Ordinal Membership is Trichotomy

Theorem
Let $A$ and $B$ be ordinals. Then $( A \in B \lor A = B \lor B \in A )$.

Proof
By Relation between Unequal Ordinals, $( A \not = B \implies ( A \subset B \lor B \subset A ) )$.

By Modus Tollendo Ponens, this is equivalent to $(A = B) \lor \left( A \subset B \lor B \subset A \right)$.

Therefore, $( A \subset B \lor A = B \lor B \subset A )$. By Ordinal Proper Subset Membership, $( A \in B \lor A = B \lor B \in A )$.

Source

 * :$7.10$