Page 86 - HOW TO PROVE IT: A Structured Approach, Second Edition

P1: PIG/ P2: Oyk/
0521861241c02 CB996/Velleman October 19, 2005 23:48 0 521 86124 1 Char Count= 0

72 Quantificational Logic
(ii) ∃a(x = ya) ∧∃b(x = zb) ∧¬∃w(w < x ∧ (w is a multiple of
both y and z)).
(iii) ∃a(x = ya) ∧∃b(x = zb) ∧¬∃w(w < x ∧∃c(w = yc) ∧
∃d(w = zd)).
2. (a) ∀x(x + 0 = x).
(b) ∀x∃y(x + y = 0).
2
(c) ∀x(x < 0 →¬∃y(y = x)).
(d) We translate this gradually:
(i) ∀x(x > 0 → x has exactly two square roots).
(ii) ∀x(x > 0 →∃y∃z(y and z are square roots of x and y = z and
nothing else is a square root of x)).
2 2 2
(iii) ∀x(x > 0 →∃y∃z(y = x ∧ z = x ∧ y = z ∧¬∃w(w =
x ∧ w = y ∧ w = z))).

Exercises

∗
1. Negate these statements and then reexpress the results as equivalent
positive statements. (See Example 2.2.1.)
(a) Everyone who is majoring in math has a friend who needs help with
his homework.
(b) Everyone has a roommate who dislikes everyone.
(c) A ∪ B ⊆ C \ D.
2
(d) ∃x∀y[y > x →∃z(z + 5z = y)].
2. Negate these statements and then reexpress the results as equivalent
positive statements. (See Example 2.2.1.)
(a) There is someone in the freshman class who doesn’t have a roommate.
(b) Everyone likes someone, but no one likes everyone.
(c) ∀a ∈ A∃b ∈ B(a ∈ C ↔ b ∈ C).
2
(d) ∀y > 0∃x(ax + bx + c = y).
3. Are these statements true or false? The universe of discourse is N.
2
2
2
(a) ∀x(x < 7 →∃a∃b∃c(a + b + c = x)).
2
(b) ∃!x((x − 4) = 9).
2
(c) ∃!x((x − 4) = 25).
2
2
(d) ∃x∃y((x − 4) = 25 ∧ (y − 4) = 25).
∗ 4. Show that the second quantifier negation law, which says that ¬∀xP(x)
is equivalent to ∃x¬P(x), can be derived from the first, which says that
¬∃xP(x) is equivalent to ∀x¬P(x). (Hint: Use the double negation law.)
5. Show that ¬∃x ∈ AP(x) is equivalent to ∀x ∈ A¬P(x).
∗
6. Show that the existential quantifier distributes over disjunction. In other
words, show that ∃x(P(x) ∨ Q(x)) is equivalent to ∃xP(x) ∨∃xQ(x).