CPSC 121: Models of Computation. Module 6: Rewriting predicate logic statements
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1 CPSC 121: Models of Computation
2 Pre-class quiz #7 is due Wednesday October 16th at 17:00. Assigned reading for the quiz: Epp, 4th edition: 4.1, 4.6, Theorem Epp, 3rd edition: 3.1, 3.6, Theorem Rosen, 6th edition: 1.6, (theorem 2 only). Rosen 7th edition: 1.7, 1.8, 4.1 (theorem 2 only). Assignment #3 is due Wednesday October 23 rd at 17:00. 2
3 Pre-class quiz #8 tentatively due Monday October 28th at 17:00 Epp, 4th edition: 12.2, pages 791 to 795. Epp, 3rd edition: 12.2, pages 745 to 747, 752 to 754 Rosen, 6th edition: 12.2 pages 796 to 798, 12.3 Rosen, 7th edition: 13.2 pages 858 to 861,
4 By the start of class, you should be able to: Determine the negation of any quantified statement. Given a quantified statement and an equivalence rule, apply the rule to create an equivalent statement (particularly the De Morgan s and contrapositive rules). Prove and disprove quantified using the challenge method (Epp, 4th edition, page 119). Apply universal instantiation, universal modus ponens, and universal modus tollens to predicate logic that correspond to the rules premises to infer implied by the premises. 4
5 Quiz 6 feedback: Once again well done overall. Two questions seemed more difficult (next slides). Not much to say about the open-ended question: most people gave correct strategies. Great! 5
6 Which are logically equivalent to x B y C, P(x) Q(y) a) ~ ( x B y C, P(x) ^ ~Q(y)) b) y B x C, P(y) Q(x) 6
7 Which of the following could be the result of applying one of the negation equivalence laws (p ~p F, p ~p T) to the statement: x D, Q(x) [( y D, P(x, y)) ~( y D, P(x, y))] a) x D, Q(x) F b) None of these, because the law doesn't match the statement (specifically, the right side of the statement). 7
8 How can we convince ourselves that an algorithm does what it's supposed to do CPSC 121: the BIG questions: We continue discussing how to prove various types of predicate logic that arise when we discuss algorithm correctness. 8
9 By the end of this Module, you should be able to: Explore alternate forms of predicate logic using the logical equivalences you have already learned plus negation of quantifiers (a generalized form of the De Morgan s Law). 9
10 Summary Thinking of quantifiers differently. Transformations: allowed or forbidden The challenge method. 10
11 Suppose D contains values x1, x2,..., xn What does x D, P(x) really mean It's the same as P(x1) ^ P(x2) ^... ^ P(xn). Similarly, x D, P(x) P(x1) v P(x2) v... v P(xn) Thinking of quantifiers this way explains Negation Universal instantiation Universal Modus Ponens, Tollens 11
12 Negation: ~ x D, P(x) ~(P(x1) ^ P(x2) ^... ^ P(xn)) ~P(x1) v ~P(x2) v... v ~P(xn) x D, ~P(x) ~ x D, P(x) ~(P(x1) v P(x2) v... v P(xn)) ~P(x1) ^ ~P(x2) ^... ^ ~P(xn) x D, ~P(x) 12
13 What can we do with the negation in: ~ c R+ n0 N n N, n n0 f(n) cg(n) a) It cannot be moved inward. b) It can only move across one quantifier because the generalized De Morgan s law can only handle one quantifier. c) It can only be moved across all three quantifiers because a negation can't appear between quantifiers. d) It could be moved across one, two or all three quantifiers. e) None of the above. 13
14 Exercise: Let A be the set of amoebae, and Parent(x,y) be true if amoeba x is amoeba y's parent. Use logical equivalences to show that these two translations of an amoeba has only one parent are logically equivalent: (1) x A, y A, Parent(y, x) ( z A, Parent(z, x) y = z). (2) x A, y A, Parent(y, x) (~ z A, Parent(z, x) y z). 14
15 Summary Thinking of quantifiers differently. Transformations: allowed or forbidden The challenge method. 15
16 Universal instantiation: ( x D, P(x)) ^ (a D) P(a) Proving it is a valid inference: Suppose x D, P(x) is true. Hence P(x1) ^ P(x2) ^... ^ P(xn) holds. If a = xi is an element of D, then by specialization we have P(xi). 16
17 Universal modus ponens: Suppose x D, P(x) Q(x) is true. P(xi) holds where xi is an element of D. Hence (P(x1) Q(x1)) ^... ^ (P(xn) Q(xn)). By specialization P(xi) Q(xi) holds. By modus ponens, we deduce Q(xi). The same reasoning explains why universal modus tollens is valid. 17
18 Applying logical equivalences to predicate logic: Suppose we have x D, P(x) Q(x) We know that P(x) Q(x) ~P(x) v Q(x) We might want to write x D, ~P(x) v Q(x) Is this valid 18
19 Which propositional logic equivalences apply to predicate logic a) Modus ponens, modus tollens, and De Morgan's (not all equivalences!) b) ~(P(x) Q(x)) P(x) ~Q(x) c) Commutative, Associative, and the definition of conditional d) All propositional logic equivalences apply to predicate logic. e) None of the above. 19
20 Applying rules of inference to predicate logic: Suppose we have x D, P(x) Q(x) We know that x D, P(x) We might want to write x D, Q(x) Is this valid 20
21 Which rules of inference apply to predicate logic a) Modus ponens and modus tollens only. b) All rules except elimination. c) All rules apply, but only if they follow universal quantifiers, not existential quantifiers. d) All rules apply, but only if they follow existential quantifiers, not universal quantifiers. e) All rules apply, no matter what quantifiers are used. f) None of the above. 21
22 Consider existential instantiation: x D, P(x) a D P(a) a) This argument is valid: P(a) is true. b) This argument is invalid: P(a) is false. c) This argument is invalid: P(a) might be false. d) This argument is invalid for another reason. 22
23 What about existential modus ponens x D, P(x) Q(x) P(a) Q(a) a) This argument is valid, and Q(a) is true. b) The argument is valid, but the 1st premise can not be true; so Q(a) might be false. c) This argument is invalid because Q(a) is false. d) The argument is invalid for another reason. 23
24 Summary Thinking of quantifiers differently. Transformations: allowed or forbidden The challenge method. 24
25 The Challenge method: A predicate logic statement is like a game with two players. you (trying to prove the statement true) your adversary (trying to prove it false). The two of you pick values for the quantified variables working from the left to right (i.e. inwards). You pick the values of existentially quantified variables. Your adversary picks the values of universally quantified variables 25
26 The Challenge method (continued): If there is a strategy that allows you to always win, then the statement is true. If there is a strategy for your adversary that allows him/her to always win, then the statement is false. What does it mean to have a winning strategy at Nim 26
27 Example 1: x Z, n Z+, 2x < n How would we say this in English How would we prove this theorem Example 2: n N, x N, n < 2x How would we say this in English How would we prove this theorem 27
28 Example 3: x N, n N, n < 2x How would we say this in English How would we prove this theorem How do we prove a statement is false 28
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