Module - 2 Discrete Mathematics and Graph Theory

MAT-1014 Discrete Mathematics and Graph Theory
Faculty: Dr.D.Ezhilmaran
Teaching Research Associate: M.Adhiyaman
Department of Mathematics, School of Advanced Sciences, VIT- Univeristy,
Vellore, Tamilnadu, India
ezhilmaran.d@vit.ac.in
January 31, 2017
Faculty: Dr.D.Ezhilmaran Teaching Research Associate: M.Adhiyaman (VIT)Discrete Mathematics January 31, 2017 1 / 18

Overview
1 Module −2 Predicate Calculus
The Predicate Calculus
Inference Theory of the Predicate Calculus

Predicate Calculus
Consider the statement
p : x is a prime number (the statement is not a proposition)
The truth value of p depends on the value of x.
p is true when x = 3, and false when x = 10.
In this section we extend the system of logic to include such an above
statements.

Definition 1. (predicates).
A predicate refers to a property that the subject of the statement can
have. A predicate is a sentence that contains a finite number of specific
values are substituted for the variables.
That is, let P(x) be a statement involving variable x and a set D. We call
P as a propositional function if for each x in D, P(x) is a proposition.

Deﬁnition 2. (universe of discourse)
The set D is called the domain of discourse (oruniverse of discourse) of P.
It is the set of all possible values which can be assigned to variables in
statements involving predicates.
Example: Let p(x) denote the statement x ≥ 4. What are the truth values
of p(5) (T) and p(2) (F).
Example: Let g(x, y) denote the statement g.c.d(x, y) = 1. What are the
truth values of g(3, 5) (T) and g(2, 8) (F)

Definition 3. (universal quantifier)
Consider the proposition
All odd prime numbers are greater than 2. The word all in this proposition
is a logical quantifier. The proposition can be translated as follows:
For every x, if x is an odd prime then x is greater than 2
Similarly, the proposition:
Every rational number is a real number may be translated as.
For every x, if x is a rational number, then x is a real number.
The phrase for every x is called a universal quantifier.

In symbols it is denoted by (∀x) or (x).
The phrases for every x, for all x and for each x have the same meaning
and we can symbolize each by (x).
If P(x) denotes a predicate (propositional function), then the universal
quantiﬁcation for P(x), is the statement.
(x) P(x) is true.

Example :
(a) Let A = {x : x is a natural number less than 9}
Here P(x) is the sentence x is a natural number less than 9. The common
property is a natural number less than 9. P(1) is true, therefore, 1 ∈ A
and P(12) is not true, therefore 12 /∈ A.
(b) The proposition (∀N) (n + 4 > 3) is true.
Since {n|n + 4 > 3} = {1, 2, 3, . . . } = N.
(c) The proposition (∀N) (n + 2 > 8) is false.
Since {n|n + 2 > 8} = {7, 8, 9, . . . } = N.

Definition 4. (existential quantifier).
In some situations we only require that there be at least one value for each
the predicate is true. This can be done by prefixing P(x) with the phrase
there exists an. The phrase there exists an is called an existential
quantifier.

The existential quantification for a predicate is the statement There exists
a value of x for which P(x).
The symbol, ∃ is used to denote the logical quantifier there exists. The
phrases There exists an x, There is a x, for some x and for at least one x
have the same meaning.
The existential quantifier for P(x) is denoted by (∃ x) P(x)

Example :
(a) The proposition there is an integer between 1 and 3 may be written as
(∃ an integer) (the integer is between 1 and 3)
(b) The proposition (∃N) (n + 4 < 7) is true.
Since {n|n + 4 < 7} = {1, 2} = φ.
(c) The proposition (∃N) (n + 6 < 4) is false.
Since {n|n + 6 < 4} = φ.

IV. Problems:
(i) Show that (x)(H(x) −→ M(x)) ∧ H(a) =⇒ M(a).
Solution:
Step 1 (x)(H(x) −→ M(x)) Rule P
Step 2 H(a) −→ M(a) Rule US
Step 3 H(a) Rule P
Step 4 M(a) {2, 3} and apply Modus Ponens

(ii) Show that
(x)(P(x) −→ Q(x)) ∧ (x)(Q(x) −→ R(x)) =⇒ (x)(P(x) −→ R(x)).
Solution:
Step 1 (x)(P(x) −→ Q(x)) Rule P
Step 2 P(a) −→ Q(a) Rule US
Step 3 (x)(Q(x) −→ R(x)) Rule P
Step 4 Q(a) −→ R(a) Rule US
Step 5 P(a) −→ R(a) {2,4},I7
Step 6 (x)P(x) −→ R(x) Rule UG

(iii) Show that (∃x)(P(x) ∧ Q(x)) =⇒ (∃x)P(x) ∧ (∃x)Q(x).
Solution:
Step 1 (∃x)(P(x) ∧ Q(x)) Rule P
Step 2 P(a) ∧ Q(a) Rule ES
Step 3 P(a) I1
Step 4 Q(a) I1
Step 5 (∃x)P(x) {3},EG
Step 6 (∃x)Q(x) {4},EG
Step 7 (∃x)P(x) ∧ (∃x)Q(x) {5,6}, I3

(iv) Show that (x)(P(x) ∨ Q(x)) =⇒ (x)P(x) ∨ (∃x)Q(x).
Solution: Proof by indirect method
Step 1 ¬((x)P(x) ∨ (∃x)Q(x)) Rule P
Step 2 ¬(x)P(x) ∧ ¬(∃x)Q(x) Rule T
Step 3 ¬(x)P(x) I1
Step 4 ¬(∃x)Q(x) I1
Step 5 (∃x)¬P(x) 3,Rule T
Step 6 (x)¬Q(x) 4,Rule T
Step 7 ¬P(a) 5,ES
Step 8 ¬Q(a) 6,US

Step 9 ¬P(a) ∧ ¬Q(a) {7,8},I3
Step 10 ¬(P(a) ∨ Q(a)) Rule T
Step 11 (x)(P(x) ∨ Q(x)) Rule P
Step 12 P(a) ∨ Q(a) US
Step 13 ¬(P(a) ∨ Q(a)) ∧ (P(a) ∨ Q(a)) {10,12}, I3
Step 14 F Rule T

(v) Show that from
(a) (∃x)(F(x) ∧ S(x)) −→ (y)(M(y) −→ W (y))
(b) (∃y)(M(y) ∧ ¬W (y))
the conclusion (x)(F(x) −→ ¬S(x)).
Solution:
Step 1 (∃y)(M(y) ∧ ¬W (y)) Rule P
Step 2 (M(a) ∧ ¬W (a)) ES
Step 3 ¬(M(a) −→ W (a)) Rule T
Step 4 (∃y)¬(M(y) −→ W (y)) EG
Step 5 ¬(y)(M(y) −→ W (y)) Rule T

Step 6 (∃x)(F(x) ∧ S(x)) −→ (y)(M(y) −→ W (y)) Rule P
Step 7 ¬(∃x)(F(x) ∧ S(x)) {5,6}, I6
Step 8 (x)¬(F(x) ∧ S(x)) Rule T
Step 9 ¬(F(a) ∧ S(a)) US
Step 10 F(a) −→ ¬S(a) Rule T
Step 11 (x)(F(x) −→ ¬S(x)) UG

Module - 2 Discrete Mathematics and Graph Theory

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