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 $\mathbf a = \map \Card A$, $\mathbf b = \map \Card B$ and $\mathbf c = \map \Card C$ 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:
 * $\forall \tuple {a, c} \in A \times C: \map g {a, c} = \tuple {\map f a, c}$

Let $\tuple {a_1, c_1} \in A \times C$ and $\tuple {a_2, c_2} \in A \times C$ such that:
 * $\map g {a_1, c_1} = \map g {a_2, c_2}$

That is:
 * $\tuple {\map f {a_1}, c_1} = \tuple {\map f {a_2}, c_2}$

By Equality of Ordered Pairs:
 * $\map f {a_1} = \map f {a_2}, c_1 = c_2$

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

By Equality of Ordered Pairs:
 * $\tuple {a_1, c_1} = \tuple {a_2, c_2}$

demonstrating that $g$ is an injection.

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