PREDICATES

 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 object Ex: John is a bachelor 

2-Place Predicate:

2 objects Ex: Rohit is elder than Rahul 

3-Place Predicate:

3 objects Ex: Santosh is brother of Rohit and Rahul 

n-Place Predicate:

n objects  Predicate connectives: 

Negation of a Predicate:

not Ex: G(s): Shivani is a good girl ¬ G(s) : Shivani is not a good girl.

 Conjunction: and Ex: B(s): Shiva is a boy T(s): Shiva is a student B(s) ∧ T(s) : Shiva is a boy and shiva is a student  Disjunction: or Ex: M(s): sravan is a man L(s): sravan is a lord M(s) v L(s): Sravan is a man or sravan is a lord  Implication: if…….then

Ex: C(s): Santosh is a scholar P(r): Rahul is a player C(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 variable Ex: M(r): Ravi is a man  Compound statement function of one variable: One or more simple statement functions with logical connectives Ex: 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 variables Ex: 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 (

p

1 

p

2 

p

3 ..........

.



p n

) is called an argument

p

1 ,

p

2 ,

p

3 ,.....

p n



c

are called premises of the argument and c is called conclusion of the argument We write the above argument in the following form .

p 1 p 2 .

p 3 p n 

c



Consistency of Premises:

 The premises p 1 , p 2 , p 3 ,…..p

n of an argument are said to be inconsistent if their conjunction (p 1 ∧ p 2 ∧ p 3 ∧ …. ∧ p n ) is false in every possible situation  The premises p 1 , p 2 , p 3 ,…..p

n of an argument are said to be consistent if their conjunction (p 1 ∧ p 2 ∧ p 3 ∧ …. ∧ p n ) is true in at least one possible situation

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

p

T T F F q T F T F

¬p

F F T T

(p v q)

T T T F

(p v q)

∧

¬p

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

p

T T F F

q

T F T F

¬p

F F T T

( ¬ p

∧

q)

F F T F

P

∧

( ¬ p

∧

q)

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

 Valid and Invalid Arguments: An argument is said to be valid if whenever each of premises p 1 , p 2 , p 3 ,…..p

n is true.

is true then the conclusion c In other words the argument (

p

1 

p

2 

p

3 ..........

.



p n

) 

c

is valid When (

P

1 

p

2 

p

3 .....



p n

) 

c

 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

 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 , r If

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 , r If

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)

p ∧

(p → q)

⇒ q is true then q is true 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 n 2 is an odd integer”.

Assume that n is an odd integer Then n= 2k+1 for integer k n 2 = (2k+1) 2 = 4k 2 +4k+1 n 2 is not divisible by2. this means that n 2 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 l Assume that both k and l are odd then K=2m+1 , l=2n+1 K + 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 by2 Therefore (k + l) is even integer and kl is odd integer

 Indirect Proof:  A conditional p ¬ q

→ →

q and its contra positive is ¬ p is logically equivalent 1. Given conditional p

→

q and write its contra positive 2. Assume ¬ q is true 3. Prove that ¬ p is true Conclusion : p

→

q is true.

Ex 1: Let n be an integer. Prove that if n 2 is odd then n is odd.

Soln: here the conditional p

→

q where P: n 2 is odd and q: n is odd Assume that ¬ q is true that is Assume that n is not an odd integer Then n=2k where k is an integer n2 =(2k) 2 = 2(2k 2 ) So that n 2 is not odd That is p is false That is ¬ p is true This proves that ¬ q

→

¬ p is true Therefore 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 integer Assume that ¬ q is true that is Assume 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 2 i.e., a is not an even integer and b is not an even integer P is false so ¬ p is true This 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)

→

¬ p Suppose (¬ q ∧ ¬ r) is true Then ¬ q is true and ¬ r is true That is x ≤ 5 and y ≤ 5 This gives x ≤ 25, so that ¬ p is true Contra positive is true So 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 n 2 = m 2 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 false 2. 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 n 2 is odd then n is odd” Soln: The given statement is in the form of p where p: n 2 is odd q: n is odd

→

q Assume that p

→

q is false that is p is true and q is false Now q is false means n is even n=2k for some integer k n 2 = (2k) 2 = 4k 2 From this n 2 is even ,that is p is false This contradicts the assumption that p is true Therefore 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 where p:n is an even integer q: n 2

→

is an even q integer Assume that p

→

q is false that is p is true and q is false From q is false n 2 is odd n 2 = n*n not divisible by 2 So n is not divisible by 2 That is n is not an even integer ,that is p is false This contradicts the assumption that p is true Therefore 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 false

→

q is false that is p is true and q is i.e., m+7 is an even integer Then m+7 = 2k for some integer k m = 2k-7 m= (2k-8)+1 Which shows that m is odd This 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 ≥50 Given statement is p

→ (

q v r) Assume that p is true and q v r is false i.e., ¬

(

q v r) = ¬q ∧ ¬ r This means that x<50 and y<50 This yields x+y <100 Thus p is false This contradicts the assumption that p is true Therefore p

→ (

q v r) is true.

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

rational number Soln: let p: