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


If $C \subset D$, then we have $A - D \subset A-C$.
We have $B - A \subset B$, hence $A- B \subset A - (B-A)$.
