Page 348 - HOW TO PROVE IT: A Structured Approach, Second Edition

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334 Appendix 1: Solutions to Selected Exercises
6. Hint: Begin by showing that ∃x(P(x) ∨ Q(x)) is equivalent to
¬∀x¬(P(x) ∨ Q(x)).
8. (∀x ∈ AP(x)) ∧ (∀x ∈ BP(x))
is equivalent to ∀x(x ∈ A → P(x)) ∧∀x(x ∈ B → P(x))
which is equivalent to ∀x[(x ∈ A → P(x)) ∧ (x ∈ B → P(x))]
which is equivalent to ∀x[(x /∈ A ∨ P(x)) ∧ (x /∈ B ∨ P(x))]
which is equivalent to ∀x[(x /∈ A ∧ x /∈ B) ∨ P(x)]
which is equivalent to ∀x[¬(x ∈ A ∨ x ∈ B) ∨ P(x)]
which is equivalent to ∀x[x ∈ (A ∪ B) → P(x)]
which is equivalent to ∀x ∈ (A ∪ B) P(x).
11. A \ B = ∅ is equivalent to ¬∃x(x ∈ A ∧ x /∈ B)
which is equivalent to ∀x¬(x ∈ A ∧ x /∈ B)
which is equivalent to ∀x(x /∈ A ∨ x ∈ B)
which is equivalent to ∀x(x ∈ A → x ∈ B)
which is equivalent to A ⊆ B.

Section 2.3
1. (a) ∀x(x ∈ F →∀y(y ∈ x → y ∈ A)).
(b) ∀x(x ∈ A →∃n ∈ N(x = 2n + 1)).
2
(c) ∀n ∈ N∃m ∈ N(n + n + 1 = 2m + 1).
(d) ∃x(∀y(y ∈ x →∃i ∈ I(y ∈ A i )) ∧∀i ∈ I∃y(y ∈ x ∧ y /∈ A i )).
4. ∩F = {red, blue} and ∪F = {red, green, blue, orange, purple}.
8. (a) A 2 ={2, 4}, A 3 ={3, 6}, B 2 ={2, 3}, B 3 ={3, 4}.
(b) ∩ i∈I (A i ∪ B i ) ={3, 4} and (∩ i∈I A i ) ∪ (∩ i∈I B i ) ={3}.
(c) They are not equivalent.
11. One example is A ={1, 2} and B ={2, 3}.
13. (a) B 3 ={1, 2, 3, 4, 5} and B 4 ={1, 2, 4, 5, 6}.
(b) ∩ j∈J B j ={1, 2, 4, 5}.
(c) ∪ i∈I (∩ j∈J A i, j ) ={1, 2, 4}.
(d) x ∈∩ j∈J (∪ i∈I A i, j ) means ∀ j ∈ J∃i ∈ I(x ∈ A i, j ) and x ∈
∪ i∈I (∩ j∈J A i, j ) means ∃i ∈ I∀ j ∈ J(x ∈ A i, j ). They are not
equivalent.

Chapter 3

Section 3.1

1. (a) Hypotheses: n is an integer larger than 1 and n is not prime. Conclu-
n
sion: 2 − 1 is not prime. The hypotheses are true when n = 6, so