Background
Let all functions under discussion be of the type $(\mathbb{N_0})^n \to \mathbb{N_0}$ where $n\in\mathbb{N_+}$ can vary.
Loosely speaking, an initial function is one of three types of functions: the successor function $s(x) = x+1$, constant functions, and projections onto arguments; these latter two can be of any non-zero arity.
Definition. Primitive recursive functions (PRF-s) are functions which are either initial functions or are constructed from other PRF-s by applying the scheme of function composition or the scheme of primitive recursion a finite number of times.
For $n=1$, the scheme of primitive recursion is as follows. First, $f(0) := k$ is set to be some constant, $k\in\mathbb{N_0}$. Then for all $x\in\mathbb{N_0}$, we set $f(x + 1) := h(x, f(x))$ where $h$ is already a known PRF. Note that $h$ depends on $f(x)$ specifically. This can be generalised for any $n\in\mathbb{N_+}$ if necessary.
The question is also posed for $n=1$ for simplicity.
Question (TL; DR)
What I would like to do is use this slight modification of primitive recursion. The $f(0) := k$ part remains unchanged. Now, for all $x\in\mathbb{N_0}$, let $$f(x + 1) := h(x, f(g(x))),$$
i.e. instead of $f(x)$, $h$ depends on $f(g(x))$. Additionally assume that $g$ is a PRF and always strictly less then the successor of its input: for all $x\in\mathbb{N_0}$, $g(x) < x + 1$. (This is what is meant by "decreasing" in the title).
- Q: How to prove that functions obtained by this kind of modified primitive recursion are also PRFs? Can it be done?
Additional
It makes some intuitive sense to me that it should be a true statement, but I have been unable to prove this rigorously. You may assume I am familiar with basic properties of PRFs (that one may add fictitious variables, rearragne variables; superposition principle), and that some basic functions (sum, product, non-negative difference and so on) are PRFs. But I am not sure these are of any help here.
Hints are also welcome.