How do the authors deduce this weaker version of Transfinite Recursion? My question is about a proof of Theorem 4.12 given in the text Introduction to Set Theory by Hrbacek and Jech.
The authors first prove Theorems 4.9, 4.10, and 4.11. For reference, I quote these theorems below.


Let $V$ be the class of all sets, $\operatorname{Ord}$ be the class of all ordinals, and $G:V\to V$ be a class function.

Theorem 4.9: Transfinite Recursion Theorem

There exists a class function $F:\operatorname{Ord}\to V$ such that $F(\alpha)=G(F\restriction \alpha)$ for all $\alpha\in\operatorname{Ord}$.

Theorem 4.10: Transfinite Recursion Theorem, Parametric Version

There exists a class function $F:\operatorname{Ord}\to V$ such that $F(z,\alpha)=G(z,F_z\restriction \alpha)$ for all $\alpha\in\operatorname{Ord}$ and $z\in V$ where $F_z\restriction \alpha:=\{\langle\beta,F(z,\beta)\rangle\mid\beta<\alpha\}$.

Theorem 4.11:

Let $G_1,G_2,G_3$ be class functions from $V$ to $V$. There exists a class function $F:\operatorname{Ord}\to V$ such that
(1) $F(0)=G_1(\emptyset)$
(2) $F(\alpha+1)=G_2(F(\alpha))$ for all $\alpha\in\operatorname{Ord}$
(3) $F(\alpha)=G_3(F\restriction\alpha)$ for all limit $\alpha\neq 0$


Then they say A parametric version of Theorem 4.11 is straightforward and we leave it to the reader. From my attempt, the parametric version of Theorem 4.11 is as follows (Honestly, I'm not sure if my attempt is correct or not)

Let $G_1,G_2,G_3$ be class functions from $V$ to $V$. There exists a class function $F$ such that, for all $z\in V$
$F(z,0)=G_1(z,\emptyset)$
$F(z,\alpha+1)=G_2(z,F_z(\alpha))$ for all $\alpha\in\operatorname{Ord}$
$F(z,\alpha)=G_3(z,F_z\restriction\alpha)$ for all limit $\alpha\neq 0$

Next they make use of this parametric version of Theorem 4.11 in the proof of Theorem 4.12.
Theorem 4.12:

For any set $a$, there is a unique infinite sequence $(f_n\mid n\in \Bbb N)$ such that
(1) $f_0=a$
(2) $f_{n+1}=G(f_n,n)$ for all $n\in \Bbb N$

The proof for Theorem 4.12 given in the text is as follows:

Let $G:V\to V$ be a class function. We want to find, for every set $a$, a sequence $(f_n\mid n\in \Bbb N)$ such that $f_0=a$ and $f_{n+1}=G(f_n,n)$ for all $n\in \Bbb N$. By the parametric version of the Transfinite Recursion Theorem 4.11, there is class function $F$ such that $F(0)=a$ and $F(n+1)=G(F(n),n)$ for all $n\in \Bbb N$. Now we apply the Axiom of Replacement: There exists a sequence $(f_n\mid n\in \Bbb N)$ that is equal to $F\restriction \omega$ and the Theorem follows.


I don't understand how the parametric version of Theorem 4.11 is used to derive the class function $F$ in the proof of Theorem 4.12. Can someone please help me fill in the blanks?
 A: We do not actually need the parametric version of Theorem 4.11. In fact, Theorem 4.11 implies Theorem 4.12. This is what your textbook says. 
Let $G_1$, $G_2$ and $G_3$ is a class function such that
$G_1(0)=(\{(0,a)\}, 0)$, 
$$G_2(x) = \begin{cases}(G(f,n),n+1) & \text{if $x=(f,n)$ for some $n$ and $f$,}\\ \varnothing & \text{otherwise,}\end{cases}$$
and $G_3(x)=\varnothing$. You can see that $F(n) = (f_n,n)$. That is, the sequence of the first component of $F(n)$ is our desired sequence.
A: On the basis of Hanul Jeon's answer. I present a detailed solution here. All credits are given to Hanul Jeon.

We define $G_1,G_2$ as follows
$G_1(x) = \begin{cases}(a, 0) & \text{if } x=0\\ \varnothing & \text{otherwise}\end{cases}$
$G_2(x) = \begin{cases}(G(f,n),n+1) & \text{if $x=(f,n)$ for some $f\in V$ and $n\in\Bbb N$}\\ \varnothing & \text{otherwise}\end{cases}$
By Theorem 4.11, there exists a function $F$ such that $F(0)=G_1(0)$ and $F(\alpha+1)=G_2(F(\alpha))$ for all $\alpha\in\operatorname{Ord}$. It follows that $F(0)=(a,0)$ and $F(n+1)=G_2(F(n))$ for all $n\in\Bbb N$.
We prove by induction that $F(n)$ is of the form $(f_n,n)$ for all $n\in\Bbb N$. It's trivial that $F(0)=(a, 0)$ is of the form $(f_0,0)$. Assume that $F(n)$ is of the form $(f_n,n)$. Then $F(n+1)=G_2(F(n))=$ $G_2(f_n,n)=(G(f_n,n),n+1)=(f_{n+1},n+1)$ where $f_{n+1}=G(f_n,n)$.
Then we extract the sequence $(f_n\mid n\in\Bbb N)$ from $(F(n)\mid n\in\Bbb N)$. It's clear that $f_0=0$ and that $f_{n+1}=G(f_n,n)$ for all $n\in\Bbb N$.
