Ordinal Addition is Associative

Theorem
Let $x$, $y$, and $z$ be ordinals.


 * $\displaystyle ( x + y ) + z = x + ( y + z )$

Proof
By Transfinite Induction on $z$.

Basis for the Induction
This proves the basis for the induction.

Induction Step
This proves the induction step.

Limit Case
By the hypothesis, we have that $z \in K_{II}$

So it is sufficient to prove that:


 * $\displaystyle \bigcup_{w \in z} ( x + ( y + w ) ) = \bigcup_{n \in ( y + z )} ( x + n )$

Take any $w \in z$. Then, $y + w < y + z$, so setting $n = y + w$, we have that:


 * $\displaystyle x + ( y + w ) \le x + n$

By Supremum Inequality for Ordinals, we have that:


 * $\displaystyle \bigcup_{w \in z} ( x + ( y + w ) ) \le \bigcup_{n \in ( y + z )} ( x + n )$

To prove the other direction, take any $n \in ( y + z )$

So by Equality of Sets, we may conclude:


 * $\displaystyle \bigcup_{n \in ( y + z )} ( x + n ) = \bigcup_{w \in z} ( x + ( y + w ) )$

This proves the limit case.

Also see

 * Natural Number Addition is Associative/Proof 2