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346 Appendix 1: Solutions to Selected Exercises
(c) Reflexive closure and symmetric closure are both D r . Transitive
closure is R × R.
3. (a) Suppose that R is an asymmetric relation on A. Then the statement
∀x ∈ A∀y ∈ A((xRy ∧ yRx) → x = y) is vacuously true, because
xRy ∧ yRx is always false.
(b) Suppose that R is a strict partial order, and suppose that for some
x, y ∈ A, (x, y) ∈ R and (y, x) ∈ R. Then by transitivity of
R, (x, x) ∈ R, which contradicts the fact that R is irreflexive. There-
fore, R is asymmetric.
5. (a) Hint: Let F ={T ⊆ A × A | T ⊆ R and T is irreflexive }. Then you
must prove that S ∈ F and ∀T ∈ F(T ⊆ S). For the first of these, you
must prove that S ⊆ R and S is irreflexive. Both of these follow easily
from the definition of S. For the second, let T ∈ F be arbitrary and
prove T ⊆ S. Since T ∈ F, you know that T ⊆ R and T is irreflexive.
Let (x, y) be an arbitrary element of T, and use these facts about T,
together with the definition of S,toprove (x, y) ∈ S.
(b) Suppose R is a partial order on A. We already showed in part (a) that
S is irreflexive. To show that it is transitive, suppose (x, y) ∈ S and
(y, z) ∈ S. Then by the definition of S, (x, y) ∈ R and (y, z) ∈ R,so
since R is transitive, (x, z) ∈ R.If x = z then we have (x, y) ∈ R and
(y, x) ∈ R, so by the antisymmetry of R, x = y. But then (x, y) ∈ i A ,
which contradicts the fact that (x, y) ∈ S = R \ i A . Therefore x = z,
so (x, z) /∈ i A and hence (x, z) ∈ S.
7. (a) Let S be the reflexive closure of R.
(→) Suppose R is reflexive. By clause 1 in the definition of reflex-
ive closure (Definition 4.5.1), R ⊆ S, and by clause 3 (with T = R),
S ⊆ R. Therefore R = S.
(←) Suppose R = S. By clause 2 in the definition of reflexive clo-
sure, R is reflexive.
(b) Yes; the proofs are very similar.
9. Hint: Let T ={(x, y) ∈ S | x ∈ Dom(R) and y ∈ Ran(R)}. Prove that
R ⊆ T and T is transitive.
12. (a) S 1 ∪ S 2 = (R 1 ∪ i A ) ∪ (R 2 ∪ i A ) = (R 1 ∪ R 2 ) ∪ i A = R ∪ i A = S.
(b) It is possible to give a proof that is similar to the proof in part (a),
using formulas for S 1 , S 2 , and S. However, we will take a different
approach. First, note that R 1 ⊆ R and R 2 ⊆ R. It follows, by exercise
11, that S 1 ⊆ S and S 2 ⊆ S,so S 1 ∪ S 2 ⊆ S. For the other direction,
notethat R = R 1 ∪ R 2 ⊆ S 1 ∪ S 2 ,andbyexercise13(b)ofSection4.3,
S 1 ∪ S 2 is symmetric. Therefore, by definition of symmetric closure,
S ⊆ S 1 ∪ S 2 .