I am studying a Theory of Computability.
Before the theorem I should provide you with the definition of limited sum and product, respectfully:
$\alpha(x,y)=\sum_{z<y}f(x,z)$ which is defined with relations:
$\alpha(x,0)=0$ and $\alpha(x,y+1)=\alpha(x,y)+f(x,y)$
and now product:
$\beta(x,y)=\sqcap_{z<y}f(x,z)$ which is defined with relations:
$\beta(x,0)=1$ and $\beta(x,y+1)=\beta(x,y)*f(x,y)$
Given the following theorem:
Theorem: Let $f(x,y)$ be a total computable function. Then the functions $\alpha(x,y)$ and $\beta(x,y)$ are also computable.
Proof for $\alpha$:
We note that the functions $\alpha$ and $\beta$ are obtained through primitive recursion from computable functions, hence they are computable.
We have: $\alpha(x,0)=0=0(x)$
$0(x)$ is a computable function.
Further: $\alpha(x,y+1)=\alpha(x,y)+f(x,y)=g(x,y,\alpha(x,y))$ where:
$g(x,y,z)=z+f(x,y)=\cup_{n+2}^{n+2} (x,y,z)+f(x,y)$ is computable as a sum of computable functions.
Now, I want to prove the computability of $\beta$ analogously to $\alpha$. My question is if this approach is good and is there anything that is incorrect or missing?
$\beta(x,0)=1$, which is a computable function since it is constant.
Further: $\beta(x,y+1)=\beta(x,y)*f(x,y)=g(x,y,\beta(x,y))$ where:
$g(x,y,z)=z*f(x,y)=?$ I am kinda stuck here...
