User:EntropyReversal/Sandbox

= Prime Enumeration Function is Primitive Recursive =

This is a sandbox edit of Prime.mover's article Prime Enumeration Function is Primitive Recursive so I can learn formatting and experiment with the wording.

Short Version
The prime enumeration function is primitive recursive.

Long Version
If the prime enumeration function is function $p: \N \to \N$ such that:
 * $p \left({n}\right) = \begin{cases}

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

Then the prime enumeration function is primitive recursive.

Main Proof
If the prime enumeration function is function $p$

Then we can define $p: \N \to \N$ such that:
 * $p \left({n}\right) = \begin{cases}

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

If we define $p$ as above

Then $p$ can be defined by primitive recursion from a constant and two primitive recursive functions (see subproof below).

If $p$ can be defined by primitive recursion from a constant and two primitive recursive functions

Then $p$ is primitive recursive (by definition).

Therefore...

If the prime enumeration function is $p$

Then $p$ is primitive recursive.

Therefore...

The prime enumeration function is primitive recursive.

Claim
Function $p$ can be defined by primitive recursion from a constant and two primitive recursive functions.

Proof
Function $p$ can be defined by primitive recursion from a constant and two primitive recursive functions

Because...

Function $p$ can be defined by primitive recursion from the constant $1$ and the primitive recursive functions $g$ and $h$

Because...

We can define $p$ recursively by:


 * $p \left({n + 1}\right) = \text{the smallest } y \in \N \text { such that } y \text { is prime and } p \left({n}\right) \le y $.

and conclude:


 * $p \left({0}\right) = 1$


 * $p \left({n+1}\right) = g \left({p \left({n}\right), h \left({n}\right)}\right)$

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

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

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

where:
 * $\chi_\Bbb P \left({y}\right)$ is the characteristic function of the set of prime numbers $\Bbb P$


 * $\mu y \left({\mathcal R}\right)$ means the smallest $y \in \N$ such that the relation $\mathcal R$ holds.

Now consider the relation $\mathcal S$ given by:
 * $\mathcal S \left({m, y}\right) \iff \chi_\Bbb P \left({y}\right) = 1$.

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

Then we have:
 * $\chi_\mathcal S \left({m, y}\right) = \chi_{\operatorname{eq}} \left({\chi_\Bbb P \left({y}\right), 1}\right)$.

So $\chi_\mathcal S$ 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 $\mathcal R$ by:
 * $\mathcal R \left({m, y}\right) \iff \mathcal S \left({m, y}\right) \land m < y \iff \chi_\Bbb P \left({y}\right) = 1 \land m < y$.

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:
 * $g \left({m, z}\right) = \mu y \le z \left({\chi_\Bbb P \left({y}\right) = 1 \land m < y}\right)$

which is primitive recursive by Bounded Minimization is Primitive Recursive.

We note that $g \left({p \left({n}\right), z}\right) = p \left({n + 1}\right)$ as long as $p \left({n + 1}\right) \le z$.

Next, let $h: \N \to \N$ be defined as $h \left({n}\right) = \exp \left({2, \exp \left({2, n}\right)}\right)$.

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 $p \left({n+1}\right) \le 2^{2^n} = h \left({n}\right)$.

It follows that:
 * $p \left({n+1}\right) = g \left({p \left({n}\right), h \left({n}\right)}\right)$

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

So using the definition of $p$ as given above, we have:
 * $p \left({0}\right) = 1$


 * $p \left({n+1}\right) = g \left({p \left({n}\right), h \left({n}\right)}\right)$.

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 $p \left({n+1}\right)$ because it is convenient in this context. Smaller ones would of course do the same job.