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.
 A: By the definition of cartesian product the set $(A\cap B)\times C = \{(x,y)|x\in A\cap B, y\in C\}$. Here we show that the above set is equivalent to the set $(A\times C)\cap (B\times C)$.
The definition of $(A\times C) = \{(x,y)|x\in A, y\in C\}$ and the definition of $(B\times C) = \{(x,y)|x\in B, y\in C\}$. Thus, there intersection is equal to $(A\times C)\cap (B\times C)=\{(x,y)|(x,y)\in (A\times C)$ and $(B\times C)\}$. Note that $\forall (x,y)\in (A\times C)\cap (B\times C)$ we have $y\in C$. Thus we may consider only those $x$. So we simplfy the set to $(A\times C)\cap (B\times C)=\{(x,y)|x\in A$ and $B, y\in C\}$. But this is simply the first set we considered! Therefore $(A\times C)\cap (B\times C)=(A\cap B)\times C$.
A: $(x,y)\in (A\cap B)\times C \iff x\in A\cap B$ and $y\in C \iff (x,y) \in A\times C$ and $ (x,y) \in B\times C \iff (x,y) \in (A\times C)\cap(B\times C)$.
A: 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
$$ (\forall a) \: a \in (A \times C) \cap (B \times C) \leftrightarrow a \in (A \cap B) \times C.$$
