Chapter 6 - Set Theory - Exercise Set 6.2 - Page 365: 10

Proof: Suppose A, B, and C are any sets and (A – B) ⋂ (C – B). We must show that (A – B) ⋂ (C – B) ⊆(A⋂C) – B, and that (A⋂C) – B ⊆ (A – B) ⋂ (C – B). Suppose x is any element in (A – B) ⋂ (C – B). By definition of intersection, x ϵ (A – B) and x ϵ (C – B). Case 1 (x ϵ A – B): By definition of set difference, x ϵ A and x  B. Since x ϵ (A – B) and x ϵ (C – B) and x  B, then x ϵ A and x ϵ C. By definition of intersection, x ϵ A and x ϵ C implies that x ϵ A⋂C. Therefore, x ϵ (A⋂C) and x  B. By definition of set difference, x ϵ (A ⋂ C) – B. Case 2 (x ϵ C – B): By definition of set difference, x ϵ C and x  B. Since x ϵ (A – B) and x ϵ (C – B) and x  B, then x ϵ A and x ϵ C. By definition of intersection, x ϵ A and x ϵ C implies that x ϵ A⋂C. Therefore, x ϵ (A⋂C) and x  B. By definition of set difference, x ϵ (A⋂C) – B. In cases 1 and 2, x ϵ (A⋂C) – B. Therefore, (A − B) ∩ (C − B) = (A ∩ C) – B. Suppose x is any element in (A⋂C) – B. By definition of set difference, x ϵ A⋂C and x  B. Case 3 (x ϵ A⋂C): By definition of intersection, x ϵ A and x ϵ C. Since x ϵ A and x ϵ C and x  B, then x ϵ A and x  B, and x ϵ C and x  B. By definition of set difference, x ϵ A – B and x ϵ C – B. By definition of intersection, x ϵ A – B and x ϵ C – B implies that x ϵ (A – B) ⋂ (C – B). Case 4 (x  B): Since x ϵ A⋂C and x  B, then x ϵ A and x  B, and x ϵ C and x  B. By definition of set difference, x ϵ A – B and x ϵ C – B. By definition of intersection, x ϵ A – B and x ϵ C – B implies that x ϵ (A – B) ⋂ (C – B). In cases 3 and 4, x ϵ (A – B) ⋂ (C – B). Therefore, (A⋂C) – B ⊆ (A – B) ⋂ (C – B). Since (A – B) ⋂ (C – B) ⊆(A⋂C) – B and (A⋂C) – B ⊆ (A – B) ⋂ (C – B), then (A − B) ∩ (C − B) = (A ∩ C) – B by definition of set equality.

Proof: Suppose A, B, and C are any sets and (A – B) ⋂ (C – B). We must show that (A – B) ⋂ (C – B) ⊆(A⋂C) – B, and that (A⋂C) – B ⊆ (A – B) ⋂ (C – B). Suppose x is any element in (A – B) ⋂ (C – B). By definition of intersection, x ϵ (A – B) and x ϵ (C – B). Case 1 (x ϵ A – B): By definition of set difference, x ϵ A and x  B. Since x ϵ (A – B) and x ϵ (C – B) and x  B, then x ϵ A and x ϵ C. By definition of intersection, x ϵ A and x ϵ C implies that x ϵ A⋂C. Therefore, x ϵ (A⋂C) and x  B. By definition of set difference, x ϵ (A ⋂ C) – B. Case 2 (x ϵ C – B): By definition of set difference, x ϵ C and x  B. Since x ϵ (A – B) and x ϵ (C – B) and x  B, then x ϵ A and x ϵ C. By definition of intersection, x ϵ A and x ϵ C implies that x ϵ A⋂C. Therefore, x ϵ (A⋂C) and x  B. By definition of set difference, x ϵ (A⋂C) – B. In cases 1 and 2, x ϵ (A⋂C) – B. Therefore, (A − B) ∩ (C − B) = (A ∩ C) – B. Suppose x is any element in (A⋂C) – B. By definition of set difference, x ϵ A⋂C and x  B. Case 3 (x ϵ A⋂C): By definition of intersection, x ϵ A and x ϵ C. Since x ϵ A and x ϵ C and x  B, then x ϵ A and x  B, and x ϵ C and x  B. By definition of set difference, x ϵ A – B and x ϵ C – B. By definition of intersection, x ϵ A – B and x ϵ C – B implies that x ϵ (A – B) ⋂ (C – B). Case 4 (x  B): Since x ϵ A⋂C and x  B, then x ϵ A and x  B, and x ϵ C and x  B. By definition of set difference, x ϵ A – B and x ϵ C – B. By definition of intersection, x ϵ A – B and x ϵ C – B implies that x ϵ (A – B) ⋂ (C – B). In cases 3 and 4, x ϵ (A – B) ⋂ (C – B). Therefore, (A⋂C) – B ⊆ (A – B) ⋂ (C – B). Since (A – B) ⋂ (C – B) ⊆(A⋂C) – B and (A⋂C) – B ⊆ (A – B) ⋂ (C – B), then (A − B) ∩ (C − B) = (A ∩ C) – B by definition of set equality.