I understand how this is true, as the entirety of the elements in sets A and B multiplied by C would be equivalent to the combination of A x C and B x C. My problem is that I am having trouble expressing this in proper predicate or first-order logic, or any type of logic for that matter.
Any insight on how to properly prove/express this statement would be greatly appreciated.
On
By first principles:
$$ \begin{align} (x,y) \in (A \cup B) \times C &\iff (x \in (A \cup B)) \land (y \in C) \\ &\iff ((x \in A) \lor (x \in B)) \land (y \in C) \\ &\iff ((x \in A) \land (y \in C)) \lor ((x \in B)) \land (y \in C)) \\ &\iff ((x,y) \in A \times C) \lor ((x,y) \in B \times C) \\ &\iff (x,y) \in (A \times C) \cup (B \times C) \end{align} $$
The key to proving that two sets $A$ and $B$ are equal is showing that $x \in A$ if and only if $x \in B$.
First, we show that $x \in (A \cup B) \times C \implies (A \times C) \cup (B \times C)$. By definition, $x \in (A \cup B) \times C$ means that $x$ is an ordered pair $x = (x_1,x_2)$ such that $x_1 \in A \cup B$ and $x_2 \in C$. If we apart that union, we have $x_1 \in A$ or $x_1 \in B$. All together, we have $$ (x_1 \in A \text{ or } x_1 \in B) \text{ and } x_2 \in C $$ This is logically equivalent to $$ (x_1 \in A \text{ and } x_2 \in C) \text{ or } (x_1 \in B \text{ and } x_2 \in C) $$ In practice, one usually does not need to prove this equivalence explicitly (especially on a test question, for example). However, it may be shown that these are equivalent with a truth table.
Of course, $(x_1 \in A \text{ and } x_2 \in C) \text{ or } (x_1 \in B \text{ and } x_2 \in C)$ means that the pair $x = (x_1,x_2)$ is an element of $(A \times C) \cup (B \times C)$, by definition.
It remains to be shown that these steps can be reversed, i.e. that $(A \times C) \cup (B \times C) \implies x \in (A \cup B) \times C$.