Counting Theorem

Theorem
Every woset is order isomorphic to a unique ordinal.

Proof

 * First we prove existence.

Let $\left({S, \preceq}\right)$ be a woset.

From Condition for Woset to be Isomorphic to Ordinal‎, it is enough to show that for every $a \in S$, the segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.

Let:
 * $E = \left\{{a \in S: S_a \text{ is not isomorphic to an ordinal}}\right\}$

We will show that $E = \varnothing$.

So, suppose that $E \ne \varnothing$.

Let $a$ be the minimal element of $E$.

This is bound to exist by definition of woset.

So, if $x \prec a$, it follows that $S_x$ is isomorphic to an ordinal.

But for $x \prec a$, we have $S_x = \left({S_a}\right)_x$ from definition of an ordinal.

So every segment of $S_a$ is isomorphic to an ordinal.

Hence from Condition for Woset to be Isomorphic to Ordinal‎, $S_a$ itself is isomorphic to an ordinal.

This contradicts the supposition that $a \in E$.

Hence $E = \varnothing$ and existence has been proved.


 * Uniqueness follows from Isomorphic Ordinals are Equal.

Hence the result.