Ordinal Subtraction when Possible is Unique

Theorem
Let $x$ and $y$ be ordinals such that $x \le y$. Then, there is a unique ordinal $z$ such that $(x+z) = y$.


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

Proof
By transfinite induction on $y$.

Inductive Case
The last step is justified because of $( x + z^+ ) = y^+$, guaranteeing existence, and $( x + w ) = ( x + z^+ ) \implies w = z^+$ by Ordinal Addition is Left Cancellable, guaranteeing uniqueness.

Limit Case
Set $A = \{ w : x < w \land w < y \}$. The above statement shows that $A$ is nonempty.


 * $\displaystyle \forall w \in A: \exists ! z: ( x + z ) = w$

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

Finally, we must prove that $\bigcup_{w \in A} ( x + F(w) ) = ( x + \bigcup_{w \in A} F(w) )$. It suffices to prove that $\bigcup_{w \in A} F(w)$ is a limit ordinal.


 * $\displaystyle \bigcup_{w \in A} F(w) \not = \varnothing$, because if $w = x^+$, then $F(w) = 1$, and Union of Ordinals is Least Upper Bound.


 * $\displaystyle \bigcup_{w \in A} F(w) \not = z^+$, because

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

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

Then, $( x + z ) = ( x + \bigcup_{w \in A} F(w) ) \implies z = \bigcup_{w \in A} F(w)$ by Ordinal Addition is Left Cancellable.