Set of Prime Numbers is Primitive Recursive

Theorem
The set $\Bbb P$ of prime numbers is primitive recursive.

Proof
A prime number is defined as an element of $\N$ with exactly two positive divisors.

So, we have that $n > 0$ is prime $\map \tau n = 2$, where $\tau: \N \to \N$ is the divisor counting function.

Thus we can define the characteristic function of the set of prime numbers $\Bbb P$ as:
 * $\forall n > 0: \map {\chi_\Bbb P} n := \map {\chi_{\operatorname {eq} } } {\map \tau n, 2}$

Now we let $g: \N^2 \to \N$ be the function given by:
 * $\displaystyle \map g {n, z} = \begin{cases}

0 & : z = 0 \\ \displaystyle \sum_{y \mathop = 1}^z \map {\operatorname {div} } {n, y} & : z > 0 \end{cases}$

As:
 * $\operatorname {div}$ is primitive recursive
 * $\displaystyle \sum_{y \mathop = 1}^z$ is primitive recursive

it follows that $g$ is primitive recursive.

Then for $n > 0$:
 * $\displaystyle \map g {n, n} = \sum_{y \mathop = 1}^n \map {\operatorname {div} } {n, y} = \map \tau n$

and from Divisor Counting Function is Primitive Recursive we have that $g$ is primitive recursive.

Then let $h: \N \to \N$ be the function defined as:
 * $\map h n = \map g {n, n}$

which is also primitive recursive.

So we have, for all $n \in \N$:
 * $\map {\chi_\Bbb P} n = \map {\chi_{\operatorname {eq} } } {\map h n, 2}$

Hence the result.