Predicate Logic and Propositional Logic in Discrete Structure

Predicate Logic
Predicate & Quantifier
Dr. S. Jeevadoss
Assistant Professor
Department of Applied Mathematics & Computational Sciences
PSG College of Technology, Coimbatore.
February 19, 2021
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From Proposition to Predicate
Propositional logic studied earlier cannot express the meanings
of statements in mathematics and in natural language.
Predicate logic extends (is more powerful than) propositional
logic.
3 + 2 = 5 is a proposition. But is X + 2 = 5 a proposition?
Because it has a variable X in it, we cannot say it is TRUE or
FALSE.
So, it is not a proposition. It is called a predicate.
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Predicate vs. Proposition
Example
1 Every computer connected to the university network is
functioning properly.
2 Someone in this room is sleeping now.
3 Not all birds fly
4 Everyone in this class will pass the midterm.
Example
1 There exists an integer x such that 3x2 = 0.
2 The square of every real number is greater than 0.
3 If p is a prime and a is any integer not divisible by p, then
ap−1 − 1 is divisible by p.
4 There are infinitely many prime numbers.
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Predicate vs. Proposition
Definition
A predicate is a sentence with variables and becomes a
proposition, when specific values are substituted for the
variables.
A predicate is a function. It takes some variable(s) as
arguments; it returns either True or False (but not both) for
each combination of the argument values.
Note:
In contrast, a proposition is not a function. It does not have any
variable as argument. It is either True or False (but not both).
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Predicates
Let P(x) be the statement involving the variable x.
P denotes the predicate and x is variable
P(x) is also said to be the value of the propositional function
P at x
Let P(x) denote the sentence “x is greater than 3”.
It has two parts.
1 The variable “x”: The subject of the sentence.
2 The predicate “is greater than 3” : The property that the
subject can have.
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Predicates -Example 1
Example
Let P(x) denotes “x > 3”.
P(5) is TRUE
P(2) is FALSE
P(3) is FALSE
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Predicates-Example 2
Example
Let P(x, y, z) denote that x + y = z and D be the integers for all
three variables.
P(−4, 6, 2) is TRUE
P(5, 2, 10) is FALSE
P(1, 2.5, 3.5) is not a proposition
P(2, y, 6) is not a proposition
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Predicates-Example 3
Example
A(x) = Computer x is under attack by a hacker.
Suppose only CS2 and MATH1 are currently under attack. What
are the truth values of
A(CS1) ≡
A(CS2) ≡
A(MATH1) ≡
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Domain of discourse
Definition (Domain)
The variables are always associated with a universe (or domain)
of discourse, which tells us what combinations of the argument
values are allowed.
The domain of a predicate variable is the set of all values that
may be substituted in place of the variable.
The domain is often denoted by U (the universe) or D (the
domain).
Suppose P(x) is a predicate, where the universe of discourse
for x is {1, 2, 3}. Then P(x) is not a proposition, but P(1) is a
proposition as well as P(5) is not a proposition.
In general, a predicate is not a proposition. But when you assign
values to all its argument variables, you get a proposition.
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Domain - Example-1
Example
P(x, y) :“x + 2 = y” is a predicate.
It has two variables x and y;
Universe of Discourse: x is in {1, 2, 3}; y is in {4, 5, 6}.
1 P(1, 4) : 1 + 2 = 4 is a proposition (it is FALSE);
2 P(2, 4) : 2 + 2 = 4 is a proposition (it is TRUE);
3 P(2, 3) : meaningless (in this example), because 3 is not in
the specified universe of discourse for y.
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Domain - Practice Problem-1
Example
1 P(x) : “(x > 3) ∨ (x < 3) ”.
What are the truth values of P(2), P(3), and P(4)?
2 Q(x, y) : x = y + 3.
What are the truth values of the propositions
Q(1, 2) =
Q(3, 0) =
Example
1 Every computer connected to the university network is
functioning properly.
2 The square of every real number is greater than 0.
3 There are infinitely many prime numbers.
4 Not all birds fly
5 Everyone in this class will pass the midterm.
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Domain - Practice Problem-2
Question
Suppose Q(x, y) : “x + y > 4,” where the universe of discourse is
all integer pairs.
Which of the following are predicates? Which are propositions?
