Ordinals are Totally Ordered

Theorem
Let $A$ and $B$ be ordinals.

Then either $A \subseteq B$ or $B \subseteq A$.

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

That is, by De Morgan's laws:
 * $\left({\neg \left({A \subseteq B}\right)}\right) \land \left({\neg \left({B \subseteq A}\right)}\right)$

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

Now from Intersection Subset, we have:
 * $A \cap B \subset A$
 * $A \cap B \subset B$

Note that by Intersection with Subset is Subset, equality does not hold.

Because the intersection of two ordinals is an ordinal‎, $A \cap B$ is an ordinal.

So by Ordinal Proper Subset Membership, we have:
 * $A \cap B \in A$
 * $A \cap B \in B$

But then $A \cap B \in A \cap B$, which contradicts the definition that the $\in$-relation is the strict well-ordering on (the ordinal) $A \cap B$.