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