Page 123 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Quantifiers 109
Example 3.3.1. Suppose A, B, and C are sets, and A \ B ⊆ C. Prove that
A \ C ⊆ B.

Scratch work
Givens Goal
A \ B ⊆ C A \ C ⊆ B
As usual, we look first at the logical form of the goal to plan our strategy. In
this case we must write out the definition of ⊆ to determine the logical form
of the goal.
Givens Goal
A \ B ⊆ C ∀x(x ∈ A \ C → x ∈ B)
Because the goal has the form ∀xP(x), where P(x) is the statement x ∈
A \ C → x ∈ B, we will introduce a new variable x into the proof to stand for
an arbitrary object and then try to prove x ∈ A \ C → x ∈ B. Note that xis
a new variable in the proof. It appeared in the logical form of the goal as a
bound variable, but remember that bound variables don’t stand for anything in
particular. We have not yet used x as a free variable in any statement, so it has
not been used to stand for any particular object. To make sure x is arbitrary
we must be careful not to add any assumptions about x to the givens column.
However, we do change our goal:
Givens Goal
A \ B ⊆ C x ∈ A \ C → x ∈ B
According to our strategy, the final proof should look like this:

Let x be arbitrary.
[Proof of x ∈ A \ C → x ∈ B goes here.]
Since x was arbitrary, we can conclude that ∀x(x ∈ A \ C → x ∈ B),
so A \ C ⊆ B.

The problem is now exactly the same as in Example 3.2.3, so the rest of the
solution is the same as well. In other words, we can simply insert the proof we
wrote in Example 3.2.3 between the first and last sentences of the proof written
here.

Solution

Theorem. Suppose A, B, and C are sets, and A \ B ⊆ C. Then A \ C ⊆ B.
Proof. Let x be arbitrary. Suppose x ∈ A \ C. This means that x ∈ A and
x /∈ C. Suppose x /∈ B. Then x ∈ A \ B, so since A \ B ⊆ C, x ∈ C. But