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:

$$ $$ $$