Which function results from primitive recursion of the functions g and h?



Which function results from primitive recursion of the functions $ g$ and $ h$ ?

  1. $ f_1=PR(g,h)$ with $ g=succ\circ zero_0, h=zero_2$
  2. $ f_2=PR(g,h)$ with $ g=zero_0, h=f_1\circ P_1^{(2)}$
  3. $ f_3=PR(g,h)$ with $ g=P_1^{(2)}, h=P_2^{(4)}$
  4. $ f_4=PR(g,h)$ with $ g=f_3\left(f_1(x),succ(x),f_2(x)\right)$

(1.) $ g:N^0\to N$ , $ h:N^2\to N$
$ f(0)=1$
$ f(0+1)=h(0,f(0))=h(0,1)=0$
$ f(1+1)=h(1,f(1))=h(1,0)=0$
$ \forall n\in N_{>0}:f(n+1)=h(n,f(n))=0$ , $ f_1$ is defined as $ f_1:N^1\to N$ with $ f_1(x)=\begin{cases}1, x=0\ 0, x>0\end{cases}$

(2.) $ g:N^0\to N$ , $ h:N^2\to N$
$ f(0)=0$
$ f(0+1)=h(0,f(0))=h(0,0)=1$ $ f(1+1)=h(1,f(1))=h(1,1)=0$ $ \forall n\in N_{>0}: f(n+1)=h(n,f(n))=0$ , $ f_2$ is defined the same as $ f_1$ , $ f_1(x)=f_2(x)$

(3.) $ g:N^2\to N$ , $ h:N^4\to N$
$ f(x,y,0)=x$
$ f(x,y,0+1)=h(x,y,0,f(x,y,0))=h(x,y,0,x)=y$ $ f(x,y,1+1)=h(x,y,1,f(x,y,1))=h(x,y,1,y)=y$ $ \forall z \in N_{>0}: f(x,y,z+1)=h(x,y,z,f(x,y,z))=y$ , $ f_3$ is defined as $ f_3:N^3\to N$ with $ f_3(x,y,z)=\begin{cases}x, z=0\ y, z>0\end{cases}$

Is this correct up to here? It looks way too easy, that’s why I’m not sure.