Proof Methods Methods of proof Direct Direct Contrapositive Contradiction pc p→ c ¬ p ∨ c ¬ c → ¬ pp ∧ ¬ c TT T T T F Section 1.6 & 1.7 T F F F F T FT T T T F FF T T T F MSU/CSE 260 Fall 2009 1 MSU/CSE 260 Fall 2009 2 How are these questions related? Proof Methods h1 ∧ h2 ∧ … ∧ hn ⇒ c ? 1. Does p logically imply c ? Let p = h ∧ h ∧ … ∧ h . The following 2. Is the proposition (p → c) a tautology? 1 2 n propositions are equivalent: 3. Is the proposition (¬ p ∨ c) is a tautology? 1. p ⇒ c 4. Is the proposition (¬ c → ¬ p) is a tautology? 2. (p → c) is a tautology. Direct 5. Is the proposition (p ∧ ¬ c) is a contradiction? 3. (¬ p ∨ c) is a tautology. Direct 4. (¬ c → ¬ p)is a tautology. Contrapositive 5. (p ∧ ¬ c) is a contradiction. Contradiction MSU/CSE 260 Fall 2009 3 MSU/CSE 260 Fall 2009 4 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Formal Proofs Formal Proof A proof is equivalent to establishing a logical To prove: implication chain h1 ∧ h2 ∧ … ∧ hn ⇒ c p1 Premise p Tautology Given premises (hypotheses) h1 , h2 , … , hn and Produce a series of wffs, 2 conclusion c, to give a formal proof that the … . p1 , p2 , pn, c . p k, k’, Inf. Rule hypotheses imply the conclusion, entails such that each wff pr is: r . establishing one of the premises or . a tautology, or pn _____ h1 ∧ h2 ∧ … ∧ hn ⇒ c an axiom/law of the domain (e.g., 1+3=4 or x > x+1 ) ∴ c justified by definition, or logically equivalent to or implied by one or more propositions pk where 1 ≤ k < r. MSU/CSE 260 Fall 2009 5 MSU/CSE 260 Fall 2009 6 Example Example h c Prove the theorem: Prove the theorem: “If integer n is odd, then n2 is odd.” “If integer n is odd, then n2 is odd.” Informal proof: Formal Proof: It is given that n is an odd integer. 1. n is odd Premise (h) 2. ∃ k (n = 2k + 1) Definition of “odd” (Universe is Integers) Thus n = 2k + 1, for some integer k. 3. n = 2c + 1, for some integer c Step 2, specialization Thus n2 = (2k + 1)2 = 4k2 + 4k + 1 4. (n = 2c + 1) → (n2 = (2c + 1)2) Laws of arithmetic = 2(2k2 + 2k) + 1 5. n2 = (2c + 1)2 Steps 3 & 4, modus ponens 2 Therefore, n is odd. = 4c2 + 4c + 1 Laws of arithmetic = 2(2c2 + 2c) + 1 Laws of arithmetic 6. ∃ k (n2 = 2k + 1) Step 5, generalization 7. n2 is odd Definition of “odd” MSU/CSE 260 Fall 2009 7 MSU/CSE 260 Fall 2009 8 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Formal Proofs….. Does (p → q) ∧ (q → r) ∧ p ⇒ r ? Truth Table Method Example: Given: p → q, q → r, p. p q r p → q q → r p r Prove: r T T T T T T T We want to establish the logical implication: T T F T F T F (p → q) ∧ (q → r) ∧ p ⇒ r. T F T F T T T We can use either of the following approaches Truth Table T F