Page 125 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Quantifiers 111
In a proof in which several of these strategies have been combined, there might
be several of these summing up sentences, one after another, at the end of the
proof. Mathematicians often condense this summing up into one sentence, or
even skip it entirely. When you are reading a proof written by someone else,
you may find it helpful to fill in these skipped steps.

Example 3.3.2. Suppose A and B are sets. Prove that if A ∩ B = A then
A ⊆ B.

Scratch work
Our goal is A ∩ B = A → A ⊆ B. Because the goal is a conditional statement,
we add the antecedent to the givens list and make the consequent the goal. We
will also write out the definition of ⊆ in the new goal to show what its logical
form is.
Givens Goal
A ∩ B = A ∀x(x ∈ A → x ∈ B)

Now the goal has the form ∀x(P(x) → Q(x)), where P(x) is the statement
x ∈ A and Q(x) is the statement x ∈ B. We therefore let x be arbitrary, assume
x ∈ A, and prove x ∈ B:
Givens Goal
A ∩ B = A x ∈ B
x ∈ A
Combining the proof strategies we have used, we see that the final proof will
have this form:
Suppose A ∩ B = A.
Let x be arbitrary.
Suppose x ∈ A.
[Proof of x ∈ B goes here.]
Therefore x ∈ A → x ∈ B.
Since x was arbitrary, we can conclude that ∀x(x ∈ A → x ∈ B), so
A ⊆ B.
Therefore, if A ∩ B = A then A ⊆ B.
As discussed earlier, when we write up the final proof we can skip the sentence
“Let x be arbitrary,” and we can also skip some or all of the last three sentences.
We have now reached the point at which we can analyze the logical form of
the goal no further. Fortunately, when we look at the givens, we discover that
the goal follows easily. Since x ∈ A and A ∩ B = A, it follows that x ∈ A ∩ B,