Page 162 - HOW TO PROVE IT: A Structured Approach, Second Edition

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148 Proofs
For the ← direction, suppose y = x 0 . We also have P(x 0 ) as a given, and
plugging in y for x 0 in this given we get P(y).
2 → 3. Statement 2 is an existential statement, so we let x 0 be some object
such that ∀y(P(y) ↔ y = x 0 ). The goal, statement 3, is a conjunction, so we
treat it as two separate goals.

Givens Goals
∀y(P(y) ↔ y = x 0 ) ∃xP(x)
∀y∀z((P(y) ∧ P(z)) → y = z)

To prove the first goal we must choose a value for x, and of course the
obvious value is x = x 0 again. Thus, we must prove P(x 0 ). The natural way to
use our only given is to plug in something for y; and to prove the goal P(x 0 ),
the obvious thing to plug in is x 0 . This gives us P(x 0 ) ↔ x 0 = x 0 . Of course,
x 0 = x 0 is true, so by the ← direction of the biconditional, we get P(x 0 ).
For the second goal, we let y and z be arbitrary, assume P(y) and P(z), and
try to prove y = z.

Givens Goal
∀y(P(y) ↔ y = x 0 ) y = z
P(y)
P(z)

Plugging in each of y and z in the first given we get P(y) ↔ y = x 0 and
P(z) ↔ z = x 0 . Since we’ve assumed P(y) and P(z), this time we use the →
directions of these biconditionals to conclude that y = x 0 and z = x 0 . Our goal
y = z clearly follows.
3 → 1. Because statement 3 is a conjunction, we treat it as two separate
givens. The first is an existential statement, so we let x 0 stand for some object
such that P(x 0 ) is true. To prove statement 1 we again let x = x 0 ,sowehave
this situation:
Givens Goal
P(x 0 ) P(x 0 ) ∧∀y(P(y) → y = x 0 )
∀y∀z((P(y) ∧ P(z)) → y = z)
We already know the first half of the goal, so we only need to prove the
second. For this we let y be arbitrary, assume P(y), and make y = x 0 our goal.
Givens Goal
P(x 0 ) y = x 0
∀y∀z((P(y) ∧ P(z)) → y = z)
P(y)