Page 152 - HOW TO PROVE IT: A Structured Approach, Second Edition

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138 Proofs
reasoning for the second case is quite similar, using the second given instead
of the first.

Solution
Theorem. Suppose that A, B, and C are sets. If A ⊆ C and B ⊆ C then A ∪
B ⊆ C.
Proof. Suppose A ⊆ C and B ⊆ C, and let x be an arbitrary element of A ∪ B.
Then either x ∈ A or x ∈ B.
Case 1. x ∈ A. Then since A ⊆ C, x ∈ C.
Case 2. x ∈ B. Then since B ⊆ C, x ∈ C.
Since we know that either x ∈ A or x ∈ B, these cases cover all the possi-
bilities, so we can conclude that x ∈ C. Since x was an arbitrary element of
A ∪ B, this means that A ∪ B ⊆ C.

Note that the cases in this proof are not exclusive. In other words, it is possible
for both x ∈ A and x ∈ B to be true, so some values of x might fall under both
cases. There is nothing wrong with this. The cases in a proof by cases must
cover all possibilities, but there is no harm in covering some possibilities more
than once. In other words, the cases must be exhaustive, but they need not be
exclusive.
Proof by cases is sometimes also helpful if you are proving a goal of the
form P ∨ Q. If you can prove P in some cases and Q in others, then as long as
your cases are exhaustive you can conclude that P ∨ Q is true. This method
is particularly useful if one of the givens also has the form of a disjunction,
because then you can use the cases suggested by this given.

To prove a goal of the form P ∨ Q:
Break your proof into cases. In each case, either prove P or prove Q.

Example 3.5.2. Suppose that A, B and C are sets. Prove that A \ (B \ C) ⊆
(A \ B) ∪ C.

Scratch work
Because the goal is ∀x(x ∈ A \ (B \ C) → x ∈ (A \ B) ∪ C), we let x be ar-
bitrary, assume x ∈ A \ (B \ C), and try to prove x ∈ (A \ B) ∪ C. Writing
these statements out in logical symbols gives us:
Givens Goal
x ∈ A ∧¬(x ∈ B ∧ x /∈ C) (x ∈ A ∧ x /∈ B) ∨ x ∈ C
We split the given into two separate givens, x ∈ A and ¬(x ∈ B ∧ x /∈ C),
and since the second is a negated statement we use one of DeMorgan’s laws to