Page 119 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Negations and Conditionals 105
Finally, our last step is to use modus tollens. We now know Q → R and
¬R, so by modus tollens we can conclude ¬Q. This is our goal, so the proof
is done.

Solution

Theorem. Suppose P → (Q → R). Then ¬R → (P →¬Q).
Proof. Suppose ¬R. Suppose P. Since P and P → (Q → R), it follows that
Q → R. But then, since ¬R, we can conclude ¬Q. Thus, P →¬Q. Therefore
¬R → (P →¬Q).

Sometimes if you’re stuck you can use rules of inference to work backward.
For example, suppose one of your givens has the form P → Q and your goal
is Q. If only you could prove P, you could use modus ponens to reach your
goal. This suggests treating P as your goal instead of Q. If you can prove P,
then you’ll just have to add one more step to the proof to reach your original
goal Q.

Example 3.2.5. Suppose that A ⊆ B, a ∈ A, and a /∈ B \ C. Prove that a ∈ C.
Scratch work

Givens Goal
A ⊆ B a ∈ C
a ∈ A
a /∈ B \ C

Our third given is a negative statement, so we begin by reexpressing it
as an equivalent positive statement. According to the definition of the dif-
ference of two sets, this given means ¬(a ∈ B ∧ a /∈ C), and by one of
DeMorgan’s laws, this is equivalent to a /∈ B ∨ a ∈ C. Because our goal is
a ∈ C,itisprobablymoreusefultorewritethisintheequivalentform a ∈ B →
a ∈ C:

Givens Goal
A ⊆ B a ∈ C
a ∈ A
a ∈ B → a ∈ C

Now we can use our strategy for using givens of the form P → Q. Our goal
is a ∈ C, and we are given that a ∈ B → a ∈ C. If we could prove that a ∈ B,