Presentation is loading. Please wait.

Presentation is loading. Please wait.

Xuan Guo xguo9@student.gsu.edu Lab 4 Xuan Guo xguo9@student.gsu.edu.

Similar presentations


Presentation on theme: "Xuan Guo xguo9@student.gsu.edu Lab 4 Xuan Guo xguo9@student.gsu.edu."— Presentation transcript:

1 Xuan Guo xguo9@student.gsu.edu
Lab 4 Xuan Guo

2 Contents Using Rules of Inference to Build Arguments
Section 1.6, exercise 5 & 17 Proof by contraposition & Proofs by contradiction Section 1.7, exercise 17 & 27

3 Exercise 5 Use rules of inference to show that the hypotheses “Randy works hard,” “If Randy works hard, then he is a dull boy,” and “If Randy is a dull boy, then he will not get the job” imply the conclusion “Randy will not get the job.”

4 Let w be “Randy works hard,”
Let d be “Randy is a dull boy,” Let j be “Randy will get the job.” The hypotheses are w, w→d, and d→¬j The conclusion is ¬j

5 Solution of Exercise 5 Let w be “Randy works hard,”
Let d be “Randy is a dull boy,” Let j be “Randy will get the job.” The hypotheses are w, w→d, and d→¬j The conclusion is ¬j Step Reason w Premise w→d Premise d modus ponens using 1) and 2) d→¬j Premise ¬j modus ponens using 3) and 4)

6 Exercise 27 Use rules of inference to show that if ∀x(P(x)→(Q(x)∧S(x))) and ∀x(P(x)∧R(x)) are true, then ∀x(R(x)∧S(x)) is true

7 Solution of exercise 27 Step Reason 1.∀x(P(x)∧R(x)) Premise
2.P(a)∧R(a) Universal instantiation from (1) 3.P(a) Simplification from (2) 4.∀x(P(x)→ (Q(x)∧S(x))) Premise 5.Q(a)∧S(a) Universal modus ponens from (3)and (4) 6.S(a) Simplification from (5) 7.R(a) Simplification from (2) 8.R(a)∧S(a) Conjunction from (7) and (6) 9.∀x(R(x)∧S(x)) Universal generalization from (5)

8 Exercise 17 Show that if n is an integer and n3+5 is odd, then n is even using a) a proof by contraposition. b) a proof by contradiction.

9 Exercise 17 Show that if n is an integer and n3+5 is odd, then n is even using a) a proof by contraposition. Proof p → q by Contraposition: Assume ¬q and show ¬p is true also. If we give a direct proof of ¬q → ¬p then we have a proof of p → q.

10 Solution to exercise 17 Show that if n is an integer and n3+5 is odd, then n is even using a) a proof by contraposition p: n3+5 is odd, q: n is even ¬p: n3+5 is even, ¬q: n is odd (¬q → ¬p) Solution: Assume that n is odd, so n=2k+1 for some integer k. Then n3+5=2(4k3+6k2+3k+3). Because n3+5 is two times some integer, it is even. QED

11 Exercise 17 Show that if n is an integer and n3+5 is odd, then n is even using b) a proof by contradiction. Proof by Contradiction p: Suppose that we can find a contradiction q such that ¬p → q is true. Because q is false, but ¬p → q is true, we can conclude that ¬p is false, which means that p is true.

12 Solution to Exercise 17 Show that if n is an integer and n3+5 is odd, then n is even using b) a proof by contradiction. p: n is even Solution: Suppose that n3+5 is odd and n is odd. Because n is odd and the product of two odd numbers is odd, it follows that n2 is odd and then that n3 is odd. But then 5 =(n3+5)−n3 would have to be even because it is the difference of two odd numbers. Therefore, the supposition that n3+5 and n were both odd is wrong. QED

13 Exercise 27 Prove that if n is a positive integer, then n is odd if and only if 5n+6 is odd

14 Exercise 27 Prove that if n is a positive integer, then n is odd if and only if 5n+6 is odd p: n is odd, q: 5n+6 is odd If and only if Directly proof: p → q Proof by Contradiction: ¬p → q is true, q is false, so p is true

15 Solution to Exercise 27 Prove that if n is a positive integer, then n is odd if and only if 5n+6 is odd Solution: First, assume that n is odd, so that n=2k+1 for some integer k. Then 5n+6=5(2k+1)+6=10k+11=2(5k+5)+1. Hence, 5n+6 is odd. To prove the converse, suppose that n is even, so that n=2k for some integer k. Then 5n+6=10k+6=2(5k+3), so 5n+6 is even. Hence, n is odd if and only if 5n+6 is odd. QED


Download ppt "Xuan Guo xguo9@student.gsu.edu Lab 4 Xuan Guo xguo9@student.gsu.edu."

Similar presentations


Ads by Google