TITLE: Proof that sets are equal
QUESTION [10 upvotes]: I want to prove that  $(A\setminus B)\setminus(A \setminus C)=(A  \cap C)\setminus B $, so  I did this: 
$ x \in (A \setminus B)\setminus(A \setminus C) \iff $ 
$ (x \in A \setminus B)  \wedge   (x \notin A \setminus C)    \iff  $
$ (x \in A \wedge x \notin B) \wedge (x \notin A \vee x \in C)   \iff   $
$ (x \in A \wedge x \notin B \wedge x \notin A)  \vee (x \in A \wedge x \notin B \wedge x \in C)   \iff $ 
(Now,  the statement in the first bracket is always false, so the truth value of the whole disjunction  in the line above depends only on the second bracket)
$x \in A \wedge x \notin B \wedge x \in C   \iff $
$ x \in  (A  \cap C) \setminus B $
Is the method correct? Thanks in advance.

REPLY [3 votes]: It's correct. A (not so) different way could be to set $U=A\cup B\cup C$ and denoting complementation with respect to $U$ by $X^c$:
\begin{align}
(A\setminus B)\setminus(A\setminus C)
&=(A\cap B^c)\cap(A\cap C^c)^c \\
&=(A\cap B^c)\cap(A^c\cup C)\\
&=(A\cap B^c\cap A^c)\cup(A\cap B^c\cap C)\\
&=\emptyset\cup(A\cap B^c\cap C)\\
&=(A\cap C)\cap B^c\\
&=(A\cap C)\setminus B
\end{align}
One could also go in a different direction:
$$
(A\setminus B)\setminus(A\setminus C)=(A\cap B^c\cap C)=(A\setminus B)\cap C
$$