Multiplication of Integers is Primitive Recursive

Theorem
Let $t : \N^2 \to \N$ be defined as:
 * $\map t {m, n} = p$

where:
 * $m$ is the code number for the integer $k$
 * $n$ is the code number for the integer $\ell$
 * $p$ is the code number for the integer $k \times \ell$

Then $t$ is a primitive recursive function.

Proof
We have:
 * $\size {k \times \ell} = \size k \times \size \ell$

By: we have that:
 * Multiplication is Primitive Recursive
 * Absolute Value of Integer is Primitive Recursive
 * $\map a {m, n} = \size {k \times \ell}$

is primitive recursive.

Additionally, we have that:
 * $k \times \ell > 0$

either:
 * $k > 0 \land \ell > 0$

or:
 * $k < 0 \land \ell < 0$

By the above statement is a primitive recursive relation.
 * Set of Strictly Positive Integers is Primitive Recursive
 * Set of Strictly Negative Integers is Primitive Recursive
 * Set Operations on Primitive Recursive Relations

Let $c^+ : \N \to \N$ be defined as:
 * $\map {c^+} n = k_{\mathop + n}$

where $k_{\mathop + n}$ is the code number for the integer $+n : \Z$.

Let $c^- : \N \to \N$ be defined as:
 * $\map {c^-} n = k_{\mathop - n}$

where $k_{\mathop - n}$ is the code number for the integer $-n : \Z$.

By both $c^+$ and $c^-$ are primitive recursive.
 * Code Number for Non-Negative Integer is Primitive Recursive
 * Code Number for Non-Positive Integer is Primitive Recursive

We now have:
 * $\map t {m, n} = \begin{cases}

\map {c^+} {\map a {m, n}} & : \paren {k > 0 \land \ell > 0} \lor \paren {k < 0 \land \ell < 0} \\ \map {c^-} {\map a {m, n}} & : \text{otherwise} \end{cases}$ which is primitive recursive by:
 * Definition by Cases is Primitive Recursive/Corollary