Proof Using cartesian products Suppose that $A$, $B$, and $C$ are sets. Prove that $(A\cap B)\times C =(A\times C)\cap(B\times C)$. Prove the statement both ways or use only if and only if statements.

The two sets $(A \cap B) \times C$ and $(A \times C) \cap (B \times C)$ are equal iff every element of one set is element of the other set and vice versa.
Consider $(A \times C) \cap (B \times C)$ first:
$$ a \in (A \times C) \cap (B \times C) \leftrightarrow a \in (A \times C) \wedge a \in (B \times C). $$
From the definition of the Cartesian product we have
$$ a \in (A \times C) \leftrightarrow (\exists x \in A)(\exists y \in C) \: a = (x,y). $$
Thus
$$ a \in (A \times C) \wedge a \in (B \times C) \leftrightarrow (\exists (x \in A) \wedge (x \in B) )(\exists y \in C) \: a = (x,y). $$
Where the first quantifier resolves to
$$ (\exists (x \in A) \wedge (x \in B) ) \leftrightarrow (\exists x \in A \cap B). $$
Again, from the definition of the Cartesian product,
$$ (\exists x \in A \cap B)(\exists y \in C) \: a = (x,y) \leftrightarrow a \in (A \cap B) \times C, $$
which yields
$$ (\forall a) \: a \in (A \times C) \cap (B \times C) \leftrightarrow a \in (A \cap B) \times C.$$
