I was given the following problem:
For sets A, B, C, and D. Prove or disprove that $(A-B)-(C-D) = (A-C)-(B-D)$.
My proof by counterexample was:
Let $A=\{1,2\}, B=\{2,3\}, C=\{3,5\}, D=\{2,4\}$. Then, $(A-B)-(C-D) = \{1\}$, but $(A-C)-(B-D) = \{1,2\}$. Thus, the statement is false.
But if I try to use the definition of set difference $A-B = \{x \in A \wedge x \notin B\}$. Something goes wrong.
$(A-B)-(C-D)=\{(x \in A \wedge x \notin B) \wedge \neg(x \in C \wedge x\notin D)\} = \{(x \in A \wedge x \notin B) \wedge (x \notin C \vee x\in D)\} = \{x \in A \wedge x \notin B \wedge x \notin C \vee x\in D\} = \{1,2,4\}$.
Best Answer
One way also to prove or disprove the statement $(A-B)-(C-D)=(A-C)-(B-D)$ for sets A, B, C and D is showing that $(A-B)-(C-D)\subset(A-C)-(B-D)$ and $(A-C)-(B-D)\subset(A-B)-(C-D)$. If you can show that they are subsets of each other, then they are equal. But if one is not a subset of the other, then we can say that they are not equal.