Page 143 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Conjunctions and Biconditionals 129
The second given starts with an existential quantifier, so we use it immedi-
ately and let x 0 stand for some object for which the statement P(x 0 ) is true.
But now plugging in x 0 for x in the first given we can conclude that ¬P(x 0 ),
which gives us the contradiction we need.
(←) For this direction of the biconditional we should assume ¬∃xP(x) and
try to prove ∀x¬P(x). Because this goal starts with a universal quantifier, we
let x be arbitrary and try to prove ¬P(x). Once again, we now have a negated
goal that can’t be reexpressed, so we use proof by contradiction:
Givens Goal
¬∃xP(x) Contradiction
P(x)

Our first given is also a negated statement, and this suggests that we could
get the contradiction we need by proving ∃xP(x). We therefore set this as our
goal.
Givens Goal
¬∃xP(x) ∃xP(x)
P(x)
To keep from confusing the x that appears as a free variable in the second
given (the arbitrary x introduced earlier in the proof) with the x that appears
as a bound variable in the goal, you might want to rewrite the goal in the
equivalent form ∃yP(y). To prove this goal we have to find a value of y that
makes P(y) come out true. But this is easy! Our second given, P(x), tells us
that our arbitrary x is the value we need.

Solution
Theorem. ∀x¬P(x) ↔¬∃xP(x).
Proof. (→) Suppose ∀x¬P(x), and suppose ∃xP(x). Then we can choose
some x 0 such that P(x 0 ) is true. But since ∀x¬P(x), we can conclude that
¬P(x 0 ), and this is a contradiction. Therefore ∀x¬P(x) →¬∃xP(x).
(←) Suppose ¬∃xP(x). Let x be arbitrary, and suppose P(x). Since we
have a specific x for which P(x) is true, it follows that ∃xP(x), which is a
contradiction. Therefore, ¬P(x). Since x was arbitrary, we can conclude that
∀x¬P(x), so ¬∃xP(x) →∀x¬P(x).

Sometimes in a proof of a goal of the form P ↔ Q the steps in the proof of
Q → P are the same as the steps used to prove P → Q, but in reverse order.
In this case you may be able to simplify the proof by writing it as a string of
equivalences, starting with P and ending with Q. For example, suppose you
found that you could prove P → Q by first assuming P, then using P to infer