# If A, B, C, and D are sets, can I proof that (A-B)-(C-D)=(A-C)-(B-D) with membership table?

I tried to figure the problem out with membership table that, A-B=A $$\cap$$ ~B, but it wasn't proved that (A-B)-(C-D)=(A-C)-(B-D).

$$(A-B)-(C-D)=(A\cap B^{\complement})\cap(C\cap D^{\complement})^{\complement}=(A\cap B^{\complement})\cap (C^{\complement}\cup D)=$$$$(A\cap B^{\complement}\cap C^{\complement})\cup(A\cap B^{\complement}\cap D)$$
Interchanging $$B$$ and $$C$$ in this result tells us that: $$(A-C)-(B-D)=(A\cap C^{\complement}\cap B^{\complement})\cup(A\cap C^{\complement}\cap D)$$
This shows that the sets are not equal, unless $$A\cap B^{\complement}\cap D=A\cap C^{\complement}\cap D$$