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

Theorem
Let $(\prec, A)$ 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


 * $\displaystyle \forall x \in A: \forall y \in B: ( x \prec y \implies x \in B )$. That is, $B$ must be $\prec$-transitive.

Then, $A = B$ or $\exists x \in A: B = A_x$.

Proof
If $A \not = B$, then $B \subsetneq A$. Therefore, by Set Difference with Proper Subset, $A \setminus B \not = \varnothing$.

One direction of inclusion is proven.

By the hypothesis,

But $x \in A \land x \not \in B$, so

Therefore, $B \subseteq A_x$, so $B \subseteq A \cap A_x$.