Page 135 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Proofs Involving Quantifiers 121
you continue to read the proof. At this point in the proof, the list would look
like this:
Givens Goal
a, b, and c are integers a | c
a | b
b | c

Because the new goal means ∃k ∈ Z(ka = c), the proof will probably pro-
ceed by finding an integer k such that ka = c. As with many proofs of existential
statements, the first step in finding such a k involves looking more closely at the
givens. The next sentence of the proof uses the given a | b to conclude that we
can choose an integer m such that ma = b. The proof doesn’t say what rule of
inference justifies this. It is up to you to figure it out by working out the logical
form of the given statement a | b, using the definition of divides. Because this
given means ∃k ∈ Z(ka = b), you should recognize that the rule of inference
being used is existential instantiation. Existential instantiation is also used in
the next sentence of the proof to justify choosing an integer n such that nb = c.
The equations ma = b and nb = c can now be added to the list of givens.
Some steps have also been skipped in the last sentence of the proof. We
expected that the goal a | c would be proven by finding an integer k such that
ka = c. From the equation c = nma and the fact that nm is an integer, it follows
that k = nm will work, but the proof doesn’t explicitly say that this value of
k is being used; in fact, the variable k is not mentioned at all in the proof. Of
course, the variable k is not mentioned in the statement of the theorem either.
It is not uncommon for a proof of an existential statement to be written in this
way, especially when, as in this case, the goal is not written out explicitly in the
statement of the theorem as an existential statement. In this case, the existential
nature of the goal became apparent only when we filled in the definition of
divides.

Exercises

Note: Exercises marked with the symbol d can be done with Proof Designer.
p
For more information about Proof Designer, see Appendix 2.
1. In exercise 7 of Section 2.2 you used logical equivalences to show that
∗
∃x(P(x) → Q(x)) is equivalent to ∀xP(x) →∃xQ(x). Now use the
methods of this section to prove that if ∃x(P(x) → Q(x)) is true, then
∀xP(x) →∃xQ(x) is true. (Note: The other direction of the equivalence
is quite a bit harder to prove. See exercise 29 of Section 3.5.)