Relation between Two Ordinals

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

Then either $S \subseteq T$ or $T \subseteq S$.

Proof
Aiming for a contradiction, suppose that the claim is false.

That is, by De Morgan's laws: Conjunction of Negations:
 * $\left({\neg \left({S \subseteq T}\right)}\right) \land \left({\neg \left({T \subseteq S}\right)}\right)$

where $\neg$ denotes logical not and $\land$ denotes logical and.

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

Also, by Intersection with Subset is Subset, we have $S \cap T \ne S$ and $S \cap T \ne T$.

That is, $S \cap T \subsetneq S$ and $S \cap T \subsetneq T$.

Note that by Intersection of Two Ordinals is Ordinal‎, $S \cap T$ is an ordinal.

So by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, we have:
 * $S \cap T \in S$
 * $S \cap T \in T$

But then $S \cap T \in S \cap T$, which contradicts the fact that the $\in$-relation is, by the definition of an ordinal, the strict well-ordering on (the ordinal) $S \cap T$.