Page 151 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Disjunctions 137
Case 2. Q is true.
[Proof of goal goes here.]
Since we know P ∨ Q, these cases cover all the possibilities. Therefore
the goal must be true.

Example 3.5.1. Suppose that A, B, and C are sets. Prove that if A ⊆ C and
B ⊆ C then A ∪ B ⊆ C.
Scratch work
We assume A ⊆ C and B ⊆ C and prove A ∪ B ⊆ C. Writing out the goal
using logical symbols gives us the following givens and goal:
Givens Goal
A ⊆ C ∀x(x ∈ A ∪ B → x ∈ C)
B ⊆ C
To prove the goal we let x be arbitrary, assume x ∈ A ∪ B, and try to prove
x ∈ C. Thus, we now have a new given x ∈ A ∪ B, which we write as x ∈
A ∨ x ∈ B, and our goal is now x ∈ C.

Givens Goal
A ⊆ C x ∈ C
B ⊆ C
x ∈ A ∨ x ∈ B

Because the goal cannot be analyzed any further at this point, we look more
closely at the givens. The first given will be useful if we ever come across an
object that is an element of A, since it would allow us to conclude immediately
that this object must also be an element of C. Similarly, the second given will
be useful if we come across an element of B. Keeping in mind that we should
watch for any elements of A or B that might come up, we move on to the third
given. Because this given has the form P ∨ Q, we try proof by cases. For the
first case we assume x ∈ A, and for the second we assume x ∈ B. In the first
case we therefore have the following givens and goal:
Givens Goal
A ⊆ C x ∈ C
B ⊆ C
x ∈ A

We’ve already decided that if we ever come across an element of A,wecan
use the first given to conclude that it is also an element of C. Since we now
have x ∈ A as a given, we can conclude that x ∈ C, which is our goal. The