Product of Subset with Union/Proof 1

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Theorem

Let $\struct {G, \circ}$ be an algebraic structure.

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

Then:

$X \circ \paren {Y \cup Z} = \paren {X \circ Y} \cup \paren {X \circ Z}$
$\paren {Y \cup Z} \circ X = \paren {Y \circ X} \cup \paren {Z \circ X}$

where $X \circ Y$ denotes the subset product of $X$ and $Y$.


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.

$\blacksquare$