Product of Subset with Union/Proof 1

Proof
Let $x \circ t \in X \circ \paren {Y \cup Z}$.

We have $x \in X, t \in Y \cup Z$ by definition of subset product.

By definition of set union, it follows that $t \in Y$ or $t \in Z$.

So we also have $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.

That is:
 * $x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$

and so:
 * $X \circ \paren {Y \cup Z} \subseteq \paren {X \circ Y} \cup \paren {X \circ Z}$

Now let $x \circ t \in \paren {X \circ Y} \cup \paren {X \circ Z}$.

By definition of set union, it follows that $x \circ t \in X \circ Y$ or $x \circ t \in X \circ Z$.

So $x \in X$, and $y \in Y$ or $y \in Z$.

That is, $x \in X$, and $y \in Y \cup Z$ by definition of set union.

Hence:
 * $x \circ t \in X \circ \paren {Y \cup Z}$

and so:
 * $\paren {X \circ Y} \cup \paren {X \circ Z} \subseteq X \circ \paren {Y \cup Z}$

That is:
 * $X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$

The result:
 * $\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$

follows similarly.