Answer
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.
Work Step by Step
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.