We denote S ≡ T .
Is ∀x(P (x) ∧ Q(x)) logically equivalent to ∀xP (x) ∧ ∀xQ(x) ?
Is ∀x(P (x) ∨ Q(x)) logically equivalent to ∀xP (x) ∨ ∀xQ(x) ?

Prove that ∀x(P (x) ∧ Q(x)) is logically equivalent to

∀xP (x) ∧ ∀xQ(x) (where the same domain is used throughout).


Use two steps:


Use two steps:
I If ∀x(P (x) ∧ Q(x)) is true, then ∀xP (x) ∧ ∀xQ(x) is true.


Use two steps:
I Proof: Suppose ∀x(P (x) ∧ Q(x)) is true.
Then if a is in the domain, P (a) ∧ Q(a) is true, and so P (a) is true
and Q(a) is true.
So, if a in in the domain P (a) is true, which is the same as ∀xP (x) is
true; and similarly, we get that ∀xQ(x) is true.
This means that ∀xP (x) ∧ ∀xQ(x) is true.


Use two steps:
and Q(a) is true.
I If ∀xP (x) ∧ ∀xQ(x) is true, then ∀x(P (x) ∧ Q(x)) is true.


Use two steps:
and Q(a) is true.
I If ∀xP (x) ∧ ∀xQ(x) is true, then ∀x(P (x) ∧ Q(x)) is true.
I Proof: Suppose that ∀xP (x) ∧ ∀xQ(x) is true.
It follows that ∀xP (x) is true and ∀xQ(x) is true.
So, if a is in the domain, then P (a) is true and Q(a) is true.
It follows that if a is in the domain P (a) ∧ Q(a) is true.
This means that ∀x(P (x) ∧ Q(x)) is true.

Prove that ∀x(P (x) ∨ Q(x)) is not logically equivalent to

∀xP (x) ∨ ∀xQ(x).


∀xP (x) ∨ ∀xQ(x).
It is enough to give a counterexample to the assertion that they have
the same truth value for all possible interpretations.


∀xP (x) ∨ ∀xQ(x).
It is enough to give a counterexample to the assertion that they have
the same truth value for all possible interpretations.
Under the following interpretation:
domain: set of people in the world
P (x) =“x is male”.
Q(x) =“x is female”.
We have:
∀x(P (x) ∨ Q(x)) (every person is a male or a female) is true;
while ∀xP (x) ∨ ∀xQ(x) (every person is a male or every person is a
female) is false.

Negating Quantified Expressions: De Morgan Laws
¬∀xP (x) ≡ ∃x¬P (x)

Proof:
¬∀xP (x) is true if and only if ∀xP (x) is false.
Note that ∀xP (x) is false if and only if there exists an element in the
domain for which P (x) is false.
But this holds if and only if there exists an element in the domain for
which ¬P (x) is true.
The latter holds if and only if ∃x¬P (x) is true.

De Morgan Laws for quantifiers (continued)
¬∃xP (x) ≡ ∀x¬P (x)

Proof:
¬∃P (x) is true if and only if ∃xP (x) is false.
Note that ∃xP (x) is false if and only there exists no element in the
domain for which P (x) is true.
But this holds if and only if for all elements in the domain we have P (x) is
false;
which is the same as for all elements in the domain we have ¬P (x) is true.
The latter holds if and only if ∀x¬P (x) is true.

Practice Exercises
1 What are the negatons of the following statements:

“There is an honest politician.”
“All americans eat cheeseburgers.”
2 What are the negations of ∀x(x2 > x) and ∃x(x2 = 2)?
3 Show that ¬∀x(P (x) → Q(x)) and ∃x(P (x) ∧ ¬Q(x)) are logically
equivalent.
Solutions in the textbook’s page 41.

Example from Lewis Caroll’s book Symbolic Logic

Consider these statements (two premises followed by a conclusion):
“All lions are fierce.”
“Some lions do not drink coffee.”
“Some fierce creatures do not drink coffee.”
Assume that the domain is the set of all creatures and P (x) =“x is a
lion”, Q(x) =“x is fierce”, R(x) =“x drinks coffee”.
Exercise: Express the above statements using P (x), Q(x) and R(x), under
the domain of all creatures.
Is the conclusion a valid consequence of the premises?

In this case, yes. (See more on this type of derivation, in a future lecture
on Rules of Inference).

Nested Quantifiers
Nested Quantifiers
Two quantifiers are nested if one is in the scope of the other.

Everything within the scope of a quantifier can be thought of as a
propositional function.
For instance,
“∀x∃y(x + y = 0)” is the same as

“∀xQ(x)”, where Q(x) is “∃y(x + y = 0)”.

Nested Quantifiers
The order of quantifiers

Let P (x, y) be the statement “x + y = y + x”.
Consider the following:
∀x∀yP (x, y) and ∀y∀xP (x, y).
What is the meaning of each of these statements?
What is the truth value of each of these statements?
Are they equivalent?
Let Q(x, y) be the statement “x + y = 0”.
Consider the following:
∃y∀xQ(x, y) and ∀x∃yQ(x, y).
What is the meaning of each of these statements?
What is the truth value of each of these statements?
Are they equivalent?

Nested Quantifiers
Summary of quantification of two variables
statement when true ? when false ?

∀x∀yP (x, y) P (x, y) is true There is a pair x, y for
∀y∀xP (x, y) for every pair x, y. which P (x, y) is false
∀x∃yP (x, y) For every x there is y There is an x such that
for which P (x, y) is true P (x, y) is false for every y
∃x∀yP (x, y) There is an x for which For every x there is a y
P (x, y) is true for every y for which P (x, y) is false.
∃x∃yP (x, y) There is a pair x, y P (x, y) is false
∃y∃xP (x, y) for which P (x, y) is true for every pair x, y.

