Is A × (B\C) = (A × B) \ (A × C) Always True For All Sets? Provide either a proof or a counterexample.
I haven't been able to find an example where these two statements are not equivalent, as trying to equate any of A, B or C to the Empty Set doesn't work, and I've also tried setting them to shared variables and separate variables. So I'm inclined to believe that this is true, however, I'm not sure how to definitively prove it. If I say, for example, "Let $x∈A, y∈B, z∈C$," that doesn't cover every possibility, such as $x∈A, x∈B$, etc. If anyone could point out even the first step to creating a proof that definitively covers these options, I'd greatly appreciate it.

If $(x,y)\in A\times(B\setminus C)$ then $x\in A$ and $y\in(B\setminus C)$.
Hence, in particular $y\in B$ so $(x,y)\in A\times B$ and $y\not\in A\times C$. Summing-up $(x,y)\in A\times B\setminus A\times C$. We conclude:
$A\times(B\setminus C)\subseteq A\times B\setminus A\times C$.
For the opposite inclusion. Assume $(x,y)\in A\times B\setminus A\times C$.
Then, of course, $x\in A$ and $y\in B$, but since $(x,y)\not\in A\times C$,
then $y\in B\setminus C$. Hence, $(x,y)\in A\times(B\setminus C)$.
Remark that the statement is false if $B=C$ or $B\subset C$. As a counter-example in this case choose $A=\{x\}$ and $B=C=\{y\}$.
Edit: After discussion with @J. Davis Taylor, I get convinced that in fact, there is no counter-example. Plz, look at the discussion below.
