# Addition is Primitive Recursive

## Theorem

The function $\Add: \N^2 \to \N$, defined as:

- $\map \Add {n, m} = n + m$

## Proof

We observe that:

- $\map \Add {n, 0} = n + 0 = n$

and that

- $\map \Add {n, m + 1} = n + \paren {m + 1} = \paren {n + m} + 1 = \map \Succ {\map \Add {n, m} }$

where $\Succ$ is the successor function, which is a basic primitive recursive function.

We are to show that $\Add$ is defined by primitive recursion.

So we need to find primitive recursive functions $f: \N \to \N$ and $g: \N^3 \to \N$ such that:

- $\map \Add {n, m} = \begin{cases} \map f n & : m = 0 \\ \map g {n, m - 1, \map \Add {n, m - 1} } & : m > 0 \end{cases}$

Because $\map \Add {n, 0} = n$, we can see that:

- $\map f n = n$.

That is, $f$ is the basic primitive recursive function $\pr_1^1: \N \to \N$.

Because $\map \Add {n, m + 1} = \map \Succ {\map \Add {n, m} }$, we can see that $g$ needs to be a function which maps the ordered triple $\tuple {n, m, \map \Add {n, m} }$ to $\map \Succ {\map \Add {n, m} }$.

A suitable function is:

- $g: \N^3 \to \N$ where $\map g {n_1, n_2, n_3} = \map \Succ {n_3}$

because then:

- $\map g {n, m, \map \Add {n, m} } = \map \Succ {\map \Add {n, m} }$

It remains to be shown that $g$ is primitive recursive.

This is achieved by using the projection function:

- $\map g {n_1, n_2, n_3} = \map \Succ {\map {\pr_3^3} {n_1, n_2, n_3} }$

Thus $g$ is obtained by substitution from the basic primitive recursive functions $\Succ$ and $\pr_3^3$ and so is primitive recursive.

So $\Add$ is obtained by primitive recursion from the primitive recursive functions $\pr_1^1$ and $g$, and so is primitive recursive.

$\blacksquare$