Prove $(A\times C)\setminus (B\times D)=(A\times (C\setminus D))\cup((A\setminus B)\times C)$ Given $A,B\subseteq X$ and $C,D\subseteq Y$, prove $(A\times C)\setminus (B\times D)=(A\times (C\setminus D))\cup((A\setminus B)\times C)$.
My attempt:
$$(A\times C)\setminus (B\times D)$$
$$(x,y)\in (A\times C) \wedge (x,y)\notin (B\times D)$$
$$(x\in A \wedge y\in C)\wedge (x\notin B \wedge y\notin D)$$
$$(x\in A \wedge y \in C \wedge y \notin D)\wedge (x\in A \wedge y\in C \wedge x\notin B)$$
$$(x\in A \wedge (y \in C \wedge y \notin D))\wedge ((x\in A \wedge x\notin B) \wedge y\in C)$$
$$(x\in A \wedge y\in (C\setminus D))\wedge (x\in (A\setminus B) \wedge y\in C)$$
$$(x,y)\in (A \times (C\setminus D))\wedge (x,y)\in ((A\setminus B)\times C)$$
$$(A\times (C\setminus D))\cup((A\setminus B)\times C)$$
Is that a valid proof? Are the rules I applied for the logic operator correct?
How could I simplify the proof or find another way to do it?
 A: There is a mistake in your proof. You should say $(x,y)\notin B\times D$ if and only if $x\notin B$ or $y\notin D$, since by definition, $(x,y)\in B\times D$ if and only if $x\in B$ and $y\in D$.
Futhermore, I'd encourage you to write out your proofs into paragraphs. Using too much logic/symbolic notation makes things difficult to read. So for instance, the first half of a proof could look like this:
Suppose $(x,y)\in(A\times C)\setminus(B\times D)$. Then $(x,y)\in A\times C$ and $(x,y)\notin B\times D$. Therefore $x\in A$ and $y\in C$, but also we must have either $x\notin B$ or $y\notin D$. If $x\notin B$ then $(x,y)\in (A\setminus B)\times C$. If $y\notin D$ then $(x,y)\in A\times(C\setminus D)$. Since one of these must be true, we see $(x,y)\in (A\times(C\setminus D))\cup((A\setminus B)\times C)$.
I'd encourage you to try the reverse direction on your own using similar ideas. Start with $(x,y)\in (A\times(C\setminus D))\cup((A\setminus B)\times C)$. By symmetry of the situation, you can just assume $(x,y)\in A\times(C\setminus D)$ (a lot of times you'll see the phrase "without loss of generality" when this is done).
