Page 110 - HOW TO PROVE IT: A Structured Approach, Second Edition

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96 Proofs
Scratch Work
Givens Goal
A ∩ C ⊆ B a /∈ A \ B
a ∈ C
Because the goal is a negated statement, we try to reexpress it:

a /∈ A \ B is equivalent to ¬(a ∈ A ∧ a /∈ B) (definition of A \ B),
which is equivalent to a /∈ A ∨ a ∈ B (DeMorgan’s law),
which is equivalent to a ∈ A → a ∈ B (conditional law).
Rewriting the goal in this way gives us:

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

We now prove the goal in this new form, using the first strategy from Sec-
tion 3.1. Thus, we add a ∈ A to our list of givens and make a ∈ B our goal:

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

The proof is now easy: From the givens a ∈ A and a ∈ C we can conclude that
a ∈ A ∩ C, and then, since A ∩ C ⊆ B, it follows that a ∈ B.

Solution
Theorem. Suppose A ∩ C ⊆ B and a ∈ C. Then a /∈ A \ B.
Proof. Suppose a ∈ A. Then since a ∈ C, a ∈ A ∩ C. But then since A ∩ C ⊆
B it follows that a ∈ B. Thus, it cannot be the case that a is an element of A
but not B,so a /∈ A \ B.

Sometimes a goal of the form ¬P cannot be reexpressed as a positive state-
ment, and therefore this strategy cannot be used. In this case it is usually best
to do a proof by contradiction. Start by assuming that P is true, and try to
use this assumption to prove something that you know is false. Often this is
done by proving a statement that contradicts one of the givens. Because you
know that the statement you have proven is false, the assumption that P was
true must have been incorrect. The only remaining possibility then is that P is
false.