Ordering of Cardinals Compatible with Cardinal Sum

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 sum 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 $C$ be chosen such that $A \cap C = \varnothing = B \cap C$.

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

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

Then the mapping $h: A \cup C \to B \cup C$ defined as:
 * $\forall x \in A \cup C: h \left({x}\right) = \begin{cases}

f \left({x}\right) & : x \in A \\ x & : x \in C \end{cases}$

Let $a_1 \in A \cup C$ and $a_2 \in A \cup C$ such that:
 * $h \left({a_1}\right) = h \left({a_2}\right)$

Then:

and:

So:
 * $h \left({a_1}\right) = h \left({a_2}\right) \implies a_1 = a_2$

demonstrating that $h$ is an injection.

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