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 1
Applying the definition of $<$, the theorem becomes:

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

We will use the principle of Mathematical Induction.

Let $P \left({a}\right)$ be the proposition that exactly one of the above is true, for all natural numbers $b$, for fixed natural number $a$.

Base case
We will prove that the proposition is true for $a = 0$.

Using Proof by Cases, we divide the proposition into two cases.

$(1)$ is true
It follows trivially from the values of $a$ and $b$.

$(2)$ is false
Aiming for a contradiction, let $c$ be an non-zero natural number such that:
 * $b + c = a$

Substituting the values of $a$ and $b$, we obtain:
 * $0 + c = 0$

By Natural Number Addition Commutes with Zero, we can simplify the equation to:
 * $c = 0$

which is in contradiction to the assumption that $c$ is non-zero.

$(3)$ is false
Aiming for a contradiction, let $c$ be an non-zero natural number such that:
 * $a + c = b$

Substituting the values of $a$ and $b$, we obtain:
 * $0 + c = 0$

By Natural Number Addition Commutes with Zero, we can simplify the equation to:
 * $c = 0$

which is in contradiction to the assumption that $c$ is non-zero.

$(1)$ is false
It follows trivially from the fact that $a = 0$.

$(2)$ is false
Aiming for a contradiction, let $c$ be an non-zero natural number such that:
 * $b + c = a$

Substituting the values of $a$, we obtain:
 * $b + c = 0$

By Non-Successor Element of Peano Structure is Unique, there exists a natural number $d$ such that:
 * $s \left({d}\right) = c$

where $s$ denotes the successor mapping.

Therefore, we have:
 * $b + s \left({d}\right) = 0$

Applying the definition of addition in Peano structure, we get:
 * $s \left({b + d}\right) = 0$

which is in contradiction with $(P4)$$:$ $0$ is not in the image of $s$.

$(3)$ is true
By Natural Number Addition Commutes with Zero, we have:
 * $a + b = b$

And the result follows.

Inductive hypothesis
It is now assumed that the proposition is true for $a = k$, where $k$ is a natural number.

That is, for all natural numbers $b$, exactly one of the following is true:
 * $(1): \quad k = b$
 * $(2): \quad k > b$
 * $(3): \quad k < b$

Then, it is to be proved that the proposition is true for $a = s \left({k}\right)$.

That is, for all natural numbers $b$, exactly one of the following is true:
 * $(1'): \quad s \left({k}\right) = b$
 * $(2'): \quad s \left({k}\right) > b$
 * $(3'): \quad s \left({k}\right) < b$

Proof 2
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 $s$.