Ordering of Cardinals Compatible with Cardinal Product

Theorem
Let $\mathbf a$, $\mathbf b$ and $\mathbf c$ be cardinals.

Then:
 * $\mathbf a \le \mathbf b \implies \mathbf a \mathbf c \le \mathbf b \mathbf c$

where $\mathbf a \mathbf c$ denotes the product of $\mathbf a$ and $\mathbf c$.

Proof
Let $\mathbf a = \operatorname{Card} \left({A}\right)$, $\mathbf b = \operatorname{Card} \left({B}\right)$ and $\mathbf c = \operatorname{Card} \left({C}\right)$ for some sets $A$, $B$ and $C$.

Let $\mathbf a \le \mathbf b$.

Then by definition of cardinal, there exists an injection $f: A \to B$.

Then the mapping $g: A \times C \to B \times C$ defined as:
 * $\forall \left({a, c}\right) \in A \times C: g \left({a, c}\right) = \left({f \left({a}\right), c}\right)$

Let $\left({a_1, c_1}\right) \in A \times C$ and $\left({a_2, c_2}\right) \in A \times C$ such that:
 * $g \left({a_1, c_1}\right) = g \left({a_2, c_2}\right)$

That is:
 * $\left({f \left({a_1}\right), c_1}\right) = \left({f \left({a_2}\right), c_2}\right)$

By Equality of Ordered Pairs:
 * $f \left({a_1}\right) = f \left({a_2}\right), c_1 = c_2$

and so as $f$ is an injection:
 * $a_1 = a_2, c_1 = c_2$

By Equality of Ordered Pairs:
 * $\left({a_1, c_1}\right) = \left({a_2, c_2}\right)$

demonstrating that $g$ is an injection.

So, by definition of product of cardinals:
 * $\mathbf a \mathbf c \le \mathbf b \mathbf c$