The Recursion Theorem (Set Theory) In the book 'Introduction to set theory' by Hrbacek and Jech, there is this theorem stated in the book:

Then in the proof, there is this part:

I don't understand the induction part. We are trying to prove that for all $n \in \mathbb{N}$, there exists an $n$-step computation. I am a bit confused what the author is trying to do.
 A: The author is using induction.  It may be unfortunate that $t$ is reused.  Rewrite the line after Clearly as $t(0)=\{(0,a)\}$ is a $0-$step computation-it is a function with domain $0$.  Now assume $t(n)$ is an $n-$step computation-a function with domain $[0,n]$.  This will assign values to all the naturals up to $n$.  We wish to extend it to a function that assigns values to all the naturals up to $n+1$.  We make it agree with the previous function on $[0,n]$, then add a value at $n+1$, which needs to be $g(t(n),n)=t(n+1)$  Now we have a function with domain $[0,n+1]$ that meets the requirement.  Since each extension was uniquely determined, there is a unique function generated.
A: As Ross Millikan pointed out Jech is proving by induction.
Following is an approach for $A = \mathbb{N}$:
Given a function $g: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N}$ and a constant $a \in \mathbb{N}$. Show that there is a unique function $f:\mathbb{N} \rightarrow \mathbb{N}$ such that
$$ (\exists! f:\{ n \in \mathbb{N}: n \le N \} \rightarrow \mathbb{N}) \: \underbrace{f(0) = a}_{(i)} \wedge \underbrace{(\forall n \in \mathbb{N} : n < N) \: f(n+1) = g(f(n),n)}_{(ii)}.$$
holds for each $N \in \mathbb{N}$. 

"$N = 0$": $f(0) = a$ is a function from $\{0\}$ to $\mathbb{N}$ and satisfies property (i). Property (ii) holds vacuously since there are no natural numbers which are smaller than 0. The function $f$ is unique since if there are two functions $f', f$ such that $f'(0) = f(0) = a$, they have to be equal. 

"$N \rightarrow N+1$": Assume the theorem holds for $N$ and define a new object $\tilde{f}$ s.t.
$$ \tilde{f}(n) = 
\begin{cases}
f(n) & \text{if } n \le N \\
g(f(N), N) & \text{if } n = N+1.
\end{cases}
 $$
It turns out that $\tilde{f}$ is a function from $\{ n \in \mathbb{N} : n \le N+1 \}$ to $\mathbb{N}$, since for each $n$ in the domain there's exactly one $\tilde{f}(n) \in \mathbb{N}$: By the induction hypothesis this is given for $n=0$ up to $N$. For $g$ being a function, it also holds for $n = N+1$. Uniqueness: the induction hypothesis requires that $f(n):\{n \in \mathbb{N} : n \le N \} \rightarrow \mathbb{N}$ is unique whereas if there are two function $g, g': \{ (f(N),N) \} \rightarrow \mathbb{N}$ such that $g(f(N),N) = g'(f(N),N) = \tilde{f}(N+1)$ they are equal. Finally, the induction hypothesis implies (i) and (ii) for $n < N$, where $\tilde{f}(N+1) = g(f(N),N)$ satisfies (ii) if $n < N+1$.
