Page 161 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Existence and Uniqueness Proofs 147
Thus, we see that ∃!xP(x) could also be written as ∃x(P(x) ∧∀y(P(y) →
y = x)). In fact, as the next example shows, several other formulas are also
equivalent to ∃!xP(x), and they suggest other approaches to proving goals of
this form.

Example 3.6.1. Prove that the following formulas are all 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).

Scratch work
If we prove directly that each of these statements is equivalent to each of the
others, then we will have three biconditionals to prove: statement 1 iff statement
2, statement 1 iff statement 3, and statement 2 iff statement 3. If we prove
each biconditional by the methods of Section 3.4, then each will involve two
conditional proofs, so we will need a total of six conditional proofs. Fortunately,
there is an easier way. We will prove that statement 1 implies statement 2,
statement 2 implies statement 3, and statement 3 implies statement 1 – just three
conditionals.Althoughwewillnotgiveaseparateproofthatstatement2implies
statement 1, it will follow from the fact that statement 2 implies statement 3 and
statement 3 implies statement 1. Similarly, the other two conditionals follow
from the three we will prove. Mathematicians almost always use some such
shortcut when proving that several statements are all equivalent. Because we’ll
be proving three conditional statements, our proof will have three parts, which
we will label 1 → 2, 2 → 3, and 3 → 1. We’ll need to work out our strategy
for the three parts separately.
1 → 2. We assume statement 1 and prove statement 2. Because statement 1
starts with an existential quantifier, we choose a name, say x 0 , for some object
for which both P(x 0 ) and ∀y(P(y) → y = x 0 ) are true. Thus, we now have the
following situation:

Givens Goal
P(x 0 ) ∃x∀y(P(y) ↔ y = x)
∀y(P(y) → y = x 0 )

Our goal also starts with an existential quantifier, so to prove it we should
try to find a value of x that makes the rest of the statement come out true.
Of course, the obvious choice is x = x 0 . Plugging in x 0 for x, we see that we
must now prove ∀y(P(y) ↔ y = x 0 ). We let y be arbitrary and prove both
directions of the biconditional. The → direction is clear by the second given.