I have the following theorem for any formula $\gamma(x,y)$:
Theorem of Transfinite Recursion: Given a well-ordered set $A$ such that for any $f$ there is a unique $y$ such that $\gamma(f,y)$ holds, then there is a unique function $F$ with domain $A$ such that $\gamma\big(F\upharpoonright\text{seg }t,F(t)\big)$ holds for all $t\in A$.
Now I'm reading through the proof of Well-Ordering Theorem $\implies$ Zorn's Lemma, for which the author has considered an arbitrary set $\mathscr A$, which can be endowed with some well-order $<$.
Then he says that the Transfinite Recursion Theorem give us the function $F:\mathscr A\to 2$ such that for every $A\in \mathscr A$
$$F(A)=\begin{cases} 1&\text{if for every } B<A \text{ such that } F(B)=1 \text{ we have } B\subseteq A\\ 0 & \text{otherwise} \end{cases}$$
How can I see that the Theorem of Transfinite Recursion permit this? What would $\gamma(x,y)$ be?
Thanks in advance.
Could it be the following?:
$\gamma(x,y)= \big(y=1 \land \varphi(x)\big)\lor\big(y=0\land \lnot\varphi(x)\big)$
where $\varphi(x)=\forall\,z\;\big(z\in\text{dom}(f)\to z\subseteq h(z)\big)$
where $h:\mathscr A\to\mathscr A$, so $h(z)=\text{the least element of }\mathscr A \text{ such that it's strictly greater than every element of }z$.
I've defined $h$ in such a way so we get $h(\text{seg }t)=t.$