CPSC 121: Models of Computation Module 6: Rewriting predicate logic - - PowerPoint PPT Presentation

CPSC 121: Models of Computation

Module 6: Rewriting predicate logic statements

CPSC 121 – 2018W T1 2

Module 6: Rewriting predicate logic statements

Pre-class quiz #7 is due Wednesday October 24th at 19:00.

Assigned reading for the quiz:

Epp, 4th edition: 4.1, 4.6, Theorem 4.4.1 Epp, 3rd edition: 3.1, 3.6, Theorem 3.4.1. Rosen, 6th edition: 1.6, 1.7. 3.4 (theorem 2 only). Rosen 7th edition: 1.7, 1.8, 4.1 (theorem 2 only).

Assignment #3 is due Friday October 26th at 19:00

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Module 6: Rewriting predicate logic statements

Pre-class quiz #8 is tentatively due Monday November 5th at 19:00

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, 13.3

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Module 6: Rewriting predicate logic statements

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 statements using the “challenge” method (Epp, 4th edition, page 119). Apply universal instantiation, universal modus ponens, and universal modus tollens to predicate logic statements that correspond to the rules’ premises to infer statements implied by the premises.

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Module 6: Rewriting predicate logic statements

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!

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How many of the following statements are log. equivalent to x B ∀ ∈ y C, ∃ ∈ P(x) → Q(y) ? ∀x B y C, ~Q(y) ∈ ∃ ∈ → ~P(x) ~ ( x B y C, P(x) ∃ ∈ ∀ ∈

^ ~Q(y))

~ ( x B y C, ~P(x) ∃ ∈ ∀ ∈

v Q(y))

∀y B x C, P(y) ∈ ∃ ∈ → Q(x) a) None of them b) One of them c) Two of them d) Three of them e) All four of them

Module 6: Rewriting predicate logic statements

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Which of the following could be the result of applying one of the negation equivalence laws to the statement:

∀x D, Q(x)→[( y D, P(x, y)) ~( y D, P(x, y))] ∈ ∃ ∈ ∧ ∀ ∈ a) ∀x D, Q(x) → F ∈ b) F c) None: we can not apply the law to just the right side of the statement. d) None: the law doesn't match the right side of the statement.

Module 6: Rewriting predicate logic statements

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Module 6: Rewriting predicate logic statements

? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?

CPSC 121: the BIG questions:

How can we convince ourselves that an algorithm does what it's supposed to do?

We continue discussing how to prove various types of predicate logic statements that arise when we discuss algorithm correctness.

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Module 6: Rewriting predicate logic statements

By the end of this module, you should be able to:

Explore alternate forms of predicate logic statements using the logical equivalences you have already learned plus negation of quantifiers (a generalized form of the De Morgan’s Law).

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Module 6: Rewriting predicate logic statements

Summary

Thinking of quantifiers differently. Transformations: allowed or forbidden? The challenge method.

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Module 6.1: Thinking of quantifiers differently

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(x ∈

1) v P(x2) v ... v P(xn)

Thinking of quantifiers this way explains

Negation Universal instantiation Universal Modus Ponens, Tollens

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Module 6.1: Thinking of quantifiers differently

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) ∈

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Module 6.1: Thinking of quantifiers differently

What can we do with the negation in:

~ c R ∃ ∈

+ n

∃ 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.

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Module 6.1: Thinking of quantifiers differently

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).

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Module 6: Rewriting predicate logic statements

Summary

Thinking of quantifiers differently. Transformations: allowed or forbidden? The challenge method.

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Module 6.2: Transformations (allowed or forbidden?)

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).

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Module 6.2: Transformations (allowed or forbidden?)

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.

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Module 6.2: Transformations (allowed or forbidden?)

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.

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Module 6.2: Transformations (allowed or forbidden?)

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.

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Module 6.2: Transformations (allowed or forbidden?)

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?

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Module 6.2: Transformations (allowed or forbidden?)

Which propositional logic equivalences apply to predicate logic?

a) Double negative, Identity, 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.

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Module 6.2: Transformations (allowed or forbidden?)

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?

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Module 6.2: Transformations (allowed or forbidden?)

Which rules of inference apply to predicate logic?

a) Modus ponens, modus tollens and elimination only. b) All rules apply, but only if they follow universal quantifiers, not existential quantifiers. c) All rules apply, but only if they follow existential quantifiers, not universal quantifiers. d) All rules apply, no matter what quantifiers are used. e) None of the above.

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Module 6.2: Transformations (allowed or forbidden?)

The only rules we really need: Universal Instantiation: for any a in D ∀x D, P(x) ∈ P(a) Universal Generalization: P(x) for an arbitrary element x of D ∀x D, P(x) ∈

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Module 6.2: Transformations (allowed or forbidden?)

The only rules we really need (continued) Existential Instantiation: ∃x D, P(x) ∈ P(w) for an unspecified (new) witness w in D Existential Generalization: P(a) for a given element a of D ∃x D, P(x) ∈

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Module 6: Rewriting predicate logic statements

Summary

Thinking of quantifiers differently. Transformations: allowed or forbidden? The challenge method.

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Module 6.3: The challenge method

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

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Module 6.3: The challenge method

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?

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Module 6.3: The challenge method

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 < 2 ∀ ∈ ∃ ∈

x

How would we say this in English? How would we prove this theorem?

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Module 6.3: The challenge method

Example 3: x N, ∃ ∈

n

N, n < 2 ∀ ∈

x

How would we say this in English? How would we prove this theorem?

How do we prove a statement is false?