# 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.