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:
 * $$\mu y \left({\overline{\sgn} \left({\chi_{\mathcal{R}} \left({n_1, n_2, \ldots, n_k, y}\right)}\right) = 0}\right)$$.

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

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 $\overline{\sgn}$ is primitive recursive.

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

Hence all of $$\chi_{T_k}$$, $$\overline{\sgn}$$ 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.