Ordering on Natural Numbers is Trichotomy

Theorem
Let $\N$ be the natural numbers.

Let $<$ be the (strict) ordering on $\N$.

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 $\N$.

Proof
Applying the definition of $<$, the theorem becomes:

Exactly one of the following is true:
 * $(1): \quad a = b$
 * $(2): \quad \exists n \in \N_{>0} : b + n = a$
 * $(3): \quad \exists n \in \N_{>0} : a + n = b$

1 implies not 2 and not 3
Assume that $a = b$.

Seeking a contradiction, assume that:


 * $\exists n \in \N_{>0} : b + n = a$

Or:


 * $\exists n \in \N_{>0} : a + n = b$

Then, by $a = b$, we can merge the two statements to become:


 * $\exists n \in \N_{>0} : a + n = a$

Then, by Natural Numbers under Addition are Cancellable:


 * $n = 0$

which contradicts with the assumption that $n \in \N_{>0}$.

2 implies not 3
Assume that $(2)$ is true, meaning that there exists $m \in \N_{>0}$ such that $b + m = a$.

Seeking a contradiction, assume that there exists $n \in \N_{>0}$ such that $a + n = b$.

Since $n \in \N_{>0}$, by Non-Successor Element of Peano Structure is Unique, there exists $p \in \N$ such that $s \left({p}\right) = n$.

Then, we have:

which is a contradiction by Peano's Axiom $(P4)$: $0$ is not in the image of $s$.