Let A, B, C be sets. Prove either using a membership table or other method that:
(A\B) x C = (AxC) \ (BxC)
$\displaystyle (x,y) \in \left( {A\backslash B} \right) \times C\quad \Rightarrow \quad x \in A \wedge x \notin B \wedge y \in C$
$\displaystyle x \in A \wedge x \notin B \wedge y \in C \Rightarrow \quad \left( {x,y} \right) \in A \times C \wedge \left( {x,y} \right) \notin B \times C$
$\displaystyle \left( {x,y} \right) \in A \times C \wedge \left( {x,y} \right) \notin B \times C \Rightarrow \quad \left( {x,y} \right) \in \left[ {\left( {A \times C} \right)\backslash \left( {B \times C} \right)} \right]
$
Can you turn that around to get both ways?
Not necessarily. See if you can do it or if you need a different proof to go backwards, make sure that it works from your beginning assumption of (AxC)\(BxC).
And to prove that two sets are equal, you prove that they are subsets of each other - not "because (x,y) is on both (A\B)xC and (AxC)\(BxC)" but maybe that's what you mean.
OK I'm giving it a try... not sure if this makes sense:
So for the second part...
If (x,y) is in [(AxC)\(BxC)] then (x,y) is not in (BxC).
This means that x is not in B or y is not in C.
Since (x,y) is in (AxC) then y is in C, so x is not in B.
Therefore, x is in A, x is not in B, and y is in C,
so (x,y) is in (A\B) x C.
Therefore [(AxC)\(BxC)] is a subset of (A\B)xC (and from before vice versa), so the two sides are equal.
Does this make sense at all??