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:
 * $\chi_< \left({n, m}\right) = \operatorname{sgn} \left({m \, \dot - \, n}\right)$

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