Ordinal Membership is Trichotomy

Theorem
Let $A$ and $B$ be ordinals. Then:
 * $\left({A = B}\right) \lor \left({A \in B}\right) \lor \left({B \in A}\right)$

where $\lor$ denotes logical or.

Proof
Becase the ordinals are totally ordered, it follows that:
 * $\left({A = B}\right) \lor \left({A \subset B}\right) \lor \left({B \subset A}\right)$

By Ordinal Proper Subset Membership, the result immediately follows.

Source

 * : $\S 7.10$