Cut-Off Subtraction is Primitive Recursive

Theorem

The cut-off subtraction function, defined as:

$\forall \tuple {n, m} \in \N^2: n \mathop {\dot -} m = \begin{cases} 0 & : n < m \\ n - m & : n \ge m \end{cases}$

is primitive recursive‎.

Proof

We see that:

$n \mathop {\dot -} \paren {m + 1} = \begin{cases} 0 & : n \mathop {\dot -} m = 0 \\ \paren {n \mathop {\dot -} m} - 1 & : n \mathop {\dot -} m > 0 \end{cases}$

Hence we can define cut-off subtraction as:

$n \mathop {\dot -} m = \begin{cases} n & : m = 0 \\ \operatorname{pred} \paren {n \mathop {\dot -} \paren {m - 1} } & : m > 0 \end{cases}$

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

Hence the result.

$\blacksquare$

Also see

• Natural Number Subtraction is not Closed

Motivation

The usual subtraction operation is not defined on $\N^2$ for all pairs $\tuple {n, m}$.

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.