Ordering on 1-Based Natural Numbers is Total Ordering

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 $<$ is a (strict) total ordering.

Proof
From Ordering on Natural Numbers is Trichotomy we have that $<$ is trichotomy.

From Ordering on Natural Numbers is Transitive we have that $<$ is transitive.

From Trichotomy is Irreflexive it follows that $<$ is irreflexive.

It follows by definition that $<$ is a strict ordering.

By the trichotomy law it follows that $<$ is a strict total ordering.