Ordinal is Member of Class of All Ordinals

Theorem
Suppose $A$ is an ordinal.

Then:
 * $A \in \operatorname{On} \lor A = \operatorname{On}$

Proof
Since Ordinal Class is Ordinal, and $A$ is an ordinal, and Ordinal Membership is Trichotomy, then $( A \in \operatorname{On} \lor A = \operatorname{On} \lor \operatorname{On} \in A )$. But $\operatorname{On}$ is a proper class by the Burali-Forti Paradox, so $( A \in \operatorname{On} \lor A = \operatorname{On}$ ).