I'm self-learning Transfinite Recursion Theorem and its variants from textbook Introduction to Set Theory by Hrbacek and Jech.
After the authors have presented proofs of theorems 4.9, 4.10, and 4.11, they conclude A parametric version of Theorem 4.11 is straightforward and we leave it to the reader.
I have tried to give it a shot, but I'm not sure if it's fine or not. Does my attempt look fine or contain flaws? I'm very pleased to receive suggestion!
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 $F$ 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 $F$ 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 $F$ 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 $\alpha\neq 0$
My attempt:
We define the class function $G:V\to V$ as follows
$G(x)=G_1(z,\emptyset)$ for $x=(z,0)$
$G(x)=G_2(z,f(\alpha))$ for $x=(z,f)$ where $f$ is a function with $\operatorname{dom}f=\alpha+1\in\operatorname{Ord}$
$G(x)=G_2(z,f)$ for $x=(z,f)$ where $f$ is a function with $\operatorname{dom}f=\alpha$ for a limit $\alpha\in\operatorname{Ord}$
$G(x)=\emptyset$ for $x$ is none of the above
By Theorem 4.10, there exists a class function $F$ such that $F(z,\alpha)=G(z,F_z\restriction \alpha)$ for all $\alpha\in\operatorname{Ord}$. Thus we have
$F(z,0)=G(z,F_z\restriction 0)=G(z,0)=G_1(z,\emptyset)$
$F(z,\alpha+1)=G(z,F_z\restriction \alpha+1)=G_2(z,F_z(\alpha))$ for all $\alpha\in\operatorname{Ord}$
$F(z,\alpha)=G(z,F_z\restriction \alpha)=G_3(z,F_z\restriction\alpha)$ for all limit $\alpha\neq 0$
This completes the proof.