Relation between Two Ordinals

Theorem
Let $S, T$ be ordinals.

If $S \ne T$ then one is an initial segment of the other.

Proof
If either $S \subset T$ or $T \subset S$ then we invoke Ordinal Subset of Ordinal is Segment, and we have finished.

So, suppose $S \not \subset T$ and $T \not \subset S$.

Now from Intersection Subset, we have $S \cap T \subset T$ and $S \cap T \subset S$.

By Intersection of Two Ordinals is Ordinal‎, $S \cap T$ is an ordinal.

So by Ordinal Subset of Ordinal is Segment, we have:
 * $S \cap T = S_a$ for some $a \in S$
 * $S \cap T = S_b$ for some $b \in T$

Then:
 * $a = S_a = S \cap T = T_b = b$

But $a \in S, b \in T$.

Thus $a = b = S \cap T$.

But $S \cap T = S_a$, so:
 * $x \in S \cap T \implies x \subset a$

In particular, this means $a \subset a$, which is a contradiction.

So either $S \subset T$ or $T \subset S$, and again we invoke Ordinal Subset of Ordinal is Segment, and the proof is complete.