1 Q(1, 2)
2 Q(1000, 2)
3 Q(x, 2)
4 Q(1000, y)
5 Q(x, y)
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Exercise
Let Q(x, y, z) denote that x − y = z and D be the integers.
Which of the following are predicates? Which are propositions?
1 P(1, 2, 3) ∧ Q(5, 4, 1)
2 P(1, 2, 4) → Q(5, 4, 0)
3 P(1, 2, 3) ↔ Q(5, 4, 0)
4 P(1, 2, 4) ∨ Q(x, 4, 0)
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Quantifiers
Quantification is another way of converting a predicate into a
proposition.
We need quantifiers to formally express the meaning of the
words “all” and “some”.
In English, the words all, some, many, none, and few are used
in quantifications.
1 Universal quantification
2 Existential quantification
3 Uniqueness quantifier
4 Quantifiers with restricted domains
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Universal quantification & Existential quantification
(The two most important quantifiers are:)
Universal quantifier, “For all”. Symbol: ∀
Existential quantifier, “There exists”. Symbol: ∃
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Universal quantification
Definition
Suppose P(x) is a predicate on some domain D. The universal
quantification of P(x) is the proposition:
“P(x) for all values of x in the domain ” We write ∀x, P(x).
∀ : Universal Quantifier
∀x P(x) ≡ T : if P(x) is true for every x in D
∀x P(x) ≡ F : if P(x) is false for at least one x in D
The truth values depends not only on P, but also on the domain
D.
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Existential quantification
Definition
Suppose P(x) is a predicate on some domain D. The existential
quantification of P(x) is the proposition:
“There exists an element x in the domain such that P(x). ” We
write ∃x, P(x).
∃ : Existential Quantifier
∃x P(x) ≡ T : if P(x) is true for any (at least one) x in D
∃x P(x) ≡ F : if P(x) is false for every x in D
The truth values depends not only on P, but also on the domain
D.
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Universal quantification & Existential quantification
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Universal quantification Example
Example
P(x) : x + 2 = 5, Domain: {1, 2, 3}.
∀x, P(x) means: “for all x in {1, 2, 3}, x + 2 = 5”.
In other words, it means: “1+2=5, 2+2=5, and 3+2=5”, which is
false. So it is a proposition.
Example
A(x) : x = 1, B(x) : x > 2, C(x) : x < 2
Domain: {1, 2, 3}.
TRUE or FLASE?
1 ∀x, (C(x) → A(x))
2 ∀x, (C(x) ∨ B(x))
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Existential quantification Example
Example
P(x) : x + 2 = 5, Domain: {1, 2, 3}.
∃x, P(x) means: “for some x in {1, 2, 3}, x + 2 = 5”.
In other words, it means: “1+2=5, 2+2=5, and 3+2=5”, which is
True. So it is a proposition.
Example
A(x) : x = 1, B(x) : x > 2, C(x) : x < 2
Domain: {1, 2, 3}.
TRUE or FLASE?
1 ∃x, (C(x) → A(x))
2 ∃x, (¬B(x))
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Translating from English into Logical Expressions
Question
Express the statements “Every student in this class has studied
calculus” and “Some student in this class has visited Mexico” using
predicates and quantifiers.
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Example
Using the predicate symbols
S(x) : x is a student,
I(x) : x is intelligent,
M(x) : x likes music,
write wffs that express the following statements. The domain is
the collection of all people.
1 All students are intelligent.
2 Some intelligent students like music.
3 Everyone who likes music is a stupid student.
4 Only intelligent students like music.
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Exercise1
A(x) : “x lives in Chennai.”
B(x) : “x is a CSE student.”
C(x) : “x has a good GPA.”
D(x) : “x majors in mathematics.”
Domain: all PSG students
1 All CSE students have good GPA
2 No CSE student lives in Chennai
3 CSE students who do not live in Chennai major in
mathematics
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Exercise2
Translate each of these statements into logical expressions using
predicates, quantifiers, and logical connectives.
P(x) : “x is perfect.”
F(x) : “x is your friend.”
Domain: all people
1 No one is perfect.
2 Not everyone is perfect.
3 All your friends are perfect.
4 At least one of your friends is perfect.
5 Everyone is your friend and is perfect.
6 Not everybody is your friend or someone is not perfect.
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Problem
Example
What is the truth value of ∀x(x2 ≥ x) if the domain consists of all
real numbers? What is the truth value of this statement if the
domain consists of all integers?
