how do you prove that $A\times (B\setminus C) = (A\times B) \setminus (A\times C)$? This is what I have come up with so far, but I am rather lost at lines 3-4...
$$\begin{align}
(a,b)\in A\times (B\setminus C)\iff& a\in A \land b\in (B\setminus C)\\
\iff& a\in A \land (b\in B \land \lnot(b\in C))\\
\iff& (a\in A \land b\in B) \land (a\in A \land b\notin C)\\
\iff& (a\in A \land b\in B)\setminus (a\in A \land b\in C)\\
\iff& (a,b)\in (A\times B)\setminus ((a,b)\in A\times C)\\
\iff& (a,b)\in (A\times B)\setminus (A\times C)
\end{align}$$
 A: I'd go with the easier path of proving the double inclusion.
Suppose $(a,b)\in A\times(B\setminus C)$. Then $a\in A$ and $b\in B\setminus C$, that is, $b\in B$ and $b\notin C$. Hence $(a,b)\in A\times B$ and $(a,b)\notin A\times C$.
Therefore $(a,b)\in(A\times B)\setminus(A\times C)$.
Suppose $(a,b)\in(A\times B)\setminus(A\times C)$. Then $(a,b)\in A\times B$ and $(a,b)\notin A\times C$. Hence $a\in A$ and $b\in B$. The condition $(a,b)\notin A\times C$ implies that either $a\notin A$ or $b\notin C$. Since $a\in A$, it follows $b\notin C$. Hence $a\in A$ and $b\in B\setminus C$.
Therefore $(a,b)\in A\times(B\setminus C)$.
A: I think the most intuitive approach is to work from both ends, simplifying the expressions so that they "meet in the middle". As you have said, we have
$$\begin{align}
(a,b)\in A\times (B\setminus C)\iff& a\in A \land b\in (B\setminus C)\\
\iff& a\in A \land (b\in B \land b\notin C).\\
\end{align}$$
On the other hand, we can state
$$\begin{align}
(a,b) \in (A \times B) \setminus (A \times C)
\iff& (a,b)\in (A\times B) \wedge \neg ((a,b)\in A\times C)\\
\iff& [a \in A \wedge b \in B] \wedge \neg [a \in A \wedge b \in C]\\
\iff& [a \in A \wedge b \in B] \wedge [a \notin A \vee b \notin C]\\
\iff& b \in B \wedge [a \in A \wedge [a \notin A \vee b \notin C]]\\
\iff& b \in B \wedge [a \in A \wedge b \notin C].\\
\end{align}$$
Now, it suffices to note that 
$$
a\in A \land (b\in B \land b\notin C) \iff b \in B \wedge [a \in A \wedge b \notin C].
$$
