Proving $A - B \subset A - (B - A)$ I believe I have been able to prove that for sets $A$ and $B$, $A - (B - A) \supset A - B$, but my proof is not particularly elegant. I was hoping someone knew of a more clever or straightforward way to show this. My proof is: 

Let $x \in A - B$. Then $x \in A$ and $x \not \in B$. So $x \not \in \{y \mid y \in B, \; y \not \in A    \}$, so $x \not \in B - A$. Since $x \in A$ and $x \not \in B-A$, $x \in A - (B - A)$, so $A - B \subset A - (B - A)$. 


Let $x$ be an arbitrary element. In the table below, let $0$ denote that $x$ is not in the  set while $1$ denote that $x$ is in the set.
$$
\begin{matrix}
A \ & B \ & A-B \ & B-A \ & A-(B-A) \\
0 \ & 0 \ & 0   \ & 0   \ & 0       \\
0 \ & 1 \ & 0   \ & 1   \ & 0       \\
1 \ & 0 \ & 1   \ & 0   \ & 1       \\
1 \ & 1 \ & 0   \ & 0   \ & 1
\end{matrix}
$$
Note that corresponding to every $1$ (in fact the only one $1$) in the column headed $A-B$, we have a $1$ in the column headed $A-(B-A)$, and there is a $1$ in the column headed $A-(B-A)$ for which the corresponding entry in the column headed $A-B$ is a $0$.
This shows that, for any sets $A$ and $B$, we always have
$$
A - B \subset A-(B-A),
$$
but there are examples when
$$
A - (B-A) \not\subset A-B.
$$
For example, let us consider the sets 
$$
A = \{ p \} = B.
$$
Then $A-B = \emptyset = B-A$, but 
$$ A-(B-A) = A - \emptyset = A \neq \emptyset. $$
In the table above, the discrepency between the columns headed $A-B$ and $A-(B-A)$ occurs in the very last line where our arbitrary element is in both the sets $A$ and $B$. And, this precisely is the hint we have used in constructing our counter-example above.
Hope this helps.
