Let A, B, and C be sets. Prove that (A-B) - C = (A-C) - (B-C) I am utilizing set identities to prove (A-C)-(B-C).
$\begin{array}{|l}(A−B)− C = \{ x | x \in ((x\in (A \cap \bar{B})) \cap \bar{C}\} \quad \text{Def. of Set Minus}
\\ 
\quad \quad \quad \quad \quad =\{ x | ((x\in A) \wedge (x\in\bar{B})) \wedge (x\in\bar{C})\} \quad \text{Def. of intersection}
\\  \quad \quad \quad \quad \quad =\{ x | (A\wedge\overline{C}\wedge\overline{B})\vee(\overline{C}\wedge\overline{B}\wedge C)\} \quad \text{Association Law}
\\
\quad \quad \quad \quad \quad =\{ x | ((x\in A) \wedge (x\in\bar{C})) \wedge ((x\in \bar{B}) \wedge (x\in\bar{C}))\} \quad \text{Idempotent Law}
\\ 
\quad \quad \quad \quad \quad =\{ x | (((x\in (A\cap\bar{C})) \cap (x\in (\bar{B} \cap\bar{C})))\} \quad \text{Def. of union}
\\ 
\quad \quad \quad \quad \quad =\{ x | (((x\in (A\cap \bar{C})) \cap \overline{(x\in (B\cup C)))} \} \quad \text{DeMorgan's Law}
\\ 
\quad \quad \quad \quad \quad =\{ x | x \in (A - C) - (B \cup C) \} \quad \text{Def. Set Minus}
\\
=(A-C)-(B-C)
\end{array}$
So it looks like I screwed up on the final step. Is there something that I am forgetting to do properly or where am I supposed to go from that final step? 
 A: \begin{align*}
(A-C)-(B-C) & = (A\cap\overline{C})-(B\cap\overline{C}) = (A\cap\overline{C})\cap(\overline{B\cap\overline{C})}\\\\
& = (A\cap\overline{C})\cap(\overline{B}\cup C) = (A\cap\overline{C}\cap\overline{B})\cup(\overline{C}\cap\overline{B}\cap C)\\\\
& = A\cap\overline{C}\cap\overline{B} = (A\cap\overline{B})\cap\overline{C} = (A-B)-C
\end{align*}
A: You might also just try to prove this in words.  Suppose $x \in (A-B)-C$.


*

*Then $x \in (A-B)$ and $x \notin C$.

*Since $x \in (A-B)$, we know $x \in A$ and $x \notin B$.

*Since $x \in A$ and $x \notin C$, we know $x \in (A-C)$.

*Since $x \notin B$, we know $x \notin (B-C)$.

*Since $x \in (A-C)$ and $x \notin (B-C)$, we know $x \in ((A-C)-(B-C))$.


Since this is true for all $x \in (A-B)-C$, we know 
$$
  (A-B)-C \subseteq ((A-C)-(B-C))
$$
Now do it the other way around.
A: Abbreviating and as $\land$ and not as $\lnot$,
$$x\in(A-B)-C\iff x\in A-B\land x\notin C \iff x\in A\land x\notin B\land x\notin C\\\iff x\in A\land x\notin C\land\lnot(x\in B\land x\notin C)\iff x\in A-C\land x\notin B-C\\\iff x\in (A-C)-(B-C).$$The third $\iff$ uses $$p\land\lnot q\land\lnot r\iff(p\land\lnot r)\land\lnot (q\land\lnot r)$$(you can verify this is a tautology), while the rest use the definition of $-$.
A: All of the above looks like hard math requiring actual thought.   I did it by means of an excel spreadsheet.  Easy, as there are only 8 possibilities.  The two highlighted columns are identical.

