True or false statement proofs I felt this statement was true, but I'm not sure if you can prove it by simply expanding the original equation of $A\times(B\cap C)$ in order to get the RHS ? 
For an sets $A$ and $B$, $A \times (B \cap C ) = (A \times B) \cap  (A \times C) $
 A: Let $(x,y) \in A \times (B \cap C)$,
then $x \in A$ and $y \in (B \cap C)$.
hence $x \in A$ and $y\in B$ and $y \in C$.
Hence $(x,y) \in A \times B $ and $(x,y) \in A \times C$
that is $(x,y) \in (A \times B) \cap (A \times C).$
For the reverse direction, read it from the bottom up.
A: I would strongly suggest to work the other way around, and start at the most complex side (which is here the right hand side).  That way, the problem becomes one of simplification.  And it is a lot easier to simplify than to 'complexify': it is a lot easier to go from $$
\tag{*}
(x \in A \land y \in B) \land (x \in A \land y \in C)
$$ to $$
\tag{**}
x \in A \land (y \in B \land y \in C)
$$ than the other way around, since starting from $\text{(**)}$ there are many ways to introduce complexity, and it could be difficult to predict which path is the correct one.  But it's easy to spot the duplication in $\text{(*)}$ and get rid of it.
Also, I would strongly suggest to use 'iff' or $\;\equiv\;$ or $\;\iff\;$, whenever a step actually preserves equivalence.  That way you don't have to prove both $\;\Rightarrow\;$ and $\;\Leftarrow\;$ separately, making the proof shorter and simpler.
Finally, specifically with set theory proofs, I find they are often easier on the element level, i.e., by first expanding the definitions first so that we can use the laws of logic rather than the slightly more complex laws of set theory.
Combining all of that could result in a proof like https://math.stackexchange.com/a/543802/11994.
