Prove or disprove: $(A-B)-C=(A-C)- B$ for sets $A$, $B$, $C$ 
Prove or disprove: For sets $A$, $B$, $C$,
$$(A-B)-C=(A-C)- B$$

Now my attempt is
$$(x\in A \;\text{and}\; x\notin B) \;\text{and}\; x\notin C \tag1$$
We can write it as
$$(x\in A  \;\text{and}\; x\notin C)  \;\text{and}\; x\notin B \tag2$$
which proves the statement.
Is it right?
 A: A valid Venn diagram-based proof:


This can equivalently be expanded to a proof of $(A\land\lnot B)\land\lnot C=(A\land\lnot C)\land\lnot B$ using a truth table.
A: Yes, that is the core of a valid proof.  All that is left is to tidy up the presentation.

The set of $(A\smallsetminus B)\smallsetminus C$ is the set of elements that are in $A$ but not in $B$ and not in $C$, which is one and the same as the set $(A\smallsetminus C)\smallsetminus B$.

$$\begin{align}(A\smallsetminus B)\smallsetminus C&=\{x:x\in A\smallsetminus B~\land~x\notin C\}\\&=\{x:(x\in A~\land~x\notin B)~\land~ x\notin C\}\\&=\{x:(x\in A~\land~x\notin C)~\land~ x\notin B\}\\&=\{x:x\in A\smallsetminus  C~\land~x\notin B\}\\&=(A\smallsetminus C)\smallsetminus B \end{align}$$

Or... whatever format you prefer.
A: Let $X$ be a set and $A,B,C \subset X.$
$A$ \ $B=A \cap B^c $(where $B^c$ is the complement of $B$ in $X$).
Then
LHS:
$(A \cap B^c)\cap C^c;$
RHS:
$(A \cap C^c) \cap B^c.$
Using associativity and commutative we are done.
