Cartesian Product Proof with Set Differences Let $A$, $B$, $C$, and $D$ be sets. Prove: 
$$
(A\setminus B)\times(C\setminus D)=(A\times C) \setminus [(A\times D)\cup (B\times C)]
$$
I've spend a lot of time on this chasing elements all over the place but I can't seem to simplify it. Everything I seem to do/able to do just makes the entire problem more complex and I feel like I'm missing something. Thanks for you help. 
 A: Consider the diagram, proof with no words (large letter on the left edge of the diagram should be C, not B.):

A: \begin{align}(x,y)\in (A\setminus B)\times(C\setminus D)&\iff x\in (A\setminus B)\wedge y\in(C\setminus D)\\&\iff x\in A\wedge x\notin B \wedge y\in C\wedge y\notin D\\&\iff(x,y)\in(A\times C)\wedge (x,y)\notin(B\times C)\wedge (x,y)\notin (A\times D)\\&\iff (x,y)\in (A\times C) \setminus [(A\times D)\cup (B\times C)] \end{align}
A: Cartesian product distributes over unions
$$ A \times (B \cup C) = A \times B \cup A \times C$$
If $B \subseteq C$, then because $B = B \setminus C \cup C$, we also have that the cartesian product distributes over set differences:
$$ A \times B = A \times (B \setminus C) \cup A \times C$$
And in this case, the union is a disjoint union, and so
$$ A \times (B \setminus C) = (A \times B) \setminus (A \times C) $$
So, we can adapt our knowledge of elementary school algebra to expand $(A \setminus B) \times (C \setminus D)$ ....
A: Here is a late alternative answer, with several features which the earlier ones lack: it goes fully back to the definitions instead of using set theory laws; it uses only small steps; and it explains all those steps.
$
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\then}{\Rightarrow}
\newcommand{\followsfrom}{\Leftarrow}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
$Looking at the original statement, it is obvious that both sides contain only pairs.  So let's start at the most complex side, the right hand side, and calculate which pairs $\;(x,y)\;$ it contains:
$$\calc
(x,y) \in (A \times C) \setminus [(A \times D)\cup (B \times C)]
\op\equiv\hint{definition of $\;\setminus\;$; definition of $\;\lor\;$}
(x,y) \in (A \times C) \;\land\; \lnot ((x,y) \in (A \times D) \lor (x,y) \in (B \times C))
\op\equiv\hint{definition of $\;\times\;$, three times}
x \in A \land y \in C \;\land\; \lnot ((x \in A \land y \in D) \lor (x \in B \land y \in C))
\op\equiv\hints{logic: use $\;x \in A\;$ on other side of first $\;\land\;$;}\hint{use $\;y \in C\;$ on other side of second $\;\land\;$}
x \in A \land y \in C \;\land\; \lnot ((\true \land y \in D) \lor (x \in B \land \true))
\op\equiv\hint{logic: simplify; DeMorgan}
x \in A \land y \in C \;\land\; x \not\in B \land y \not\in D
\op\equiv\hint{logic: reorder conjuncts; definition of $\;\setminus\;$, twice}
x \in A \setminus B \;\land\; y \in C \setminus D
\op\equiv\hint{definition of $\;\times\;$}
(x,y) \in (A \setminus B) \times (C \setminus D)
\endcalc$$
Therefore, by set extensionality, $\;(A \times C) \setminus [(A \times D)\cup (B \times C)] \;=\; (A \setminus B) \times (C \setminus D)\;$.
A: Let $\left({x, y}\right) \in \left({
A \setminus B}\right) \times \left({C \setminus D}\right)$
Then:
$x \in \left({
A \setminus B}\right) \land \displaystyle  y \in\left({C \setminus D}\right)$
$ \iff \left( x \in A \land x \notin B\right) \land \left(y \in C \land y \notin D\right)$
$ \iff  \left(x \in A \land y \in C \land y \notin D\right) \lor \left(x \in A \land x \notin B \land y \in C\right) $
$ \iff  \left(x \in A \land y \in \left(C \setminus D \right) \right) \lor \left(x \in \left(A \setminus B \right) \land y \in C \right) $
$ \iff  \left(\left({x, y}\right) \in A \times \left(C \setminus D \right) \right) \lor \left(\left({x, y}\right) \in \left(A \setminus B \right) \times C \right) $
$ \iff  \left(\left({x, y}\right) \in \left(A \times C\right) \setminus \left(A \times D \right) \right) \lor \left(\left({x, y}\right) \in \left(A \times C\right) \setminus \left(A \times B \right) \right) $
$ \iff  \left({x, y}\right) \in \left(A \times C\right) \setminus \left(A \times D \right) \cup \left(A \times C\right) \setminus \left(A \times B \right)  $
$ \iff  \left({x, y}\right) \in (A\times C) \setminus [(A\times D)\cup (B\times C)]  $
