Ordinal Subtraction when Possible is Unique

Theorem
Let $x$ and $y$ be ordinals such that $x \le y$.

Then there exists a unique ordinal $z$ such that $\left({x + z}\right) = y$.

That is:
 * $\displaystyle x \le y \implies \exists ! z \in \operatorname{On}: \left({x + z}\right) = y$

Proof
By transfinite induction on $y$.

Inductive Case
The last step is justified because of:
 * $\left({x + z^+}\right) = y^+$

which guarantees existence, and by Ordinal Addition is Left Cancellable:
 * $\left({x + w}\right) = \left({x + z^+}\right) \implies w = z^+$

which guarantees uniqueness.

Limit Case
Set $A = \left\{{w : x < w \land w < y}\right\}$.

The above statement shows that $A$ is nonempty.

Then:
 * $\displaystyle \forall w \in A: \exists ! z: \left({x + z}\right) = w$

Create a mapping $F$, that sends each $w \in A$ to the unique $z$ that satisfies $\left({x + z}\right) = w$.

Finally, we must prove that:
 * $\displaystyle \bigcup_{w \mathop \in A} \left({x + F \left({w}\right)}\right) = \left({x + \bigcup_{w \in A} F \left({w}\right)}\right)$

It suffices to prove that $\displaystyle \bigcup_{w \mathop \in A} F \left({w}\right)$ is a limit ordinal.

Let $w = x^+$.

Then $F \left({w}\right) = 1$, and Union of Ordinals is Least Upper Bound.

Thus:
 * $\displaystyle \bigcup_{w \mathop \in A} F \left({w}\right) \ne \varnothing$

Thus:
 * $\displaystyle \bigcup_{w \mathop \in A} F \left({w}\right) \ne z^+$

Therefore $\bigcup_{w \mathop \in A} F \left({w}\right)$ must be a limit ordinal.

To prove uniqueness, assume $x + z = y$.

Then by Ordinal Addition is Left Cancellable:
 * $\displaystyle x + z = x + \bigcup_{w \mathop \in A} F \left({w}\right) \implies z = \bigcup_{w \mathop \in A} F \left({w}\right)$

Also see

 * Definition:Ordinal Subtraction