TITLE: Need help to prove $(A\cup B) \setminus (C \setminus A) = A \cup (B \setminus C)$
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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.