UNIT IIPREDICATES

Predicates: Ex: x is a student

Subject Predicate

Predicate refer to a property that the subject of the statement can have.

The logic based upon the analysis of predicates in any statement is called predicate logic.

Predicates are denoted by capital letters- A,B,C,……..P,Q,R,S…..

Subjects are denoted by small letters a,b,c……. Any statement of the type p is Q where Q is a

predicate and p is the subject can be denoted by Q(p).

1-Place Predicate: one objectEx: John is a bachelor 2-Place Predicate: 2 objectsEx: Rohit is elder than Rahul 3-Place Predicate:3 objectsEx: Santosh is brother of Rohit and Rahul n-Place Predicate: n objects Predicate connectives: Negation of a Predicate: notEx: G(s): Shivani is a good girl ¬ G(s) : Shivani is not a good girl.

Conjunction: andEx: B(s): Shiva is a boy T(s): Shiva is a studentB(s) T(s) : Shiva is a boy and shiva is a student∧ Disjunction: orEx: M(s): sravan is a man L(s): sravan is a lordM(s) v L(s): Sravan is a man or sravan is a lord Implication: if…….thenEx: C(s): Santosh is a scholar P(r): Rahul is a playerC(s) → P(r): If Santosh is a scholar then rahul is a

player

Quantifiers: Existential Quantifier( ): ∃ for some , there exists Ex: x, P(x): x is a prime number∃ Universal Quantifier( ): ∀ for all, for every, for

each, for any ∀ x , P(x): x is a prime number

The statement functions and variables: Simple statement function of one variable: Predicate symbol

and one variableEx: M(r): Ravi is a man Compound statement function of one variable:One or more simple statement functions with logical

connectivesEx: G(x): x is a girl , B(x): x is a boy¬ G(x) , G(x) B(x), G(x) v B(x),G(x) →∧ B(x)Statement function of two variables:Predicate symbol and two individual variablesEx: if r represents Rohit and s represents santhosh and the

statement function of two variables is C(x , y): x is cleverer than v

C(r , s): Rohit is cleverer than santhosh C(s , r): santhosh is cleverer than Rohit

Free and Bound Variables: When a quantifier is used on the variable x or

when we assign a value to this variable , we say that this occurrence of the variable is bound. An occurrence of a variable that is not bound by a quantifier is said to be free.

Rules of Inference:-- Argument:Compound proposition of the form is called an argument are called premises of the argument and c is called conclusion of the argumentWe write the above argument in the following form

p1

p2

p3

.

.pn

cpppp n )...........( 321

npppp ,.....,, 321

c

Consistency of Premises: The premises p1, p2, p3,…..pn of an argument

are said to be inconsistent if their conjunction (p1 p∧ 2 p∧ 3 …. p∧ ∧ n) is false in every possible situation

The premises p1, p2, p3,…..pn of an argument are said to be consistent if their conjunction (p1

p∧ 2 p∧ 3 …. p∧ ∧ n) is true in at least one possible situation

Ex: Consider the Premises (p v q) and ¬p

Therefore the Premises (p v q) and ¬p are consistent Ex: Consider the Premises p and (¬p q)∧

Therefore the premises p and (¬p q) are inconsistent∧

p q ¬p (p v q) (p v q) ∧¬p

T T F T FT F F T FF T T T TF F T F F

p q ¬p (¬p ∧q)

P ∧(¬p ∧q)

T T F F FT F F F FF T T T FF F T F F

Valid and Invalid Arguments:An argument is said to be valid if whenever each of

premises p1, p2, p3,…..pn is true then the conclusion c is true.

In other words the argument is validWhen In an argument the premises are always taken to be

true where as the conclusion may be true or false Conclusion is true- valid argument Conclusion is false- invalid argument

cpppp n )...........( 321

cpppP n ).....( 321

The following rules are called rules of inference Rule of conjunctive simplification:For any propositions p, q if p q is true then p is true∧

(p q) p∧ ⇒Rule of disjunctive Amplification:For any propositions p, q if p is true then p v q is true

p (p v q) ⇒Rule of syllogism:For any propositions p, q , rIf p → q is true and q → r is true then p → r is true

(p → q) ∧ (q → r) ⇒ (p → r)

Rule of syllogism:For any propositions p, q , rIf p → q is true and q → r is true then p → r is true

(p → q) ∧ (q → r) ⇒ (p → r) Or (p → q)

(q → r) (p → r)

Modus pones (Rule of detachment):If p is true and (p → q) is true then q is true

p ∧ (p → q) q⇒Or p

(p → q) q

Modus Tollens:If (p → q) is true and q is false then p is false (p → q) ∧ ¬q ¬p⇒

Direct Proof:1.First assume that p is true 2.Prove that q is true 3. conclusion: p → q is true.Ex 1: Give a direct proof of the statement “The

square of an odd integer is an odd integer”.Soln: We have to prove that “if n is an odd integer then n2 is an odd integer”.Assume that n is an odd integerThen n= 2k+1 for integer kn2 = (2k+1)2 = 4k2+4k+1n2 is not divisible by2. this means that n2 is an odd

integer

Ex 2: Prove that for all integers k and l, if k and l are both odd then (k + l) is even and kl is odd.

