Page 140 - HOW TO PROVE IT: A Structured Approach, Second Edition

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126 Proofs
Proof. Suppose x ∈ A. Since A ⊆ B, it follows that x ∈ B, and since A and C
are disjoint, we must have x /∈ C. Thus, x ∈ B \ C. Since x was an arbitrary
element of A, we can conclude that A ⊆ B \ C.

Using our strategies for working with conjunctions, we can now work out
the proper way to deal with statements of the form P ↔ Q in proofs. Because
P ↔ Q is equivalent to (P → Q) ∧ (Q → P), according to our strategies a
given or goal of the form P ↔ Q should be treated as two separate givens or
goals: P → Q, and Q → P.

To prove a goal of the form P ↔ Q:
Prove P → Q and Q → P separately.
To use a given of the form P ↔ Q:
Treat this as two separate givens: P → Q, and Q → P.

This is illustrated in the next example, in which we use the following
definitions: An integer x is even if ∃k ∈ Z(x = 2k), and x is odd if ∃k ∈
Z(x = 2k + 1). We also use the fact that every integer is either even or odd,
but not both. We’ll see a proof of this fact in Chapter 6.

2
Example 3.4.2. Suppose x is an integer. Prove that x is even iff x is even.
Scratch work
2
The goal is (x is even) ↔ (x is even), so we prove the two goals (x is even) →
2
2
(x is even) and (x is even) → (x is even) separately. For the first, we assume
2
that x is even and prove that x is even:
Givens Goal
2
x ∈ Z x is even
x is even
Writing out the definition of even in both the given and the goal will reveal
their logical forms:

Givens Goal
2
x ∈ Z ∃k ∈ Z(x = 2k)
∃k ∈ Z(x = 2k)

Because the second given starts with ∃k, we immediately use it and let k
stand for some particular integer for which the statement x = 2k is true. Thus,