Condition for Woset to be Isomorphic to Ordinal

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

Let $\left({S, \preceq}\right)$ be such that $\forall a \in S$, the segment $S_a$ of $S$ determined by $a$ is order isomorphic to some ordinal.

Then $\left({S, \preceq}\right)$ itself is order isomorphic to an ordinal.

Proof
For each $a \in S$, let $g_a: S_a \to Z \left({a}\right)$ be an order isomorphism from $S_a$ to an ordinal $Z \left({a}\right)$.

By Isomorphic Ordinals are Equal‎ and Order Isomorphism Between Wosets is Unique, both $Z \left({a}\right)$ and $g_a$ are unique.

So this defines a mapping $Z$ on $S$.

Let the image of $Z$ be $W$:
 * $W = \left\{{Z \left({a}\right): a \in S}\right\}$

Now we define $f: S \to W$ as:
 * $f \left({a}\right) = Z \left({a}\right)$

Now we show that $x, y, \in S, x \prec y \implies Z \left({x}\right) \subset Z \left({y}\right)$.

So, let $x, y, \in S$ such that $x \prec y$. Then:
 * $(1) \qquad g_x: S_x \cong Z \left({x}\right)$

Also, since:

we have:
 * $(2) \qquad g_y \left({S_x}\right): S_x \to \left({Z \left({y}\right)}\right)_{g_y \left({x}\right)}$ is an order isomorphism.

So by Isomorphic Ordinals are Equal, we have:
 * $(3) \qquad Z \left({x}\right) = \left({Z \left({y}\right)}\right)_{g_y \left({x}\right)}$.

So, in particular, $Z \left({x}\right) \subset Z \left({y}\right)$.

By this result, $f: X \to W$ is a bijection.

Also by this result, $f: X \to \left({W, \subseteq}\right)$ is an order isomorphism.

This means that $W$ is well-ordered by $\subseteq$.

Now we show that $W$ is an ordinal.

Let $y \in S$.

Since $Z \left({y}\right)$ is an ordinal, we have:
 * $x \prec y \implies \left({Z \left({y}\right)}\right)_{g_y \left({x}\right)} = g_y \left({x}\right)$.

So by $(3)$ above, we have:
 * $(4) \qquad x \prec y \implies Z \left({x}\right) = g_y \left({x}\right)$.

Hence:

As $y$ is an arbitrary element of $S$, it follows that $Z \left({y}\right)$ is an arbitrary element of $W$.

So $W$ is an ordinal, as we wanted to prove.