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 $\mathcal{R} \left({n_1, n_2, \ldots, n_k, y}\right)$:
 * $\mu y \ \mathcal{R} \left({n_1, n_2, \ldots, n_k, y}\right)$

Consider the

From Minimization on Relation Equivalent to Minimization on Function, this is equivalent to:
 * $\mu y \left({\overline{\operatorname{sgn}} \left({\chi_{\mathcal{R}} \left({n_1, n_2, \ldots, n_k, y}\right)}\right) = 0}\right)$.

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

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.