Cut-Off Subtraction is Primitive Recursive

Theorem
The cut-off subtraction function, defined as:
 * $\forall \left({n, m}\right) \in \N^2: n \, \dot - \, m = \begin{cases}

0 & : n < m \\ n - m & : n \ge m \end{cases}$ is primitive recursive‎.

Proof
We see that: $n \, \dot - \, \left({m + 1}\right) = \begin{cases} 0 & : n \, \dot - \, m = 0 \\ \left({n \, \dot - \, m}\right) - 1 & : n \, \dot - \, m > 0 \end{cases}$

Hence we can define cut-off subtraction as:
 * $n \, \dot - \, m = \begin{cases}

n & : m = 0 \\ \operatorname{pred} \left({n \, \dot - \, \left({m - 1}\right)}\right) & : m > 0 \end{cases}$

This is a definition by primitive recursion from the primitive recursive function $\operatorname{pred}$.

Hence the result.

Comment
The usual subtraction operation is not defined on $\N^2$ for all pairs $\left({n, m}\right)$.

If $m > n$, then $n - m$, although well-defined for the integers $\Z$, has no definition in the set of natural numbers $\N$ which go no lower than $0$.

Hence the need to define this hybrid operation.