If $A,B,$ and $C$ are sets, then $A\times(B-C)$ = $(A \times B)$ $-$ $(A \times C)$. If $A,B,$ and $C$ are sets, then $A\times(B-C)$ = $(A \times B)$ $-$ $(A \times C)$.
Proof. Observe the following sequence of equalities.
$$\begin{align}
A\times(B-C) &= \{(x,y)\} : (x \in A) \wedge (y \in (B-C))\} \, (\text{Definition of Cartesian Product}) \\
&=\{(x,y) : (x \in A) \wedge \big((y \in B) \wedge (y\notin C)\big)\} \,
 (\text{Definition of } -) \\ 
&=\{(x,y) : (x \in A) \wedge (x \in A) \wedge \big((y \in B) \wedge (y\notin C)\big)\} \, (P=P \wedge P) \\
&=\{(x,y) : \big((x \in A) \wedge (y \in B)\big) \wedge \big((x \in A) \wedge (y\notin C)\big)\} \, (\text{Rearrange}) \\
&=\{(x,y) : \big((x \in A) \wedge (y \in B)\big) \wedge \big((x \in A) \wedge (y\notin C)\big)\} \, (\text{Definition of }\cap) \\
\end{align}$$
I'm stuck on the last part -- $(x \in A) \wedge (y\notin C)$ translates to $(A-C)$ but I need it to be $(A \times C)$. I can't quite figure out how to reach that.
 A: First, when you make a proof with logic, don't use the equality symbol, we use the implication symbol because the implication symbol is defined for the  logic propositions.
So, i'm gona show first that $(A \times B) -(A\times C) \subseteq A \times (B-C)$
Let $(x,y)\in ((A \times B) -(A\times C))
\implies (x,y)\in (A \times B) \land (x,y) \not \in (A \times C)$
$\implies (x\in A \land y\in B) \land \lnot (x\in A \land y\in C) 
\implies (x\in A \land y\in B) \land (x\not \in A \lor y \not \in C)$
$\implies (x \in A \land y \in B \land x\not \in A) \lor (x \in A \land y \in B \land x\not \in C) $
By TTP:
$\implies (x \in A \land y \in B \land x\not \in C) \implies x\in A \land y \in(B-C)$
$\implies (x,y)\in(A \times (B-C))$
Now, i'm gonna show that $A \times (B-C) \subseteq (A \times B) -(A\times C)$
Let $(x,y)\in (A \times (B-C))
\implies x\in A \land y\in (B-C)
\implies x\in A \land (y\in B \land y \not \in C)$
$\implies (x\in A \land y\in B \land y \not \in C) \lor (x \in A \land y \in B \land x\not \in A)$
$\implies (x\in A \land y\in B) \land (x\not \in A \lor y \not \in C)$
$\implies (x\in A \land y\in B) \land \lnot (x\in A \land y\in C)
\implies (x,y)\in (A \times B) \land (x,y) \not \in (A \times C)$
$\implies (x,y)\in ((A \times B) -(A\times C))$
Then, $(A \times B) -(A\times C) = A \times (B-C) \blacksquare$
