Prime Enumeration Function is Primitive Recursive

Theorem
Let the function $p: \N \to \N$ be the prime enumeration function, defined as:
 * $\map p n = \begin{cases}

1 & : n = 0 \\ \mbox{the } n \mbox{th prime number} & : n > 0 \end{cases}$

Then $p$ is primitive recursive.

Proof
We can define $p$ recursively by:
 * $\map p {n + 1} = \text{the smallest } y \in \N \text { such that } y \text { is prime and } \map p n \le y$

Hence we can express it as:
 * $\map p {n + 1} = \map {\mu y} {\map {\chi_\Bbb P} y = 1 \land \map p n \le y}$

where:
 * $\map {\chi_\Bbb P} y$ is the characteristic function of the set of prime numbers $\Bbb P$


 * $\map {\mu y} \RR$ means the smallest $y \in \N$ such that the relation $\RR$ holds.

Now consider the relation $\SS$ given by:
 * $\map \SS {m, y} \iff \map {\chi_\Bbb P} y = 1$.

We have a reason for making $\SS$ a binary relation, even though $m$ is not actually invoked in its definition.

Then we have:
 * $\map {\chi_\SS} {m, y} = \map {\chi_{\operatorname{eq} } } {\map {\chi_\Bbb P} y, 1}$.

So $\chi_\SS$ is primitive recursive as it is obtained by substitution from:
 * the primitive recursive function $\chi_{\operatorname{eq} }$


 * the primitive recursive function $\chi_\Bbb P$.

Then we have that $<$ is primitive recursive.

So we define the relation $\RR$ by:
 * $\map \RR {m, y} \iff \map \SS {m, y} \land m < y \iff \map {\chi_\Bbb P} y = 1 \land m < y$.

This is primitive recursive from the above and Set Operations on Primitive Recursive Relations.

Now let us define the function $g: \N^2 \to \N$ as:
 * $\map g {m, z} = \mu y \le \map z {\map {\chi_\Bbb P} y = 1 \land m < y}$

which is primitive recursive by Bounded Minimization is Primitive Recursive.

We note that $\map g {\map p n, z} = \map p {n + 1}$ as long as $\map p {n + 1} \le z$.

Next, let $h: \N \to \N$ be defined as $\map h n = \map \exp {2, \map \exp {2, n} }$.

Then $h$ is primitive recursive since it is obtained by substitution from:
 * the primitive recursive function $\exp$;
 * the constant $2$.

From Upper Bounds for Prime Numbers, we have $\map p {n + 1} \le 2^{2^n} = \map h n$.

It follows that:
 * $\map p {n + 1} = \map g {\map p n, \map h n}$

where $g$ and $h$ are both primitive recursive.

So using the definition of $p$ as given above, we have:
 * $\map p 0 = 1$


 * $\map p {n + 1} = \map g {\map p n, \map h n}$.

So $p$ is defined by primitive recursion from the constant $1$ and the primitive recursive functions $g$ and $h$.

Note
Note that we use the extravagantly large upper bound $2^{2^n}$ for the prime number $\map p {n + 1}$ because it is convenient in this context. Smaller ones would of course do the same job.