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:
 * $(1): \quad a = b$
 * $(2): \quad \exists n \in \N_{>0} : b + n = a$
 * $(3): \quad \exists n \in \N_{>0} : a + n = b$

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 a 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:
 * $(1'): \quad s \left({k}\right) = b$
 * $(2'): \quad s \left({k}\right) > b$
 * $(3'): \quad s \left({k}\right) < b$

Inductive step
We have:

Case 1: $k = b$
In this case:

$(1')$ is false
Aiming for a contradiction, let $(1')$ be true instead.

Then:

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

$(2')$ is true
This is apparent from the definition of $>$.

$(3')$ is false
Aiming for a contradiction, let $(3')$ be true instead.

Let $c$ be a non-zero natural number such that:
 * $s \left({k}\right) + c = b$

Then:

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

$(1')$ is true
This follows from the assumption.

$(2')$ is false
Aiming for a contradiction, let $(2')$ be true instead.

Let $c$ be a non-zero natural number such that:
 * $s \left({k}\right) = b + c$

Then:

which is in contradiction with our assumption that $c$ is non-zero.

$(3')$ is false
Aiming for a contradiction, let $(3')$ be true instead.

Let $c$ be a non-zero natural number such that:
 * $s \left({k}\right) + c = b$

Then:

which is in contradiction with our assumption that $c$ is non-zero.

Case 3: $k < b$ and $s \left({k}\right) \ne b$
From the definition of $<$, we have the existence of a non-zero natural number $m$ such that:
 * $k + m = b$

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

Substituting:
 * $k + s \left({p}\right) = b$

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

$(1')$ is false
It is assumed in this case that:
 * $s \left({k}\right) \ne b$

Therefore, $(1')$ is false.

$(2')$ is false
Aiming for a contradiction, let $(2')$ be true instead.

Let $c$ be a non-zero natural number such that:
 * $s \left({k}\right) = b + c$

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

Then:

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

$(3')$ is true
We have:

And the result follows.

Case 4: $k > b$
From the definition of $>$, we have the existence of a non-zero natural number $m$ such that:
 * $k = b + m$

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

Substituting:
 * $k = b + s \left({p}\right)$

By applying $s$ to both sides, we obtain:
 * $s \left({k}\right) = s \left({b + s \left({p}\right)}\right)$

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

$(1')$ is false
Aiming for a contradiction, let $(1')$ be true instead.

Then:

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

$(2')$ is true
We have:
 * $s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$

And the result follows.

$(3')$ is false
Aiming for a contradiction, let $(3')$ be true instead.

Let $c$ be a non-zero natural number such that:
 * $s \left({k}\right) + c = b$

Then:

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