Isomorphic Ordinals are Equal

Theorem

Let $A$ and $B$ be ordinals that are order isomorphic.

Then $A = B$.

Proof 1

Let $S \cong T$.

Aiming for a contradiction, suppose that $S \ne T$.

Then from the corollary to Relation between Two Ordinals, either:

$S$ is an initial segment of $T$

or:

$T$ is an initial segment of $S$.

But as $S \cong T$, from Well-Ordered Class is not Isomorphic to Initial Segment, neither $S$ nor $T$ can be an initial segment of the other.

From this contradiction it follows that $S = T$.

$\blacksquare$

Proof 2

From Well-Ordered Class is not Isomorphic to Initial Segment, neither $A$ nor $B$ can be an initial segment of the other.

By definition, every element of an ordinal is an initial segment of it.

Hence, neither $A$ nor $B$ can be an element of the other.

By Ordinal Membership is Trichotomy, it follows that $A = B$.

$\blacksquare$