Page 191 - HOW TO PROVE IT: A Structured Approach, Second Edition

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Relations 177
Proof. We will prove 1, 2, and half of 4, and leave the rest as exercises. (See
exercise 6.)

−1 −1
1. First of all, note that R −1 is a relation from B to A,so(R ) is a relation
−1 −1
from A to B, just like R. To see that (R ) = R, let (a, b) be an arbitrary
ordered pair in A × B. Then
−1 −1
(a, b) ∈ (R ) iff (b, a) ∈ R −1 iff (a, b) ∈ R.
−1
2. First note that Dom(R ) and Ran(R) are both subsets of B. Now let b be
an arbitrary element of B. Then
−1
−1
b ∈ Dom(R )iff ∃a ∈ A((b, a) ∈ R )
iff ∃a ∈ A((a, b) ∈ R)iff b ∈ Ran(R).
4. Clearly T ◦ (S ◦ R) and (T ◦ S) ◦ R are both relations from A to D. Let
(a, d) be an arbitrary element of A × D.
First, suppose (a, d) ∈ T ◦ (S ◦ R). By the definition of composition, this
means that we can choose some c ∈ C such that (a, c) ∈ S ◦ R and (c, d) ∈
T . Since (a, c) ∈ S ◦ R, we can again use the definition of composition
and choose some b ∈ B such that (a, b) ∈ R and (b, c) ∈ S. Now since
(b, c) ∈ S and (c, d) ∈ T , we can conclude that (b, d) ∈ T ◦ S. Similarly,
since (a, b) ∈ R and (b, d) ∈ T ◦ S, it follows that (a, d) ∈ (T ◦ S) ◦ R.
Now suppose (a, d) ∈ (T ◦ S) ◦ R. A similar argument, which is
left to the reader, shows that (a, d) ∈ T ◦ (S ◦ R). Thus, T ◦ (S ◦ R) =
(T ◦ S) ◦ R.

−1 −1
Commentary. Statement 1 means ∀p(p ∈ (R ) ↔ p ∈ R), so the proof
should proceed by introducing an arbitrary object p and then proving p ∈
−1 −1
−1 −1
(R ) ↔ p ∈ R. But because R and (R ) are both relations from A to
B, we could think of the universe over which p ranges as being A × B,so
p must be an ordered pair. Thus, in the preceding proof we’ve written it as
an ordered pair (a, b) from the start. The proof of the biconditional statement
−1 −1
(a, b) ∈ (R ) ↔ (a, b) ∈ R uses the method, introduced in Example 3.4.4,
of stringing together a sequence of equivalences.
The proofs of statements 2 and 4 are similar, except that the biconditional
proof for statement 4 cannot easily be done by stringing together equivalences,
so we prove the two directions separately. Only one direction was proven.
The key to this proof is to recognize that the given (a, d) ∈ T ◦ (S ◦ R)isan
existential statement, since it means ∃c ∈ C((a, c) ∈ S ◦ R and (c, d) ∈ T ), so
we should introduce a new variable c into the proof to stand for some element
of C such that (a, c) ∈ S ◦ R and (c, d) ∈ T . Similarly, (a, c) ∈ S ◦ R is an