Assume $f\colon ω × ω → ω$ is a computable function. How can we prove that there is a primitive recursive function $g\colon ω × ω → ω$ where the following holds:
$∀n [∃s(f(n, s) = 1) ↔ ∃k(g(n, k) = 1)]$
So for every $n$, there is an $s$ such that $f(n, s) = 1$ if and only if there is a $k$ such that $g(n, k) = 1$.
Been working on this problem for a while now, if anyone could please help?