Product of Subset with Union

Theorem
Let $\left({G, \circ}\right)$ be an algebraic structure.

Let $X, Y, Z \subseteq G$.

Then:


 * $X \circ \left({Y \cup Z}\right) = \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$
 * $\left({Y \cup Z}\right) \circ X = \left({Y \circ X}\right) \cup \left({Z \circ X}\right)$

Proof 1

 * Let $x \circ t \in X \circ \left({Y \cup Z}\right)$.

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:
 * $x \circ t \in \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$

and so:
 * $X \circ \left({Y \cup Z}\right) \subseteq \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$

Now let $x \circ t \in \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$.

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:
 * $x \circ t \in X \circ \left({Y \cup Z}\right)$

and so:
 * $\left({X \circ Y}\right) \cup \left({X \circ Z}\right) \subseteq X \circ \left({Y \cup Z}\right)$

That is:
 * $X \circ \left({Y \cup Z}\right) = \left({X \circ Y}\right) \cup \left({X \circ Z}\right)$

The result:
 * $\left({Y \cup Z}\right) \circ X = \left({Y \circ X}\right) \cup \left({Z \circ X}\right)$

follows similarly.

Proof 2
Consider the relation $\mathcal R \subseteq G \times G$ defined as:


 * $\forall g, h \in G: \left({g, h}\right) \in \mathcal R \iff \exists g \in X$

Then:
 * $\forall S \subseteq G: X \circ S = \mathcal R \left({S}\right)$

Then:

Next, consider the relation $\mathcal R \subseteq G \times G$ defined as:


 * $\forall g, h \in G: \left({g, h}\right) \in \mathcal R \iff \exists h \in X$

Then:
 * $\forall S \subseteq G: S \circ X = \mathcal R \left({S}\right)$

Then: