Substitution of Constant yields Primitive Recursive Function

Theorem
Let $f: \N^{k+1} \to \N$ be a primitive recursive function‎.

Then $g: \N^k \to \N$ given by:
 * $g \left({n_1, n_2, \ldots, n_k}\right) = f \left({n_1, n_2, \ldots, n_{i-1}, a, n_i \ldots, n_k}\right)$

is primitive recursive.

Proof
Let $n = \left({n_1, n_2, \ldots, n_{i-1}, n_i \ldots, n_k}\right)$.

We see that:
 * $g \left({n_1, n_2, \ldots, n_k}\right) = f \left({\operatorname{pr}^k_1 \left({n}\right), \operatorname{pr}^k_2 \left({n}\right), \ldots, \operatorname{pr}^k_{i-1} \left({n}\right), f_a \left({n}\right), \operatorname{pr}^k_i \left({n}\right), \ldots, \operatorname{pr}^k_k \left({n}\right)}\right)$

We have that:
 * $\operatorname{pr}^k_j$ is a basic primitive recursive‎ function for all $j$ such that $1 \ne j \le k$
 * $f_a$ is a primitive recursive‎ function.

So $g$ is obtained by substitution from primitive recursive‎ functions and so is primitive recursive‎.