Page 208 - HOW TO PROVE IT: A Structured Approach, Second Edition

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194 Relations
much in deciding how to prove that b = c. The only other fact we know about
b and c is that they are both smallest elements of B, which means ∀x ∈ B(bRx)
and ∀x ∈ B(cRx). The most promising way to use these statements is to plug
something in for x in each statement. What we plug in should be an element of
B, and we only know of two elements of B at this point, b and c. Plugging in
both of them in both statements, we get bRb, bRc, cRb, and cRc. Of course,
we already knew bRb and cRc, since R is reflexive. But when you see that
bRc and cRb, you should think of antisymmetry. Since R is a partial order, it
is antisymmetric, so from bRc and cRb it follows that b = c.
2. Our first goal is to prove that b is a minimal element of B, which means
¬∃x ∈ B(xRb ∧ x = b). Because this is a negative statement, it might help to
reexpress it as an equivalent positive statement:
¬∃x ∈ B(xRb ∧ x = b)iff ∀x ∈ B¬(xRb ∧ x = b)
iff ∀x ∈ B(¬xRb ∨ x = b)
iff ∀x ∈ B(xRb → x = b).

Thus, to prove that b is minimal we could let x be an arbitrary element of B,
assume that xRb, and prove x = b.
Once again, it’s a good idea to take stock of what we know at this point about
b and x. We know xRb, and we know that b is the smallest element of B, which
means ∀x ∈ B(bRx). If we apply this last fact to our arbitrary x, then as in part
1 we can use antisymmetry to complete the proof.
We still must prove that b is the only minimal element, and as in part 1
this means ∀c(c is a minimal element of B → b = c). So we let c be arbitrary
and assume that c is a minimal element of B, and we must prove that b =
c. The assumption that c is a minimal element of B means that c ∈ B and
¬∃x ∈ B(xRc ∧ x = c), but as before, we can reexpress this last statement in
the equivalent positive form ∀x ∈ B(xRc → x = c). To use this statement we
should plug in something for x, and because our goal is to show that b = c,
plugging in b for x seems like a good idea. This gives us bRc → b = c,soif
only we could show bRc, we could complete the proof by using modus ponens
to conclude that b = c. But we know b is the smallest element of B, so of course
bRc is true.
3. Of course, we start by assuming that R is a total order and b is a minimal
element of B. We must prove that b is the smallest element of B, which means
∀x ∈ B(bRx), so we let x be an arbitrary element of B and try to prove bRx.
We know from examples we’ve looked at that minimal elements in partial
orders are not always smallest elements, so the assumption that R is a total order
must be crucial. The assumption that R is total means ∀x ∈ A∀y ∈ A(xRy ∨