Small set theory question from Munkres I'm suppose to translate the following sets into $A$, $B$, and $C$ using $\cap$, $\cup$, and $-$.
$$F=\{ x : x \in A \text{ and } (x \in B \implies x \in C)\}$$
My first attempt was: $$F= A \cap (B \cap C)$$.
But my worry is about everything within the parentheses.  I'm aware that $$B \cap C=\{x : x \in B\} \text{ and } \{x : x \in C\}.$$  This is not an implication.
 A: Edit: I've rewritten my answer because I think I aimed it at a different level when I didn't really need to.

Consider the definition of $F$:
$$F = \left\{ x : x \in A \text{ and if } x \in B \text{ then } x \in C \right\}$$
If we're to have $x \in F$ then we must certainly have $x \in A$, since otherwise the condition isn't satisfied. So we know that $F \subseteq A$.
This means that we obtain $F$ by throwing out some elements of $A$ $-$ namely, those elements which don't satisfy the condition.
Which elements do we throw out? Precisely those which are in $B$ but not in $C$. Since if an element is in $B$ but not $C$ then $x \in B \to x \in C$ isn't satisfied (and these are the only elements for which this condition isn't satisfied).
So we need to throw out $B - C$. That is, $F = A - (B - C)$.
A: $p\implies q$ is equivalent to $\neg p\vee q$, so $$x\in A\wedge(x\in B \implies x\in C)$$ is equivalent to $$x\in A\wedge(x\notin B \vee x\in C)$$ Also we have that $\neg\neg p\equiv p$, so we can write it as $$x\in A\wedge \neg(\neg(x\notin B\vee x\in C))$$ and by De Morgan's laws $$x\in A\wedge\neg(x\in B \wedge x\notin C)$$ Finally $(x\in A\wedge x\notin B)\equiv x\in(A\smallsetminus B)$, and applying it twice we get $$x\in A\smallsetminus(B\smallsetminus C)$$hence $F=A\smallsetminus(B\smallsetminus C)$
