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UNIT II

UNIT II. P REDICATES. 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.

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UNIT II

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  1. UNIT II PREDICATES

  2. Predicates: Ex: xis 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).

  3. 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.

  4. 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

  5. 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

  6. 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

  7. 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.

  8. 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 argument We write the above argument in the following form p1 p2 p3 . . pn

  9. Consistency of Premises: • The premises p1, p2, p3,…..pn of an argument are said to be inconsistent if their conjunction (p1 ∧ p2 ∧ p3 ∧ …. ∧ pn) is false in every possible situation • The premises p1, p2, p3,…..pn of an argument are said to be consistent if their conjunction (p1 ∧ p2 ∧ p3 ∧ …. ∧ pn) is true in at least one possible situation

  10. 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

  11. 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 valid When • 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

  12. 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)

  13. 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) is true then q is true p ∧ (p → q) ⇒ q Or p (p → q) q

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

  15. 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 integer Then n= 2k+1 for integer k n2 = (2k+1)2 = 4k2+4k+1 n2 is not divisible by2. this means that n2 is an odd integer

  16. 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

  17. Indirect Proof: • A conditional p → q and its contra positive is ¬ q → ¬ 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.

  18. 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 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(2k2) So that n2 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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 false Now q is false means n is even n=2k for some integer k n2= (2k)2 = 4k2 From this n2 is even ,that is p is false This contradicts the assumption that p is true Therefore p → q is true.

  24. 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 odd n2 = 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.

  25. 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 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

  26. 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.

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

  28. 6. Using proof by contradiction show that is not a rational number Soln: let p:

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