Ordering Relations are Primitive Recursive

Theorem
The ordering relations on $$\N^2$$: are all primitive recursive.
 * $$n < m$$;
 * $$n \le m$$;
 * $$n \ge m$$;
 * $$n > m$$

Proof
We note that:
 * $$n < m \iff m \, \dot - \, n > 0$$;
 * $$n \ge m \iff m \, \dot - \, n = 0$$.

So it can be seen that the characteristic function of $$<$$ is given by:
 * $$\chi_{<} \left({n, m}\right) = \sgn \left({m \, \dot - \, n}\right)$$.

So $$\chi_{<}$$ is defined by substitution from:
 * the primitive recursive function $\sgn$;
 * the primitive recursive function $\dot -$.

Thus $$\chi_{<}$$ is primitive recursive.

So $$<$$ is a primitive recursive relation.

Next we see that $$n \le m \iff n < m \lor n = m$$ from Strictly Precedes.

From Equality Relation is Primitive Recursive, we have that $$=$$ is primitive recursive.

From above, we have that $$<$$ is primitive recursive.

Thus $$\le$$ is primitive recursive from Set Operations on Primitive Recursive Relations.

We could use the same reasoning for $$>$$ and $$\ge$$ but there's a different approach.

Note that $$n \le m \iff n \not > m$$, and so $$>$$ is primitive recursive from Set Operations on Primitive Recursive Relations.

Finally the same applies to $$\ge$$.

Hence the result.