Union of Cartesian products: $X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)$ How do I prove or disprove $X \times (Y \cup Z)= (X \times Y) \cup(X\times Z)$ 
for all sets $X$, $Y$, and $Z$? I'm lost on the steps here.
 A: Basically, this is a simplification problem: find out which elements are in $\;(X \times Y) \cup (X \times Z)\;$ by expanding the definitions and simplifying, then work towards $\;X \times (Y \cup Z)\;$.
So for all $\;x\;$,
\begin{align}
& p \in (X \times Y) \cup (X \times Z) \\
\equiv & \qquad \text{"definition of $\;\cup\;$"} \\
& p \in X \times Y \;\lor\; p \in X \times Z \\
\equiv & \qquad \text{"definition of $\;\times\;$, twice, writing $\;p\;$ as $\;(x,y)\;$ for some $\;x,y\;$"} \\
& (x \in X \land y \in Y) \;\lor\; (x \in X \land y \in Z) \\
\equiv & \qquad \text{"logic: extract common conjunct $\;x \in X\;$ -- the key simplification"} \\
& x \in X \land (y \in Y \lor y \in Z) \\
\equiv & \qquad \text{"definition of $\;\cup\;$"} \\
& x \in X \land (y \in Y \cup Z) \\
\equiv & \qquad \text{"definition of $\;\times\;$, switching back to $\;p\;$"} \\
& p \in X \times (Y \cup Z) \\
\end{align}
By set extensionality, this proves the statement.
A: Let's do the right side first:
Take $a \in (X \times Y) \cup (X \times Z)$. Then what are the possibilities? $a \in (X \times Y)$ or $a \in (X \times Z)$. 
Now there are two cases: suppose $a \in (X \times Y)$. If $a \in X \times Y$, then $a \in X \times (Y \cup Z)$, because the cartesian cross product is defined as $A \times B  = \{(a,b) | a \in A \land b \in B\}$, and uniting $Y$ and $Z$ does not mean that we are losing elements of $Y$. The case for $a \in X \times Z$ is similar (you should do that one). Now we have proven that if $a \in (X \times Y) \cup (X \times Z)$, then $a \in X \times (Y \cup Z)$. This implies that $(X \times Y) \cup (X \times Z) \subseteq X \times (Y \cup Z)$.
To show that the sets are equal, we'll then want to show that $ X \times (Y \cup Z) \subseteq (X \times Y) \cup (X \times Z)$. The reasoning in this case (for the entire left side) is similar to the reasoning above. I'll leave it to you, as this vaguely looks like homework.
A: $$A\times(B\cap C)\Leftrightarrow (A\cup A)\times(B\cup C)$$
$\left({x, y}\right) \in \left({A \cup A}\right) \times \left({B \cup C}\right)$ 
$\Leftrightarrow \left({x \in A \lor x \in A}\right)\land\left({y \in B \lor y \in C}\right)$
$\Leftrightarrow \left[{\left({x \in A \lor x \in A}\right) \land y \in B}\right]\lor\left[{\left({x \in A \lor x \in A}\right) \land y \in C}\right]$
$\Leftrightarrow \left({x \in A \land y \in B}\right)\lor\left({x \in A \land y \in B}\right)\lor\left({x \in A \land y \in C}\right)\lor\left({x \in A \land y \in C}\right)$
$\Leftrightarrow \left({x, y}\right) \in \left({A \times B}\right) \cup \left({A \times B}\right) \cup \left({A \times C}\right) \cup \left({A \times C}\right)$
$\Leftrightarrow \left({x, y}\right) \in \left[\left({A \times B}\right) \cup \left({A \times C}\right)\right]$