Nested Quantifiers
Translating Math Statements into Nested quantifiers
Translate the following statements:

1 “The sum of two positive integers is always positive.”
2 “Every real number except zero has a multiplicative inverse.”
(a multiplicative inverse of x is y such that xy = 1).
3 “Every positive integer is the sum of the squares of four integers.”

Nested Quantifiers
Translating from Nested Quantifiers into English
Let C(x) denote “x has a computer” and F (x, y) be “x and y are

friends.”, and the domain be all students in your school.
Translate:
1 ∀x(C(x) ∨ ∃y(C(y) ∧ F (x, y)).
2 ∃x∀y∀z((F (x, y) ∧ F (x, z) ∧ (y 6= z)) → ¬F (y, z))

Nested Quantifiers
Translating from English into Nested Quantifiers
1 “If a person is female and is a parent, then this person is someone’s

mother.”
2 “Everyone has exactly one best friend.”
3 “There is a woman who has taken a flight on every airline of the
world.”

Nested Quantifiers
Negating Nested Quantifiers
Express the negation of the following statements, so that no negation

precedes a quantifier (apply DeMorgan successively):
∀x∃y(xy = 1)
∀x∃yP (x, y) ∨ ∀x∃yQ(x, y)
∀x∃y(P (x, y) ∧ ∃zR(x, y, z))

Using Predicate Calculus
Predicate calculus in Mathematical Reasoning
Using predicates to express definitions.

D(x) =“x is a prime number”
(defined term)
P (x) =“x ≥ 2 and the only divisors of x are 1 and x”
(defining property about x)
Definition of prime number: ∀x(D(x) ↔ P (x))
Note that definitions in English form use if instead of if and only if, but we
really mean if and only if.

Predicate calculus in Mathematical Reasoning (cont’d)
Let P (n, x, y, z) be the predicate xn + y n = z n .

1 Write the following statements in predicate logic, using the domain of
positive integers:
“For every integer n > 2, there does not exist positive integers x, y
and z such that xn + y n = z n .”
2 Negate the previous statement, and simplify it so that no negation
precedes a quantifier.
3 What needs to be found in order to give a counter example to 1 ?
4 Which famous theorem is expressed in 1, who proved and when?

Predicate calculus in Program Verification: a toy example

The following program is designed to exchange the value of x and y:
temp := x
x := y
y := temp
Find preconditions, postconditions and verify its correctness.
Precondition: P (x, y) is “x = a and y = b”, where a and b are the
values of x and y before we execute these 3 statements.


temp := x
x := y
y := temp
Postconditon: Q(x, y) is “x = b and y = a”.


temp := x
x := y
y := temp
Assume P (x, y) holds before and show that Q(x, y) holds after.


temp := x
x := y
y := temp
Originally x = a and y = b, by P (x, y).


temp := x
x := y
y := temp
After step 1, x = a, temp = a and y = b.


temp := x
x := y
y := temp
After step 2, x = b, temp = a and y = b.


temp := x
x := y
y := temp
After step 3, x = b, temp = a and y = a.


temp := x
x := y
y := temp
After step 3, x = b, temp = a and y = a.
Therefore, after the program we know Q(x, y) holds.
Predicate calculus in System Specification
Use predicates and quantifiers to express system specifications:

1 “Every mail message larger than one megabyte will be compressed.”


I S(m, y):“mail message m is larger than y megabytes”


I C(m): “mail message m will be compressed”


I ∀m(S(m, 1) → C(m))


I ∀m(S(m, 1) → C(m))
2 “If a user is active, at least one network link will be available.”


I ∀m(S(m, 1) → C(m))
I A(u): “User u is active.”


I ∀m(S(m, 1) → C(m))
I S(n, x): “Network n is in state x”.


I ∀m(S(m, 1) → C(m))
I S(n, x): “Network n is in state x”.
I ∃uA(u) → ∃nS(n, available)

Predicate calculus in Logic Programming

Prolog is a declarative language based in predicate logic.
The program is expressed as Prolog facts and Prolog rules.
Execution is triggered by running queries over these relations.
mother_child(trude, sally).
father_child(tom, sally).
father_child(tom, erica).
father_child(mike, tom).
sibling(X, Y) :- parent_child(Z, X), parent_child(Z, Y).
parent_child(X, Y) :- father_child(X, Y).

parent_child(X, Y) :- mother_child(X, Y).
The result of the following query is given:

?- sibling(sally, erica).
Yes

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Predicates and Quantifiers Nested Quantifiers Using Predicate Calculus

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Given each propositional function determine its true/false value when

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Quantification is another way to create propositions from a propositional

Predicates and Quantifiers

Domain, domain of discourse, universe of discourse

Before deciding on the truth value of a quantified predicate, it is

P (x) =“x is an odd number”

∀xP (x) is false for the domain of integer numbers; but

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

∀xP (x) is false when there is an x for which P (x) is false.

If the domain is finite {x1 , x2 , . . . , xn }, ∀xP (x) is the same as

P (x1 ) ∧ P (x2 ) ∧ · · · ∧ P (xn ).

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

Universal quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura

Predicates and Quantifiers

If the domain is finite {x1 , x2 , . . . , xn }, ∃xP (x) is the same as

P (x1 ) ∨ P (x2 ) ∨ · · · ∨ P (xn ).

Predicates and Quantifiers

Existencial quantifiers: example

Let P (x) be “x2 > 1000 .

CSI2101 Discrete Structures Winter 2010: Predicate Logic Lucia Moura