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 $\paren {x + z} = y$.

That is:
 * $x \le y \implies \exists! z \in \On: \paren {x + z} = y$

Proof
By transfinite induction on $y$.

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

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

which guarantees uniqueness.

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

The above statement shows that $A$ is nonempty.

Then:
 * $\forall w \in A: \exists ! z: \paren {x + z} = w$

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

Finally, we must prove that:
 * $\displaystyle \bigcup_{w \mathop \in A} \paren {x + \map F w} = \paren {x + \bigcup_{w \in A} F \map F w}$

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

Let $w = x^+$.

Then $\map F w = 1$, and Union of Ordinals is Least Upper Bound.

Thus:
 * $\displaystyle \bigcup_{w \mathop \in A} \map F w \ne \O$

Thus:
 * $\displaystyle \bigcup_{w \mathop \in A} F \map F w \ne z^+$

Therefore $\bigcup_{w \mathop \in A} \map F w$ must be a limit ordinal.

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

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

Also see

 * Definition:Ordinal Subtraction