Page 153 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Disjunctions 139
reexpress it as the positive statement x /∈ B ∨ x ∈ C.
Givens Goal
x ∈ A (x ∈ A ∧ x /∈ B) ∨ x ∈ C
x /∈ B ∨ x ∈ C

Now the second given and the goal are both disjunctions, so we’ll try consid-
ering the two cases x /∈ B and x ∈ C suggested by the second given. According
to our strategy for proving goals of the form P ∨ Q, if in each case we can
either prove x ∈ A ∧ x /∈ B or prove x ∈ C, then the proof will be complete.
For the first case we assume x /∈ B.

Givens Goal
x ∈ A (x ∈ A ∧ x /∈ B) ∨ x ∈ C
x /∈ B
In this case the goal is clearly true, because in fact we can conclude that
x ∈ A ∧ x /∈ B. For the second case we assume x ∈ C, and once again the
goal is clearly true.
Solution

Theorem. SupposethatA,B,andCaresets.Then A \ (B \ C) ⊆ (A \ B) ∪ C.
Proof. Suppose x ∈ A \ (B \ C). Then x ∈ A and x /∈ B \ C. Since x /∈ B \
C, it follows that either x /∈ B or x ∈ C. We will consider these cases separately.
Case 1. x /∈ B. Then since x ∈ A, x ∈ A \ B,so x ∈ (A \ B) ∪ C.
Case 2. x ∈ C. Then clearly x ∈ (A \ B) ∪ C.
Since x was an arbitrary element of A \ (B \ C), we can conclude that A \
(B \ C) ⊆ (A \ B) ∪ C.

Sometimes you may find it useful to break a proof into cases even if the cases
are not suggested by a given of the form P ∨ Q. Any proof can be broken into
cases at any time, as long as the cases exhaust all of the possibilities.

2
Example 3.5.3. Prove that for every integer x, the remainder when x is divided
by 4 is either 0 or 1.
Scratch work
We start by letting x be an arbitrary integer and then try to prove that the
2
remainder when x is divided by 4 is either 0 or 1.
Givens Goal
2
2
x ∈ Z (x ÷ 4 has remainder 0) ∨ (x ÷ 4 has remainder 1)