Page 163 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Existence and Uniqueness Proofs 149
But now we know both P(y) and P(x 0 ), so the goal y = x 0 follows from the
second given.

Solution
Theorem. The following are equivalent:

1. ∃x(P(x) ∧∀y(P(y) → y = x)).
2. ∃x∀y(P(y) ↔ y = x).
3. ∃xP(x) ∧∀y∀z((P(y) ∧ P(z)) → y = z).

Proof. 1 → 2. By statement 1, we can let x 0 be some object such that P(x 0 ) and
∀y(P(y) → y = x 0 ). To prove statement 2 we will show that ∀y(P(y) ↔ y =
x 0 ). We already know the → direction. For the ← direction, suppose y = x 0 .
Then since we know P(x 0 ), we can conclude P(y).
2 → 3. By statement 2, choose x 0 such that ∀y(P(y) ↔ y = x 0 ). Then, in
particular, P(x 0 ) ↔ x 0 = x 0 , and since clearly x 0 = x 0 , it follows that P(x 0 )
is true. Thus, ∃xP(x). To prove the second half of statement 3, let y and z be
arbitrary and suppose P(y) and P(z). Then by our choice of x 0 (as something
for which ∀y(P(y) ↔ y = x 0 ) is true), it follows that y = x 0 and z = x 0 ,so
y = z.
3 → 1. By the first half of statement 3, let x 0 be some object such that P(x 0 ).
Statement 1 will follow if we can show that ∀y(P(y) → y = x 0 ), so suppose
P(y). Since we now have both P(x 0 ) and P(y), by the second half of statement
3 we can conclude that y = x 0 , as required.
Because all three of the statements in the theorem are equivalent to ∃!xP(x),
we can prove a goal of this form by proving any of the three statements in the
theorem. Probably the most common technique for proving a goal of the form
∃!xP(x) is to prove statement 3 of the theorem.
To prove a goal of the form ∃!xP(x):
Prove ∃xP(x) and ∀y∀z((P(y) ∧ P(z)) → y = z). The first of these goals
shows that there exists an x such that P(x) is true, and the second shows that it
is unique. The two parts of the proof are therefore sometimes labeled existence
and uniqueness. Each part is proven using strategies discussed earlier.
Form of final proof:
Existence: [Proof of ∃xP(x) goes here.]
Uniqueness: [Proof of ∀y∀z((P(y) ∧ P(z)) → y = z) goes here.]

Example 3.6.2. Prove that there is a unique set A such that for every set B,
A ∪ B = B.