Which function results from primitive recursion of the functions $g$ and $h$?
- $f_1=PR(g,h)$ with $g=succ\circ zero_0, h=zero_2$
- $f_2=PR(g,h)$ with $g=zero_0, h=f_1\circ P_1^{(2)}$
- $f_3=PR(g,h)$ with $g=P_1^{(2)}, h=P_2^{(4)}$
- $f_4=PR(g,h)$ 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.
