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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).

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$$(A-B)-(C-D)=(A\cap B^{\complement})\cap(C\cap D^{\complement})^{\complement}=(A\cap B^{\complement})\cap (C^{\complement}\cup D)=$$$$(A\cap B^{\complement}\cap C^{\complement})\cup(A\cap B^{\complement}\cap D)$$

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 $A\cap B^{\complement}\cap D=A\cap C^{\complement}\cap D$

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Let, AC=Ø ∧ BD≠Ø ∧ AB=Ø ∧ CB-D. Now, A-CA-(B-D). Therefore, the equality is not true. In essence, the 4 relations are your counterexample. Now create sets A,B,C, and D such that the 4 relations hold and you will disprove the equality.

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