Page 192 - HOW TO PROVE IT: A Structured Approach, Second Edition

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178 Relations
existential statement, so it suggests introducing the variable b. Once these new
variables have been introduced, it is easy to prove the goal (a, d) ∈ (T ◦ S) ◦ R.

Statement 5 of Theorem 4.2.5 perhaps deserves some comment. First of all,
−1
−1
notice that the right-hand side of the equation is R −1 ◦ S , not S −1 ◦ R ;
the order of the relations has been reversed. You are asked to prove statement 5
in exercise 6, but it might be worthwhile to try an example first. We’ve already
seen that, for the relations E and T from parts 5 and 6 of Example 4.2.2,

T ◦ E ={(s, p) ∈ S × P | the student s is enrolled in some course
taught by the professor p}.
It follows that

(T ◦ E) −1 ={(p, s) ∈ P × S | the student s is enrolled in some course
taught by the professor p}.
To compute E −1 ◦ T −1 , first note that T −1 is a relation from P to C and E −1 is
a relation from C to S,so E −1 ◦ T −1 is a relation from P to S. Now, applying
the definition of composition, we get
E −1 ◦ T −1 ={(p, s) ∈ P × S |∃c ∈ C((p, c) ∈ T −1 and (c, s) ∈ E −1 )}
={(p, s) ∈ P × S |∃c ∈ C((c, p) ∈ T and (s, c) ∈ E)}
={(p, s) ∈ P × S |∃c ∈ C(the course c is taught by the
professor p and the student s is enrolled in the course c)}
={(p, s) ∈ P × S | the student s is enrolled in some course
taught by the professor p}.

Thus, (T ◦ E) −1 = E −1 ◦ T −1 .

Exercises

1. Find the domains and ranges of the following relations.
∗
(a) {(p, q) ∈ P × P | the person p is a parent of the person q}, where
P is the set of all living people.
2
2
(b) {(x, y) ∈ R | y > x }.
2. Find the domains and ranges of the following relations.
(a) {(p, q) ∈ P × P | the person p is a brother of the person q},
where P is the set of all living people.
2
2
2
(b) {(x, y) ∈ R | y = 1 − 2/(x + 1)}.