In this proof of recursion theorem, why $h_{0}$ is a function by construction?

Question

In L.J.Halbeisen's Combinatorial Set Theory :

With the Axiom Schema of Separation we can define the following set:

$H = \{h\in\ ^{\omega}\mathscr{P}(A):h(0)=C \land\forall n\in \omega(h(n+1) = g(h(n)))\}$

... In order to show that $H$ contains at least one function $h_{0}$ we proceed as follows:
Let $h_{0} \subseteq \omega\times\mathscr{P}(A)$ be the set of ordered pairs $\langle n,D\rangle$ for which we have:

• $\langle 0,C \rangle\in h_{0}$
• $\forall n\forall D \forall n'(\langle n,D\rangle\in h_{0} \land n'\lt n \to \exists D'(\langle n',D'\rangle\in h_{0}))$
• $\forall n\forall D \forall n'\forall D'(\langle n,D\rangle\in h_{0} \land \langle n',D'\rangle\in h_{0} \land n'= n+1 \to D'=g(D))$

It remains to show that $h_{0} \in \ ^{\omega}\mathscr{P}(A)$: By construction, $h_{0}=C$. Furthermore, if $h_{0}(0)=D$, for some $n\in\omega$ and $D\subseteq A$, then, by construction, $h_{0}(n+1) = g(D)$. So, by induction on $n$ we get $h_{0}: \omega \to \mathscr{P}(A) $ and $h_{0}\in H$.

I think it is a proof of recursion theorem(to make sure $h$ exists and is unique by this recursive definition). But I do not understand why $h_{0}$ is a function on $\omega$.
In particular, I am confused with the second condition of $h_{0}$, which requires that $\forall n'( (n'\lt n)\to \langle n',D\rangle$ is defined). Only given $\langle 0,C\rangle$, how to prove there is$ \langle 1,D\rangle$ and other $n' > n$ is in $h_{0}$?