Page 120 - HOW TO PROVE IT: A Structured Approach, Second Edition

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106 Proofs
then we could use modus ponens to reach our goal. So let’s try treating a ∈ B
as our goal and see if that makes the problem easier:

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

Now it is clear how to reach the goal. Since a ∈ A and A ⊆ B, a ∈ B.

Solution
Theorem. Suppose that A ⊆ B, a ∈ A, and a /∈ B \ C. Then a ∈ C.
Proof. Since a ∈ A and A ⊆ B, we can conclude that a ∈ B. But a /∈ B \ C,
so it follows that a ∈ C.

Exercises

∗
1. This problem could be solved by using truth tables, but don’t do it that
way. Instead, use the methods for writing proofs discussed so far in this
chapter. (See Example 3.2.4.)
(a) Suppose P → Q and Q → R are both true. Prove that P → R is
true.
(b) Suppose ¬R → (P →¬Q) is true. Prove that P → (Q → R)is
true.
2. This problem could be solved by using truth tables, but don’t do it that
way. Instead, use the methods for writing proofs discussed so far in this
chapter. (See Example 3.2.4.)
(a) Suppose P → Q and R →¬Q are both true. Prove that P →¬R
is true.
(b) Suppose that P is true. Prove that Q →¬(Q →¬P) is true.
3. Suppose A ⊆ C, and B and C are disjoint. Prove that if x ∈ A then
x /∈ B.
4. Suppose that A \ B is disjoint from C and x ∈ A. Prove that if x ∈ C
then x ∈ B.
∗ 5. Use the method of proof by contradiction to prove the theorem in
Example 3.2.1.
6. Use the method of proof by contradiction to prove the theorem in
Example 3.2.5.
7. Suppose that y + x = 2y − x, and x and y are not both zero. Prove that
y = 0.