Soln: take any two integers k and lAssume that both k and l are odd thenK=2m+1 , l=2n+1K + l = 2m+1+2n+1

=2m+2n+2 = 2(m+n+1)

And kl= (2m+1)(2n+1) =4mn+2m+2n+1

(K +l) is divisible by 2 and kl is not divisible by2Therefore (k + l) is even integer and kl is odd integer

Indirect Proof: A conditional p → q and its contra positive is

¬ q → ¬ p is logically equivalent1. Given conditional p → q and write its contra

positive2. Assume ¬ q is true3. Prove that ¬ p is trueConclusion : p → q is true.

Ex 1: Let n be an integer. Prove that if n2 is odd then n is odd.

Soln: here the conditional p → q where P: n2 is odd and q: n is oddAssume that ¬ q is true that isAssume that n is not an odd integer Then n=2k where k is an integern2 =(2k)2 = 2(2k2)So that n2 is not oddThat is p is falseThat is ¬ p is trueThis proves that ¬ q → ¬ p is trueTherefore p → q is true

Ex 2:Give an indirect proof of the statement “The product of two even integers is an even integer”Soln: The given statement is“If a and b are even integers then ab is an even integer”Let p: a and b are even integers q : ab is an even integerAssume that ¬ q is true that isAssume that ab is not an even integer That means that ab is not divisible by 2 That is a is not divisible by b2 and b is not divisible by 2i.e., a is not an even integer and b is not an even integerP is false so ¬ p is trueThis proves contra positive Therefore p → q is true

Provide an indirect proof of the following statement.

“ For all positive real numbers x and y , if the product xy exceeds 25, then x > 5 or y > 5

The given statement is p →(q v r)Its contra positive is (¬ q ∧ ¬ r) → ¬ pSuppose (¬ q ∧ ¬ r) is trueThen ¬ q is true and ¬ r is trueThat is x ≤ 5 and y ≤ 5This gives x ≤ 25, so that ¬ p is trueContra positive is trueSo p →(q v r) is true

Provide an indirect proof of the following statements

1. For all integers k and l if kl is odd , then both k and l are odd

2. For all integers k and l if k+l is even , then k and l are both even or both odd.

3. Let m and n be integers. Prove that n2 = m2 if and only if m = n and m = -n

Proof by Contradiction:1. Assume that p → q is false that is p is true and q

is false2. Starting with q is false and employing the rules of

logic and other known facts ,infer that p is false. This contradicts the assumption that p is true.

3. Conclusion: Because of the contradiction arrived in the analysis we infer that p → q is true

1. Provide a Proof by contradiction of the following statement:

“For every integer n, if n2 is odd then n is odd”Soln: The given statement is in the form of p → q

where p: n2 is odd q: n is odd Assume that p → q is false that is p is true and q is

falseNow q is false means n is even n=2k for some integer kn2= (2k)2 = 4k2

From this n2 is even ,that is p is false This contradicts the assumption that p is trueTherefore p → q is true.

2.Prove the statement “The square of an even integer is even” by the method of contradiction.

Soln: The given statement is in the form of p → q where p:n is an even integer q: n2 is an even integer

Assume that p → q is false that is p is true and q is false

From q is false n2 is oddn2 = n*n not divisible by 2So n is not divisible by 2That is n is not an even integer ,that is p is false This contradicts the assumption that p is trueTherefore p → q is true.

3. Prove that if m is an even integer then m+7 is an odd integer

Soln: p: m is an even integer q: m+7 is an odd integer

Assume that p → q is false that is p is true and q is false

i.e., m+7 is an even integerThen m+7 = 2k for some integer km = 2k-7m= (2k-8)+1Which shows that m is oddThis means that p is false, which contradicts the

assumption that p is true.Therefore p → q is true

4. Prove that, for all real numbers x and y, if x+y ≥ 100 then x ≥50 or y ≥50

Soln: p: x+y ≥ 100 q: x ≥50 r: y ≥50Given statement is p → (q v r)Assume that p is true and q v r is falsei.e., ¬ (q v r) = ¬q ¬ r∧This means that x<50 and y<50This yields x+y <100Thus p is falseThis contradicts the assumption that p is trueTherefore p → (q v r) is true.

5. Prove that there is no rational number whose square is 2.

6. Using proof by contradiction show that is not a rational number

Soln: let p:

2