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.