Counting Theorem/Proof 2

Proof
Let $A$ be a properly well-ordered class.

Let $\On$ denote the class of all ordinals.

By the Comparability Theorem, either:
 * $A$ is order isomorphic to a lower section of $\On$

or:
 * $\On$ is order isomorphic to a lower section of $A$.

Let $A$ be a set.

From Well-Ordering on Set is Proper Well-Ordering, $A$ is a properly well-ordered class.

As $A$ is a set, every lower section of $A$ is a set.

$\On$ is order isomorphic to a lower section $L$ of $A$.

Then $\On$ would then be in one-to-one correspondence with that lower section.

Thus there would be a mapping $\phi: L \to \On$.

By the Axiom of Replacement, $\phi \sqbrk L = \On$ is then a set.

But from Class of All Ordinals is Proper Class, $\On$ is a proper class.

From this contradiction it follows that $\On$ cannot be order isomorphic to a lower section $L$ of $A$.

Hence $A$ is order isomorphic to a lower section of $\On$.

From Lower Section of Class of All Ordinals is Ordinal, that means $A$ is order isomorphic to an ordinal.

From Distinct Ordinals are not Order Isomorphic it follows that $A$ is order isomorphic to exactly one such ordinal.