Set of Prime Numbers is Primitive Recursive

Theorem
The set $$\mathbb{P}$$ 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 iff $$\tau \left({n}\right) = 2$$, where $$\tau: \N \to \N$$ is the tau function.

Thus we can define the characteristic function of the set of prime numbers $$\mathbb{P}$$ as:
 * $$\forall n > 0: \chi_{\mathbb{P}} \left({n}\right) = \chi_{\operatorname{eq}} \left({\tau \left({n}\right), 2}\right)$$.

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

0 & : z = 0 \\ \sum_{y = 1}^z \operatorname{div} \left({n, y}\right) & : z > 0 \end{cases}$$

Since: then $$g$$ is primitive recursive.
 * $\operatorname{div}$ is primitive recursive;
 * $\sum_{y = 1}^z$ is primitive recursive,

Then for $$n > 0$$:
 * $$g \left({n, n}\right) = \sum_{y = 1}^n \operatorname{div} \left({n, y}\right) = \tau \left({n}\right)$$

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

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

which is also primitive recursive.

So we have, for all $$n \in \N$$:
 * $$\chi_{\mathbb{P}} \left({n}\right) = \chi_{\operatorname{eq}} \left({h \left({n}\right), 2}\right)$$.

Hence the result.