Proof: Show (A-C) U (B-C) ⊆ (A U B) - C and

(A U B) - C ⊆ (A-C) U (B-C).

Let A, B, and C be arbitrary sets.

First, assume (A-C) U (B-C).

Let x ∈ (A-C) U (B-C).

Without loss of generality, suppose x ∈ (A-C).

Since x ∈ (A-C), we have x ∈ A and x ∉ C.

So, x ∈ A and by a theorem proved in class A ⊆ (A U B).

Thus, x ∈ (A U B) and x ∉ C means x ∈ (A U B) - C.

Therefore, (A-C) U (B-C) ⊆ (A U B) - C.

Proof (A U B) - C ⊆ (A-C) U (B-C).

For the reverse inclusion, let x ∈ (A U B) - C.

Then x ∈ (A U B) and x ∉ C. Without loss of generality, suppose x ∈ A.

Since x ∈ A and x ∉ C then, x ∈ (A - C).

It follows that, x ∈(A-C) U (B-C).

Hence, (A U B) - C ⊆ (A-C) U (B-C).

Therefore, by definition of set equality, (A-C) U (B-C) = (A U B) - C.