Page 155 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Disjunctions 141
2
2
Case 2. x is odd. Then x = 2k + 1 for some integer k,so x = 4k + 4k + 1.
2
Clearly in this case the remainder when x is divided by 4 is 1.
Sometimes in a proof of a goal that has the form P ∨ Q it is hard to figure
out how to break the proof into cases. Here’s a way of doing it that is often
helpful. Simply assume that P is true in case 1 and assume that it is false in
case 2. Certainly P is either true or false, so these cases are exhaustive. In the
first case you have assumed that P is true, so certainly the goal P ∨ Q is true.
Thus, no further reasoning is needed in case 1. In the second case you have
assumed that P is false, so the only way the goal P ∨ Q could be true is if Q
is true. Thus, to complete this case you should try to prove Q.

To prove a goal of the form P ∨ Q:
If P is true, then clearly the goal P ∨ Q is true, so you only need to worry
about the case in which P is false. You can complete the proof in this case by
proving that Q is true.

Scratch work
Before using strategy:

Givens Goal
— P ∨ Q
—
After using strategy:

Givens Goal
— Q
—
¬P

Form of final proof:
If P is true, then of course P ∨ Q is true. Now suppose P is false.
[Proof of Q goes here.]
Thus, P ∨ Q is true.
Thus, this strategy for proving P ∨ Q suggests that you transform the prob-
lem by adding ¬P as a new given and changing the goal to Q. It is interesting
to note that this is exactly the same as the transformation you would use if
you were proving the goal ¬P → Q! This is not really surprising, because we
already know that the statements P ∨ Q and ¬P → Q are equivalent. But we