Features of elements in ordered pair, when it doesn't belong to certain set? So, I have the equation: $$(A\times B)\setminus C=(A\setminus C)\times (B\setminus C)$$
And excercise is to provide proof of wheter this is true with all sets or not. I'm doing it by proving both sides to be subset of another, first starting with that: $$x\in (A\times B)\setminus C $$
and x is ordered pair$$x=(s,t)$$
Then from definition of \ and definiton of product:
$$s\in A \wedge t\in B \wedge (s,t)\notin C$$
Now the problem is that I can't deduce anything for what s and t are alone with set $C$. From pictures I've drawn it seems that the original equation should be true, but then with the pictures it also seems that if $(s,t)\notin C$, then it's enough that either of them isn't in C to satisfy it, which with it seems equation would not be true, as left-side would not always be subset of right-side.
Any tips of how $(s,t)\notin C$ should be dealt with, or if this is completely wrong way to try to prove this, then what could be right way to start with?
 A: Your equation is ill-typed, so it has no hope of being true in general.
Writing $(A\times B)\setminus C$ carries an implicit assumption that the elements of $C$ are pairs of $a$s and $b$s, whereas $A\setminus C$ on the other hand assumes that some $a$s themselves are elements of $C$.
This doesn't quite mean that your equation is nonsense -- because in formal, axiomatic set theory everything is a set and there are no types, so you can subtract any set from any other no matter whether they happen to have elements in common -- but in ordinary everyday mathematics an expression with such a type error in it is useless and a very good sign that some confused thinking went into coming up with it.
Armed with this understanding, it is easy to come up with a counterexample:
Take $A=\{1\}$, $B=\{2\}$, $C=\{2\}$, for example. Then
$A\times B = \{\langle 1,2\rangle\}$ and so also $(A\times B)\setminus C= \{\langle 1,2\rangle\}$ because $\langle 1,2\rangle \notin \{2\}$. However
$$(A\setminus C)\times(B\setminus C) = A\times\varnothing = \varnothing$$
and $\varnothing$ is not the same set as $\{\langle 1,2\rangle\}$.
