Condition for Existence of Cardinal Sum

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

Then:
 * $\mathbf a \le \mathbf b \iff \exists \mathbf c: \mathbf a + \mathbf c = \mathbf b$

where $\mathbf c$ is also a cardinal.

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

From Equivalence of Definitions of Dominate (Set Theory), there exists a bijection from $A$ onto a subset $E$ of $B$.

Thus:
 * $\mathop a = \operatorname{Card} \left({E}\right)$

Let $F = \complement_B \left({E}\right)$

From Set with Relative Complement forms Partition:
 * $B = E \cup F$
 * $E \cap F = \varnothing$

By definition of sum of cardinals, it follows that:

By defining $\mathbf c := \operatorname{Card} \left({F}\right)$ it follows that:
 * $\mathbf a + \mathbf c = \mathbf b$

for the $\mathbf c$ that has been demonstrated to exist.

Now suppose there exists a cardinal $\mathbf c$ such that $\mathbf a + \mathbf c = \mathbf b$.

Then by definition of sum of cardinals:
 * $\operatorname{Card} \left({B}\right) = \operatorname{Card} \left({A \cup C}\right)$

for some sets $A$, $B$ and $C$ such that $A \cap C = \varnothing$.

Let $f: A \to B$ be an injection, proved to exist by Equivalence of Definitions of Dominate (Set Theory).

Then it follows that $\mathbf a \le \mathbf b$.