Multiplication is Primitive Recursive

Theorem
The function $\operatorname{mult}: \N^2 \to \N$, defined as:
 * $\map \Mult {n, m} = n \times m$

is primitive recursive‎.

Proof
We observe that:
 * $\map \Mult {n, 0} = n \times 0 = 0$

and that
 * $\map \Mult {n, m + 1} = n \times \paren {m + 1} = \paren {n \times m} + n = \map \Add {\map \Mult {n, m}, n}$.

We are to show that $\Mult$ is obtained by primitive recursion from known primitive recursive‎ functions.

First we note that:
 * $\map \Mult {n, 0} = 0 = \map \Zero n$

where $\map \Zero n$ is the zero function, which is basic primitive recursive‎.

Next we need to find a primitive recursive‎ function $g: \N^3 \to \N$ such that:
 * $\map \Mult {n, m + 1} = \map g {n, m, \map \Mult {n, m} }$

Because $\map \Mult {n, m + 1} = \map \Add {\map \Mult {n, m}, n}$, we see that a suitable function for $g$ is:
 * $\map g {n_1, n_2, n_3} = \map \Add {n_3, n_1}$

Using projection functions, we can write this as:
 * $\map g {n_1, n_2, n_3} = \map \Add {\map {\pr_3^3} {n_1, n_2, n_3}, \map {\pr_1^3} {n_1, n_2, n_3} }$.

This shows that $g$ is obtained by substitution from $\Add$, which is primitive recursive, and the basic primitive recursive‎ functions $\pr_3^3$ and $\pr_1^3$.

So $g$ is primitive recursive‎.

So $\Mult$ is obtained by primitive recursion from the primitive recursive functions‎ $g$ and $\Zero$, and so is primitive recursive‎.