Page 160 - HOW TO PROVE IT: A Structured Approach, Second Edition

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146 Proofs
Proof. Let x be an arbitrary real number.
2
Case 1. x = 0. Let y = 1. Then xy = 0 and y − x = 1 − 0 = 1, so
2
xy = y − x.
2
Case 2. x = 0. Let y = 0. Then xy = 0 and y − x =−x = 0, so
2
xy = y − x.
2
Since these cases are exhaustive, we have shown that ∃y ∈ R(xy =
2
y − x). Since x was arbitrary, this shows that ∀x ∈ R∃y ∈ R(xy =
y − x).
29. Prove that if ∀xP(x) →∃xQ(x) then ∃x(P(x) → Q(x)). (Hint: Re-
member that P → Q is equivalent to ¬P ∨ Q).

*30. Consider the following putative theorem.
Theorem? Suppose A, B, and C are sets and A ⊆ B ∪ C. Then either
A ⊆ Bor A ⊆ C.

Is the following proof correct? If so, what proof strategies does it use?
If not, can it be fixed? Is the theorem correct?

Proof. Let x be an arbitrary element of A. Since A ⊆ B ∪ C, it follows
that either x ∈ B or x ∈ C.
Case 1. x ∈ B. Since x was an arbitrary element of A, it follows that
∀x ∈ A(x ∈ B), which means that A ⊆ B.
Case 2. x ∈ C. Similarly, since x was an arbitrary element of A,we
can conclude that A ⊆ C.
Thus, either A ⊆ B or A ⊆ C.

31. Prove ∃x(P(x) →∀yP(y)).

3.6. Existence and Uniqueness Proofs

In this section we consider proofs in which the goal has the form ∃!xP(x). As
we saw in Section 2.2, this can be thought of as an abbreviation for the formula
∃x(P(x) ∧¬∃y(P(y) ∧ y = x)). According to the proof strategies discussed
in previous sections, we could therefore prove this goal by finding a particular
value of x for which we could prove both P(x) and ¬∃y(P(y) ∧ y = x). The
last part of this proof would involve proving a negated statement, but we can
reexpress it as an equivalent positive statement:

¬∃y(P(y) ∧ y = x)
is equivalent to ∀y¬(P(y) ∧ y = x) (quantifier negation law),
which is equivalent to ∀y(¬P(y) ∨ y = x) (DeMorgan’s law),
which is equivalent to ∀y(P(y) → y = x) (conditional law).