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Example
If no domain of discourse has been identified, a proposition such
as ∀x, (x2 6= 2) is ambiguous.
If the domain of discourse is the set of real numbers, the
proposition is false, with x =
√
2 as counterexample.
If the domain is the set of naturals, the proposition is true, since
2 is not the square of any natural number.
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Precedence of Quantifiers
The quantifiers ∀ and ∃ have higher precedence than all logical
operators from propositional logic.
∀xP(x) ∧ Q(x) means (∀xP(x)) ∧ Q(x). In particular, this
expression contains a free variable.
∀x(P(x) ∧ Q(x)) means something different.
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Problem
Domain is empty. What is the truth value of ∀xP(x) and ∃xP(x)?
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Binding Variables
When a quantifier is used on the variable x, we say that this
occurrence of the variable is bound.
An occurrence of a variable that is not bound by a quantifier
or set equal to a particular value is said to be free.
The part of a logical expression to which a quantifier is applied
is called the scope of this quantifier.
A variable is free if it is outside the scope of all quantifiers in
the formula that specify this variable.
A statement with free variable is not a proposition.
Example
1 ∃x(x + y = 1)
2 ∃x(P(x) ∧ Q(x)) ∨ ∀xR(x)
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Logical Equivalences Involving Quantifiers
Definition
Statements involving predicates and quantifiers are logically
equivalent if they have the same truth value no matter which
predicates are substituted into these statements and which domain
of discourse is used for the variables in these propositional
functions.
Example
1 ∀x (P(x) ∧ Q(x)) ≡ ∀xP(x) ∧ ∀x Q(x)
2 ∃x (P(x) ∨ Q(x)) ≡ ∃xP(x) ∨ ∃x Q(x)
3 ∀x (P(x) ∨ Q(x)) 6≡ ∀xP(x) ∨ ∀x Q(x)
4 ∃x (P(x) ∧ Q(x)) 6≡ ∃xP(x) ∧ ∃x Q(x)
5 ∀x (P(x) → Q(x)) 6≡ ∀xP(x) → ∀x Q(x)
6 ∃x (P(x) → Q(x)) 6≡ ∃xP(x) → ∃x Q(x)
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Distribution of ∀ over ∧
∀x (P(x) ∧ Q(x)) ≡ ∀x P(x) ∧ ∀x Q(x)
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Distribution of ∀ over ∨
∀x (P(x) ∨ Q(x)) 6≡ ∀x P(x) ∨ ∀x Q(x)
P(x) : x is odd
Q(x) : x is even
Domain : Set of all integers.
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Distribution of ∃ over ∧
∃x (P(x) ∧ Q(x)) 6≡ ∃x P(x) ∧ ∃x Q(x)
P(x) : x is odd
Q(x) : x is even
Domain : Set of all integers.
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Quantifiers as Conjunctions& Disjunctions
If the domain is finite then universal/existential quantifiers can
be expressed by conjunctions/disjunctions.
If D consists of the integers 1,2, and 3, then sub
∀xP(x) ≡ P(1) ∧ P(2) ∧ P(3)
∃xP(x) ≡ P(1) ∨ P(2) ∨ P(3)
Even if the domains are infinite, you can still think of the
quantifiers in this fashion, but the equivalent expressions
without quantifiers will be infinitely long.
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De Morgan’s Law for Quantifiers
The rules for negating quantifiers are:
¬∀xP(x) ≡ ∃x¬P(x)
¬∃xP(x) ≡ ∀x¬P(x)
Example (Do these two statements have the same meaning?)
1 Not every PSG student majors in computer science.
2 There is PSG student who does not major in computer
science.
Example (Do these two statements have the same meaning?)
1 There is no PSG student living in Chennai
2 Every PSG student lives in a town other than Chennai
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Quantifier negation
Example (Find the negation of the following quantified
statements.)
1 Every student in your class has taken a course in calculus.
2 Some people are born lucky.
3 All dogs bark
4 Most cars are inexpensive.
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Exercise
Negation of Quantified Statements
1 Every day I bring breakfast to college.
2 (∀x ∈ Z) ( x is a multiple of 2).
3 (∃x ∈ Z)(x3 > 0).
