Page 130 - HOW TO PROVE IT: A Structured Approach, Second Edition

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116 Proofs
for which you know either P(a)or ¬Q(a), and plug this a in for x when it
appears.
We’ve already used this technique in some of our earlier proofs when dealing
with givens of the form A ⊆ B. For instance, in Example 3.2.5 we used the
givens A ⊆ B and a ∈ A to conclude that a ∈ B. The justification for this rea-
soning is that A ⊆ B means ∀x(x ∈ A → x ∈ B), so by universal instantiation
we can plug in a for x and conclude that a ∈ A → a ∈ B. Since we also know
a ∈ A, it follows by modus ponens that a ∈ B.

Example 3.3.4. Suppose F and G are families of sets and F ∩ G = ∅. Prove
that ∩F ⊆∪G.
Scratch work

Our first step in analyzing the logical form of the goal is to write out the meaning
of the subset symbol, which gives us the statement ∀x(x ∈∩F → x ∈∪G).
We could go further with this analysis by writing out the definitions of union
and intersection, but the part of the analysis that we have already done will be
enough to allow us to decide how to get started on the proof. The definitions
of union and intersection will be needed later in the proof, but we will wait
until they are needed before filling them in. When analyzing the logical forms
of givens and goals in order to figure out a proof, it is usually best to do only
as much of the analysis as is needed to determine the next step of the proof.
Going further with the logical analysis usually just introduces unnecessary
complication, without providing any benefit.
Because the goal means ∀x(x ∈∩F → x ∈∪G), we let x be arbitrary,
assume x ∈∩F, and try to prove x ∈∪G.

Givens Goal
F ∩ G = ∅ x ∈∪G
x ∈∩F
The new goal means ∃A ∈ G(x ∈ A), so to prove it we should try to find a
value that will “work” for A. Just looking at the goal doesn’t make it clear how
to choose A, so we look more closely at the givens. We begin by writing them
out in logical symbols:

Givens Goal
∃A(A ∈ F ∩ G) ∃A ∈ G(x ∈ A)
∀A ∈ F(x ∈ A)
The second given starts with ∀A, so we may not be able to use this given
until a likely value to plug in for A pops up during the course of the proof. In