Cardinal One is Cancellable for Cardinal Sum

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

Then:
 * $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 \implies \mathbf a = \mathbf b$

where $\mathbf 1$ is (cardinal) one.

Proof
Suppose $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1$.

Then from Condition for Existence of Cardinal Sum there exists some cardinal $\mathbf c$ such that
 * $\mathbf a + \mathbf 1 = \mathbf b + \mathbf 1 = \mathbf c$

By definition of cardinal there exists a set $C$ such that $\mathbf c = \operatorname{Card} \left({C}\right)$.

$C$ also has subsets $A$ and $B$ such that there also exist elements of $\alpha, \beta \in C$ such that:
 * $\complement_C \left({A}\right) = \left\{{\alpha}\right\}$
 * $\complement_C \left({B}\right) = \left\{{\beta}\right\}$

If $\alpha = \beta$ then $A = B$ and so $\mathbf a = \mathbf b$.

On the other hand, if $\alpha \ne \beta$ then $\alpha \in B$ and $\beta \in A$ and so:
 * $\left\{{\beta}\right\} \cup \left({A \cap B}\right) = A$
 * $\left\{{\alpha}\right\} \cup \left({A \cap B}\right) = B$

Since:
 * $\left\{{\beta}\right\} \cap A \cap B = \varnothing = \left\{{\alpha}\right\} \cap A \cap B$

it follows that:
 * $\mathbf a = \operatorname{Card} \left({A}\right) = \mathbf 1 + \operatorname{Card} \left({A \cap B}\right) = \operatorname{Card} \left({B}\right) = \mathbf b$