Need help with proving $A \times (B \setminus C) = (A \times B) \setminus (A \times C)$? To prove this I have to show that:
$$1. \space\space A \times (B \setminus C) \subseteq  (A \times B) \setminus (A \times C)$$
$$2. \space\space (A \times B) \setminus (A \times C) \subseteq  A \times (B \setminus C)$$
The first one went pretty well.
In the second part I tried this, but got stuck:
Let $x \in (A \times B) \setminus (A \times C)$ we need to show that $x \in A \times (B \setminus C)$
If $x \in (A \times B) \setminus (A \times C)$ It means that:
$x \in (A \times B) \space \space and \space \space x \not \in (A \times C)$   
Here I got stuck, any help is welcome.
Sorry for my bad english :)
 A: If $x \in (A \times B)$ and $x \not\in (A \times C)$, we have $x = (a,b)$ where $a \in A$ and $b \in B$. Furthermore, $b \not\in C$ as $x \not\in A \times C$. Therefore, $b \in B \backslash C$, hence $x \in A \times (B \backslash C)$. Since this is true for every $x \in (A \times B) \backslash (A \times C)$, we have proven the desired inclusion.
A: Here is another way to show this: for every $\;p\;$,
\begin{align}
& p \in (A \times B) \setminus (A \times C) \\
\equiv & \qquad \text{"definition of $\;\setminus\;$"} \\
& p \in A \times B \;\land\; \lnot (p \in A \times C) \\
\equiv & \qquad \text{"definition of $\;\times\;$, twice, writing $\;p\;$ as $\;(x,y)\;$"} \\
& x \in A \land y \in B \;\land\; \lnot (x \in A \land y \in C) \\
\equiv & \qquad \text{"logic: use $\;x \in A\;$ on other side of $\;\land\;$"} \\
& x \in A \land y \in B \;\land\; \lnot (\text{true} \land y \in C) \\
\equiv & \qquad \text{"logic: simplify"} \\
& x \in A \;\land\; y \in B \land \lnot (y \in C) \\
\equiv & \qquad \text{"definition of $\;\setminus\;$} \\
& x \in A \;\land\; y \in B \setminus C \\
\equiv & \qquad \text{"definition of $\;\times\;$, reintroducing $\;p\;$"} \\
& p \in A \times (B \setminus C) \\
\end{align}
