Proof on elementary set theory I want to show that $A\times(A\setminus B)=(A\times A)\setminus(A\times B).$
So I started off with $(A\times A)\setminus(A\times B)=\{(a,b):(a,b)\in(A\times A) \wedge (a,b)\notin(A\times B)\}=\{(a,b):a\in A\wedge b \in A \wedge a \notin A \wedge b \notin B\}.$
So from $b\in A \wedge b\notin B\Longrightarrow b\in (A\setminus B).$ But how do interpret the other terms $a\in A \wedge a \notin A$. Where have I gone wrong? I appreciate your answers.
 A: If $(x,y) \in A \times (A \setminus B)$, then $x \in A$ and $y \in A \setminus B$ (definition of Cartesian product). The latter means that $y \in A$ and $y \notin B$. This means that $(x,y) \in A \times A$ (both are in $A$) and $(x,y) \notin A \times B$, as $y \notin B$. So $(x,y) \in (A \times A)\setminus (A \times B)$.
On the other hand, if $(x,y) \in (A \times A)\setminus (A \times B)$, then $(x,y) \in A \times A$ and $(x,y) \notin A \times B$. So $x,y \in A$ from the first, and $y \in A \setminus B$ (we already know $y \in A$, and if $y \in B$, $(x,y)$ would have been in $A \times B$, which it is not, so $y \notin B$). Hence $(x,y) \in A \times (A \setminus B)$.
This shows both inclusions and hence equality.
In your proof you go from $(x,y) \notin A \times B$ to $x \notin A$ and $x \notin B$, but in order for a point not to be in a product, only one of them has to fail. So we have or instead of and. 
A: $
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\false}{\text{false}}
\newcommand{\empty}{\varnothing}
\newcommand{\diff}{\mathbin \triangle}
$Just for fun, here is the same proof as from the first answer, but with both directions proved at the same time, and written down in a slightly different format.
We start at the most complex side of the equality, and check which pairs $\;(x,y)\;$ are in that set, and then simplify:
$$\calc
(x,y) \in (A \times A) \setminus (A \times B)
\calcop\equiv{definition of $\;\setminus\;$, and of $\;\times\;$ twice}
x \in A \land y \in A \;\land\; \lnot(x \in A \land y \in B)
\calcop\equiv{logic: use $\;x \in A\;$ on other side of leftmost $\;\land\;$}
x \in A \land y \in A \;\land\; \lnot(\true \land y \in B)
\calcop\equiv{logic: simplify and reorder -- to get closer to our goal}
x \in A \;\land\; y \in A \land \lnot(y \in B)
\calcop\equiv{definitions of $\;\setminus\;$ and $\;\times\;$}
(x,y) \in A \times (A \setminus B)
\endcalc$$
By set extensionality, this proves $\;(A \times A) \setminus (A \times B) \;=\; A \times (A \setminus B)\;$.
