Isomorphic Ordinals are Equal/Proof 1
Then $A = B$.
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 $T$
- $T$ 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 $S = T$.
- 1993: Keith Devlin: The Joy of Sets: Fundamentals of Contemporary Set Theory (2nd ed.) ... (previous) ... (next): $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.10$