Recursive Function uses One Minimization

Theorem
Every recursive function can be obtained from the basic primitive recursive functions using:
 * substitution
 * primitive recursion
 * at most one minimization on a function.

Proof
Let $f: \N^k \to \N$ be any recursive function.

Consider the minimization operation on the $k + 2$-ary relation $\map \RR {n_1, n_2, \ldots, n_k, y}$:
 * $\mu y \mathrel \RR \tuple {n_1, n_2, \ldots, n_k, y}$

Consider the

From Minimization on Relation Equivalent to Minimization on Function, this is equivalent to:
 * $\map {\mu y} {\map {\overline {\operatorname{sgn} } } {\map {\chi_\RR} {n_1, n_2, \ldots, n_k, y} } = 0}$.

So we can rewrite the statement of Kleene's Normal Form Theorem as:
 * $(1) \quad \map f {n_1, n_2, \ldots, n_k} \approx \map U {\map {\mu z} {\map {\overline {\operatorname{sgn} } } {\map {\chi_\RR} {e, n_1, n_2, \ldots, n_k, z} } = 0} }$.

From the proof of that theorem, we have that $T_k$ is primitive recursive.

Hence from the definition of characteristic function, so is $\chi_{T_k}$.

We also know that $\overline {\operatorname{sgn} }$ is primitive recursive.

We also have by hypothesis that $U$ is primitive recursive.

Hence all of $\chi_{T_k}$, $\overline {\operatorname{sgn} }$ and $U$ can be defined without using minimization.

So the only minimization involved in obtaining the values of $f$ in $(1)$ is the one explicitly mentioned in $(1)$.

Hence the result.