Well-Ordered Transitive Subset is Equal or Equal to Initial Segment

Theorem
Let $\left({\prec, A}\right)$ be a well-ordered set.

For every $x \in A$, let every $\prec$-initial segment $A_x$ be a set.

Let $B$ be a subclass of $A$ such that


 * $\forall x \in A: \forall y \in B: \left({x \prec y \implies x \in B}\right)$.

That is, $B$ must be $\prec$-transitive.

Then:
 * $A = B$

or:
 * $\exists x \in A: B = A_x$

Proof
Let $A \ne B$.

Then $B \subsetneq A$.

Therefore, by Set Difference with Proper Subset:
 * $A \setminus B \ne \varnothing$

Then:

One direction of inclusion is proven.

By the hypothesis:

But $x \in A \land x \notin B$, so:

Therefore:
 * $B \subseteq A_x$

and so $B:
 * \subseteq A \cap A_x$