Need help to prove $(A\cup B) \setminus (C \setminus A) = A \cup (B \setminus C)$ Having trouble with a discrete math question involving sets. Have been asked to prove:
$$(A\cup B) \setminus (C  \setminus A) = A \cup (B \setminus C)$$
This is what I have so far:
$x \in A$ or  $x\in (B \setminus C)$
$x \in A$ or $x\in B$ and $x\not\in  C$
This where I get stuck. I can see how to combine the $x \in A$ or $x \in B$ into $x \in A\cup B$, but I do not know how to derive the other half. Please help. 
 A: If you need a purely algebraic-looking proof, I would write
$$ \begin{align} (A\cup B)\setminus(C\setminus A)
  &= (A\cup B)\setminus(C\cap A^\complement)
\\&= (A\cup B)\cap (C\cap A^\complement)^\complement
\\&= (A\cup B)\cap (C^\complement \cup A)
\\&= A\cup(B\cap C^\complement)
\\&= A\cup(B\setminus C)
\end{align}$$
A: Assume that $x \in A \cup (B - C)$. Then ($x \in A$) OR $(x \in B$ and $x \notin C$).
If $x \in A$, then $x \in A \cup B$ and $x \notin C - A$, so $x \in (A \cup B) - (C-A)$.
If $x \in B$ and $x \notin C$, then $x \in A \cup B$ and $x \notin C - A$, so $x \in (A \cup B) - (C-A)$.
This means that $A \cup (B-C) \subset (A \cup B)-(C-A)$.
A: Proof of $\subseteq$:
Let $x \in (A \cup B) - (C - A)$. Then $x \in A$ or $x \in B$, and $x \notin (C-A)$. The last part means that either $x \in C \cap A$ or $x \notin C$.
Suppose $x \in C \cap A$. Then $x \in A$, so we're done as $A \subseteq A \cup (B-C)$.
Suppose $x \notin C$. But we also know that either $x \in A$ (in which case we're done) or $x \in B$ (in which case we're done, as $x \in (B-C)$).
I'll leave $\supseteq$ for you to try along similar lines.
