Relation between Two Ordinals

Theorem
Let $$S, T$$ be ordinals.

If $$S \ne T$$ then one is a 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.