Ordering on Natural Numbers is Trichotomy

Theorem
Let $\N_{> 0}$ be the 1-based natural numbers:
 * $\N_{> 0} = \left\{{1, 2, 3, \ldots}\right\}$

Let $<$ be the (strict) ordering on $\N_{> 0}$ defined as Ordering on Natural Numbers:


 * $\forall a, b \in \N_{>0}: a < b \iff \exists c \in \N_{>0}: a + c = b$

Then exactly one of the following is true:
 * $(1): \quad a = b$
 * $(2): \quad a > b$
 * $(3): \quad a < b$

That is, $<$ is a trichotomy on $\N_{> 0}$.

Proof
Using the following axioms:

Axiom $E$ states:
 * $\forall a, b \in \N_{>0}$, either:
 * $a = b$, in which case $(1)$ holds
 * $\exists x \in \N_{> 0}: a = b + x$, in which case, by definition of the ordering defined, $a > b$, in which case $(2)$ holds
 * $\exists x \in \N_{> 0}: a + x = b$, in which case, by definition of the ordering defined, $a < b$, in which case $(3)$ holds.

Hence the result.