4 Some math majors have red hair
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Commutativity and Distributivity of quantifiers
∀x∀y P(x, y) ≡ ∀y ∀x P(x, y)
∃x∃y P(x, y) ≡ ∃y ∃x P(x, y)
∀x∃y P(x, y) 6≡ ∃y ∀x P(x, y)
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Translating from Logical Expressions into English
Example
Translate the statement ∀x(C(x) ∨ ∃y(C(y) ∧ F(x, y))) into
English, where
C(x) : “x has a computer”
F(x, y) is “x and y are friends”
the domain for both x and y is the set of all students in your
college.
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Example
Translate the statement ∀x(C(x) ∨ ∃y(C(y) ∧ F(x, y))) into
English, where
C(x) : “x has a computer”
F(x, y) is “x and y are friends”
the domain for both x and y is the set of all students in your
college.
The statement says that
For every student x in your college x has a computer or there
is a student y such that y has a computer and x and y are
friends.
In other words, every student in your school has a computer or
has a friend who has a computer.
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Example
Translate the statement “Everyone has exactly one best friend” as
a logical expression.
B(x, y) : y is the best friend of x.
∀x∃y∀z(B(x, y) ∧ ((z 6= y) → ¬B(x, z)))
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An Example from Calculus
Example
Express that the limit of a real-valued function f at point a is L.
lim
x→a
f (x) = L
In predicate logic
∀ ∃δ ∀x (| x − a | δ →| f (x) − L | )
where the domain of and δ are the positive real numbers and the
domain of x are all real numbers.
Now express its negation, i.e., lim
x→a
f (x) 6= L
∃ ∀δ ∃x (| x − a | δ ∧ | f (x) − L |≥ )
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Rules of Inference for Quantified Statements
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Combining Rules of Inference for Propositions and
Quantified Statements
These inference rules are frequently used and combine propositions
and quantified statements:
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Problem-1
Example
Show that the premises “Everyone in this discrete mathematics
class has taken a course in computer science” and “Megha is a
student in this class” imply the conclusion “Megha has taken a
course in computer science.”
D(x) : “x is in this discrete mathematics class
C(x) : “x has taken a course in computer science.
Hypotheses: ∀x (D(x) → C(x)) and D(Megha).
Conclusion: C(Megha).
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Problem-2
Example
Show that the premises:
A student in this class has not read the book.
Everyone in this class passed the first exam.
imply the conclusion
Someone who passed the first exam has not read the book.
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Problem-2
Example
A(x) : “x is in this class”
B(x) : “x read the book”
P(x) : “x passed the first exam”
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Problem-2
Example
A(x) : “x is in this class”
B(x) : “x read the book”
P(x) : “x passed the first exam”
Hypotheses: ∃x (A(x) ∧ ¬B(x)) and ∀x (A(x) → P(x)).
Conclusion: ∃x (P(x) ∧ ¬B(x)).
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Problem-2 - Solution
Hypotheses: ∃x (A(x) ∧ B(x)) and ∀x (A(x) → P(x)).
Conclusion: ∃x (P(x) ∧ ¬B(x)).
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Problem-3
Example
Every man has two legs.
John Smith is a man.
Therefore, John Smith has two legs.
Define the predicates:
M(x) : x is a man
L(x) : x has two legs
J : John Smith, a member of the domain
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Problem-4
Is the wff a valid argument? Prove or disprove.
Example
∃x R(x)
¬∃x [R(x) ∧ S(x)]
Therefore, ∃x ¬S(x)
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Exercise-1
Question
Verify the validity of the argument:
Hypotheses: ∀x (P(x) ∨ Q(x)) and
∀x ((¬P(x) ∧ Q(x)) → R(x))
Conclusion: ∀x (¬R(x) → P(x)).
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Exercise-2
Question
Show that the following argument is valid:
All Canny integers are non-Slotty
3 is Canny
All Bumpy integers are Slotty
Therefore, There is an integer that is not Bumpy
C(x) : x is a Canny integer
S(x) : x is a Slotty integer
B(x) : x is a Bumpy integer
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Exercise-3
Question
Show that the following argument is valid: “Every laptop has an
internal disk drive. Some laptops have a DVD drive. Therefore
some laptops have both an internal disk drive and a DVD drive.”
Using the following predicates:
L(x) is “x is a laptop.”
I(x) is “x has an internal disk drive.”
D(x) is “x has a DVD drive.”
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Predicate Logic and Propositional Logic in Discrete Structure

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