F F T T F A chain of logical implications F T T T T F T Note that if A ⇒ B and B ⇒ C then A ⇒ C F T F T F F F F F T T T F T F F F T T F F MSU/CSE 260 Fall 2009 9 MSU/CSE 260 Fall 2009 10 Does (p → q) ∧ (q → r) ∧ p ⇒ r ? Example Chain Method Prove: If 2 is even and if 3 is even and if the h1 =p→ q, h2 = q → r, h3 = p, c = r sum of any two even integers is even, then all We want to prove that h1 ∧ h2 ∧ h3 ⇒ c integers greater than 1 and less than 6 are even. 1. p Premise 2. p → q Premise 1. 2 is even Premise 2. 3 is even Premise 3. q Steps 1 & 2, modus ponens* 3. ∀n ∀m ((n is even) ∧ (n is even) → (n+m is even)) Premise 4. q → r Premise 4. (2 is even) ∧ (2 is even) → (4 is even) Specialization (twice), step 3, math 5. r Steps 3&4, modus ponens 5. (3 is even) ∧ (2 is even) → (5 is even) Specialization (twice), step 3, math 6. 4 is even Conj. & modus ponens, steps 1&4 7. 5 is even Conj. & modus ponens, steps 1,2,&5 *See Table 1, page 66 8. (2 is even) ∧ (3 is even) ∧ (4 is even) ∧ (5 is even) Conj.(many times), steps 1,2,6&7 So this is a bona fide theorem – the statement is true! MSU/CSE 260 Fall 2009 11 MSU/CSE 260 Fall 2009 12 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Proving conclusions of the form p → q Example: Direct proof Direct: Assume p, in addition to the given Prove hypothetical syllogism. hypotheses, and conclude q. Prove: (p → q) ∧ (q → s) ⇒ (p → s) Contrapositive: Assume ¬ q, in addition to the given hypotheses, and conclude ¬ p. 1. p Assumption Contradiction: Assume both p and ¬ q, in 2. p → q Premise addition to the given hypotheses, and conclude False. 3. q 1, 2, modus ponens 4. q → s Premise Vacuous: Assume the given hypotheses, and conclude ¬ p. 5. s 3, 4, modus ponens Trivial: Assume the given hypotheses, and 6. p → s 1, 5, direct method of proof conclude q. MSU/CSE 260 Fall 2009 13 MSU/CSE 260 Fall 2009 14 Example: Contrapositive proof Example: Contradiction proof Prove hypothetical syllogism. Prove hypothetical syllogism. Prove: (p → q) ∧ (q → s) ⇒ (p → s) Prove: (p → q) ∧ (q → s) ⇒ (p → s) 1. p Assumption 1. ¬ s Assumption 2. ¬ s Assumption 2. q → s Premise 3. q → s Premise 3. ¬ q 1, 2, modus tollens 4. ¬ q 2, 3, modus tollens 4. p → q Premise 5. p → q Premise 5. ¬ p 3, 4, modus tollens 6. ¬ p 4, 5, modus tollens 6. p → s 1, 5, contrapositive method 7. p ∧ ¬ p 1, 6, conjunction 8. False 7, logical equivalence 9. p → s 1, 2, 8, contradiction method MSU/CSE 260 Fall 2009 15 MSU/CSE 260 Fall 2009 16 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Example: Vacuous proof Example: Trivial proof Prove: If p → r and q →¬r, then p ∧ q → s Prove: If p → r and ¬ r, then q →¬p