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 $A - B \subset A - (B - A)$. 

 A: If $C \subset D$, then we have $A - D \subset A-C$.
We have $B - A \subset B$, hence $A- B \subset A - (B-A)$.
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.
A: $\begin{align}A-(B-A)&=A\cap \overline{(B-A)}=A\cap(\overline{B\cap\overline{A}})=A\cap(A\cup \overline{B})=(A\cap A)\cup(A\cap \overline{B})\\&=A\cup(A-B)\\&=A\end{align}$
Thus it is clear from second line that $(A-B)$ is a subset of this set.
Note that in the end, the conclusion is no more than $A-B\subset A$ since the above set is just $A$.
A: That's straightforward element chasing and in my opinion the way you did it IS  the best proof.
We can go to "concepts" but this assume the reader has a strong intuition which... I'm learning not all do ... and what good is a proof if it's hard to follow:
For any sets $W,K, M$ where $K\subset M$ then $W-M \subset W-K$. Is that clear?
If not.  $W-M = $ all elements of $W$ minus set of all elements of $M$ which "clearly" a subset of all elements of $W$ minus only some of the elements of $M$ which is what $W-K$ is as $K\subset M$ so $K$ is only some of the elements of $K$.
(Of course, the emotional appeal to "clear" obviousness is precisely what we invented cold hard mathematics to avoid....)
So as $B-A\subset B$ then $A- B \subset A- (B-A)$.
=======
Actually let's make that better:
Lemma:  If $M \subset K$ then $K^c = \{x\not \in K$ where it is understood there is some universal collection of all valid elements$\} \subset M^c =\{x\not \in M\}$.
Pf: If $x \not \in K$ then $x \not \in M$ as $M \subset K$.  So $K^c \subset M^c$.
Lemma 2: If $M \subset K$ then $W- K\subset W-M$.
Pf: If $x \in W-K$ then $x\in W$ and $x \not \in K$.  But then $x \not \in M$ because $M \subset K$. And $x\in W$.  So $x\in W-M$ so $W-K \subset W-K$.
Lemma 3 (just to be thourough):  $B-A \subset B$.
Pf:  Well, doi!  if $x\in B-A$ then $x \in B$ and $x\not \in A$.  SO $x \in B$.
and that's it.
$B-A \subset B$ so
$A-B \subset A-(B-A)$.
....
Hmmm, I changed my mind.  That is more elegant and more basic.... albeit it much longer....
That's my proof not and I'm sticking with it....until I change my mind again.
