Page 164 - HOW TO PROVE IT: A Structured Approach, Second Edition

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150 Proofs
Scratch work
Our goal is ∃!AP(A), where P(A) is the statement ∀B(A ∪ B = B). Accord-
ing to our strategy, we can prove this by proving existence and uniqueness
separately. For the existence half of the proof we must prove ∃AP(A), so we
try to find a value of A that makes P(A) true. There is no formula for finding
this set A, but if you think about what the statement P(A) means, you should
realize that the right choice is A =∅. Plugging this value in for A, we see that
to complete the existence half of the proof we must show that ∀B(∅ ∪B = B).
This is clearly true. (If you’re not sure of this, work out the proof!)
For the uniqueness half of the proof we prove ∀C∀D((P(C) ∧ P(D)) →
C = D). To do this, we let C and D be arbitrary, assume P(C) and P(D), and
prove C = D. Writing out what the statements P(C) and P(D) mean, we have
the following givens and goal:

Givens Goal
∀B(C ∪ B = B) C = D
∀B(D ∪ B = B)

To use the givens, we should try to find something to plug in for B in each of
them. There is a clever choice that makes the rest of the proof easy: We plug in
D for B in the first given, and C for B in the second. This gives us C ∪ D = D
and D ∪ C = C. But clearly C ∪ D = D ∪ C. (If you don’t see why, prove it!)
The goal C = D follows immediately.

Solution
Theorem. There is a unique set A such that for every set B, A ∪ B = B.
Proof. Existence: Clearly ∀B(∅ ∪B = B), so ∅ has the required property.
Uniqueness: Suppose ∀B(C ∪ B = B) and ∀B(D ∪ B = B). Applying the
first of these assumptions to D we see that C ∪ D = D, and applying the second
toCweget D ∪ C = C. But clearly C ∪ D = D ∪ C,so C = D.
Sometimes a statement of the form ∃!xP(x) is proven by proving statement
1 from Example 3.6.1. This leads to the following proof strategy.

To prove a goal of the form ∃!xP(x):
Prove ∃x(P(x) ∧∀y(P(y) → y = x)), using strategies from previous
sections.

Example 3.6.3. Prove that for every real number x, if x = 2 then there is a
unique real number y such that 2y/(y + 1) = x.