Answer:
See proof below
Explanation:
Remember that if we want to prove that two propositions P and Q are equivalent, we have to prove that P implies Q and Q implies P.
(1)→(2) Suppose that A \ B = B \ A, we will prove that A = B. Let x∈A, and for the sake of a contradiction suppose that x∉B. Then x∈A \ B by definition of difference of sets. Since A \ B = B \ A and x∈A \ B then x∈B \A, that is, x∈B and x∉A, which is a contradiction to the first assumption regarding x. Therefore x∈B, for all x∈A, which implies that A⊆B.
We can use a similar argument to prove that B⊆A, and both inclusions imply that A=B as we wanted to prove. Indeed, let x∈B, and suppose that x∉A Thus x∈B \ A. But A \ B = B \ A, then x∈A \B, that is, x∈A and x∉B, which is absurd. Then x∈A, for all x∈B, that is, B⊆A.
(2)→(1). Suppose that A = B. We will prove that A \ B = B \ A. By definition, A \ B is the set of elements of A that do not belong to B. However, A=B, so A \ B is the set of elements of A that do not belon to A. There no exists such an element, since this condition is contradictory, thus A \ B=∅, the empty set. Similarly, B \ A = B \ B = ∅. The empty set is unique, therefore A \ B=∅= B \ A.
The previous parts show that (1) and (2) are logically equivalent.