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.