Page 209 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Ordering Relations 195
yRx), so to use it we should plug in something for x and y. The only likely
candidatesforwhattopluginarebandourarbitraryobjectx,andpluggingthese
in we get xRb ∨ bRx. Our goal is bRx, so this certainly looks like progress.
If only we could rule out the possibility that xRb, we’d be done. So let’s see if
we can prove ¬xRb.
Because this is a negative statement, we try proof by contradiction. Suppose
xRb. What given statement can we contradict? The only given we haven’t used
yet is the fact that b is minimal, and since this is a negative statement, it is the
natural place to look for a contradiction. To contradict the fact that b is minimal,
we should try to show that ∃x ∈ B(xRb ∧ x = b). But we’ve already assumed
xRb, so if we could show x = b we’d be done.
You should try proving x = b at this point. You won’t get anywhere. The
fact is, we started out by letting x be an arbitrary element of B, and this means
that it could be any element of B, including b. We then assumed that xRb,but
since R is reflexive, this still doesn’t rule out the possibility that x = b. There
really isn’t any hope of proving x = b. We seem to be stuck.
Let’s review our overall plan for the proof. We needed to show ∀x ∈ B(bRx),
so we let x be an arbitrary element of B, and we’re trying to show bRx.We’ve
now run into problems because of the possibility that x = b. But if our ultimate
goal is to prove bRx, then the possibility that x = b really isn’t a problem after
all. Since R is reflexive, if x = b then of course bRx will be true!
Now, how should we structure the final write-up of the proof? It appears that
our reasoning to establish bRx will have to be different depending on whether
or not x = b. This suggests proof by cases. In case 1 we assume that x = b,
and use the fact that R is reflexive to complete the proof. In case 2 we assume
that x = b, and then we can use our original line of attack, starting with the
fact that R is total.
Proof.

1. Suppose b is a smallest element of B, and suppose c is also a smallest
element of B. Since b is a smallest element, ∀x ∈ B(bRx), so in particular
bRc. Similarly, since c is a smallest element, cRb. But now since R is
a partial order, it must be antisymmetric, so from bRc and cRb we can
conclude b = c.
2. Let x be an arbitrary element of B and suppose that xRb. Since b is the
smallest element of B, we must have bRx, and now by antisymmetry it
follows that x = b. Thus, b must be a minimal element.
To see that it is the only one, suppose c is also a minimal element. Since
b is the smallest element of B, bRc. But then since c is minimal, we must
have b = c. Thus b is the only minimal element of B.