Equivalently, prove: Equivalently, prove: (p → r ) ∧ (q →¬r ) ⇒ (p ∧ q → s) (p → r ) ∧ ( ¬ r ) ⇒ ( q →¬p) 1. p → r Premise 1. p → r Premise 2. ¬ p ∨ r 1, Implication 2. ¬ r Premise 3. q →¬r Premise 3. ¬ p 1, 2, modus tollens 4. ¬ q ∨¬r 3, Implication 4. q →¬p 3, Trivially 5. ¬ p ∨¬q 2, 4, Resolution 6. ¬ (p ∧ q ) 5, DeMorgan 7. p ∧ q → s 6, Vacuously MSU/CSE 260 Fall 2009 17 MSU/CSE 260 Fall 2009 18 Non‐examples: Vacuous, trivial proofs Example Recall hypothetical syllogism. Let Prove: (p → q) ∧ (q → s) ⇒ (p → s) h1 =q∨ d h = (q ∨ d ) → ¬ p Why didn’t we give example vacuous or trivial 2 → ∧ proofs of hypothetical syllogism? h3 = ¬ p (a ¬ b) h4 = (a ∧ ¬ b) → (r ∨ s) c=r∨ s we want to establish h1 ∧ h2 ∧ h3 ∧ h4 ⇒ c. MSU/CSE 260 Fall 2009 19 MSU/CSE 260 Fall 2009 20 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Solution 1 Solution 2 Let Let h =q∨ dh= (q ∨ d ) → ¬ p 1 2 h1 =q∨ dh2 = (q ∨ d ) → ¬ p h = ¬ p → (a ∧ ¬ b) h = (a ∧ ¬ b) → (r ∨ s) 3 4 h3 = ¬ p → (a ∧ ¬ b) h4 = (a ∧ ¬ b) → (r ∨ s) c=r∨ s c=r∨ s, we want to establish h ∧ h ∧ h ∧ h ⇒ c. 1 2 3 4 we want to establish h1 ∧ h2 ∧ h3 ∧ h4 ⇒ c. 1. (q ∨ d ) → ¬ p Premise 1. q ∨ d Premise 2. ¬ p → (a ∧ ¬ b)Premise 2. (q ∨ d ) → ¬ p Premise 3. (q ∨ d ) → (a ∧ ¬ b)1&2, Hypothetical Syllogism 3. ¬ p 1&2, and modus ponens 4. (a ∧ ¬ b) → (r ∨ s)Premise 4. ¬ p → (a ∧ ¬ b)Premise 5. (q ∨ d ) → (r ∨ s)3&4, HS 5. (a ∧ ¬ b)3&4, modus ponens 6. q ∨ d Premise 6. (a ∧ ¬ b) → (r ∨ s) Premise 7. r ∨ s 5&6, Modus Ponens 7. r ∨ s 5&6, modus ponens MSU/CSE 260 Fall 2009 21 MSU/CSE 260 Fall 2009 22 Example Is [(¬ (p ∧ q)) → (¬ p ∨ q)] ≡ (¬ p ∨ q) ? Truth Table Method Question: Is [(¬ (p ∧ q)) → (¬ p ∨ q)] ≡ (¬ p ∨ q) ? pq¬(p ∧ q)(¬p ∨ q)LHSRHSAnswer Different ways to answer the above question 1. By means of the Truth Table. TT FTT T YES 2. By means of derivation. 3. By formulating it as logical equivalence, that is, as a TF TFF F YES “proof”. FT TTT T YES FF TTT T YES MSU/CSE 260 Fall 2009 23 MSU/CSE 260 Fall 2009 24 © 2006 by A-H. Esfahanian. All Rights Reserved. 1- Is [(¬ (p ∧ q)) → (¬ p ∨ q)] ≡ (¬ p ∨ q) ? Is [ (¬ (p ∧ q)) → (¬ p ∨ q)] ≡ (¬ p ∨ q) ? Derivation Method Logical Equivalence Method (¬ (p ∧ q)) → (¬ p ∨ q) ≡ ¬(¬(p ∧ q)) ∨ (¬ p ∨ q) To show S ≡ R: show that S ⇒ R and R ⇒ S ≡ (p ∧ q) ∨ (¬ p ∨ q) ≡ ((p ∧ q) ∨ ¬ p) ∨ q ) ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ ≡ (¬ p ∨ (p ∧ q)) ∨ q In this case, we have ≡ ((¬ p ∨ p) ∧ (¬ p ∨ q)) ∨ q ≡ ((T) ∧ (¬ p ∨ q)) ∨q S = [¬ (p ∧ q)) → (¬ p ∨ q)] and ≡ (¬ p ∨ q) ∨q R = (¬ p ∨ q) ≡ (¬ p) ∨ (q ∨q ) ≡ (¬ p) ∨ (q) ≡ (¬ p ∨ q) MSU/CSE 260 Fall 2009 25 MSU/CSE 260 Fall 2009 26 Prove [(¬ (p ∧ q)) → (¬ p ∨ q)] ⇒ (¬ p ∨ q) Example 1.