Rewriting $\{x\ \mid\ x \in A \textrm{ and } (x \in B \implies x \in C)\}$ I'm reviewing some basic set theory, and for some reason the following problem (Munkres 1.7) is tripping me up.

Rewrite in terms of $A, B,$ and $C$ the following set using the symbols $\cup, \cap,$ and $-$.
  $$F = \{x\ \mid\ x \in A \textrm{ and } (x \in B \implies x \in C)\}.$$

Now, I know quite well what all of the symbols mean, but I'm having trouble ascertaining what the parenthetical term comes out to.
The statement $x \in B \implies x \in C$ is true for all $x \in B \cap C$, obviously. The intersection of this and $A$ gets us $A\cap B \cap C$.
The contrapositive says that $x \notin C \implies x \notin B$. So this is true for all elements that are not in $C$ and also not in $B$.
Therefore, I should have
$$\begin{align*}
(A - (A\cap B)) \cup (A\cap B \cap C) &= ((A - A) \cup (A - B))\cup (A\cap B \cap C) \\
 &= (A-B) \cup (A \cap B \cap C).
\end{align*}$$
Am I correct?
 A: For every $x$ it holds that:
$$\begin{align} x\in A\land (x\in B\longrightarrow x\in C ) &\iff x\in A\land (x\not \in B\,\lor x\in C) \\
&\iff (x\in A \land x\not \in B)\lor (x\in A\land x\in C) \\
&\iff x\in (A- B)\cup(A\cap C)\end{align}$$
However what you got is also true.
A: $P \Rightarrow Q$ is equivalent to $\neg P \vee Q$. So therefore each of the following lines are equivalent to the next
$(x \in A) \wedge (x \in B  \Rightarrow x \in C)$
$(x \in A) \wedge (x \notin B \vee x \in C)$
$A \cap (\bar{B} \cup C)$
A: Here's another route you can go: We can use the fact that $P \rightarrow Q \equiv \lnot P \lor Q \equiv \lnot(P \land \lnot Q)$, by DeMorgan's
$$\begin{align} x\in F &\iff (x \in A) \land (x \in B  \Rightarrow x \in C)\\ \\
 & \iff (x \in A) \land \lnot(x \in B \land x \notin C) \\ \\ & \iff (x \in A) \land (x \notin (B - C))\\ \\ &\iff x \in A - (B - C)\end{align}$$
$$\therefore \{x\ \mid\ x \in A \textrm{ and } (x \in B \implies x \in C)\} = A - (B - C)$$
A: You are, but it can be simplified. Formula $x \in B \Rightarrow x \in C$ is false only in $B - C$, while is true at $C$ and $\overline{B}$. Given that everything has to be contained in $A$, you can simplify $\overline{B}$ to $A-B$ and get $$(A-B) \cup (A \cap C).$$
I hope this helps $\ddot\smile$
