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 $a$.
=== Basis for the Induction ===
We will prove that the proposition is true for $a = 0$.
Using Proof by Cases, we divide the proposition into two cases.
==== Case 1: $b = 0$ ====
===== $(1)$ is true =====
It follows trivially from the values of $a$ and $b$.
===== $(2)$ is false =====
 $c$ is 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 a contradiction of the assumption that $c$ is non-zero.
===== $(3)$ is false =====
 $c$ is 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 a contradiction of the assumption that $c$ is non-zero.
==== Case 2: $b \ne 0$. ====
===== $(1)$ is false =====
It follows trivially from the fact that $a = 0$.
===== $(2)$ is false =====
 $c$ is 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 a contradiction of $(P4)$: $0$ is not in the image of $s$.
===== $(3)$ is true =====
By Natural Number Addition Commutes with Zero, we have:
:$a + b = b$
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 =====
 $(1')$ is true.
Then:
which is a contradiction of $(P4)$: $0$ is not in the image of $s$.
===== $(2')$ is true =====
This is apparent from the definition of $>$.
===== $(3')$ is false =====
 $(3')$ is true.
Let $c$ be a non-zero natural number such that:
:$s \left({k}\right) + c = b$
Then:
which is a contradiction of $(P4)$: $0$ is not in the image of $s$.
==== Case 2: $k < b$ and $s \left({k}\right) = b$ ====
===== $(1')$ is true =====
This follows from the assumption.
===== $(2')$ is false =====
 $(2')$ is true.
Let $c$ be a non-zero natural number such that:
:$s \left({k}\right) = b + c$
Then:
which is a contradiction of our assumption that $c$ is non-zero.
===== $(3')$ is false =====
 $(3')$ is true.
Let $c$ be a non-zero natural number such that:
:$s \left({k}\right) + c = b$
Then:
which is a contradiction of 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 =====
 $(2')$ is true.
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 a contradiction of $(P4)$: $0$ is not in the image of $s$.
===== $(3')$ is true =====
We have:
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 =====
 $(1')$ is true.
Then:
which is a contradiction of $(P4)$: $0$ is not in the image of $s$.
===== $(2')$ is true =====
We have:
:$s \left({k}\right) = b + s \left({s \left({p}\right)}\right)$
The result follows.
===== $(3')$ is false =====
 $(3')$ is true.
Let $c$ be a non-zero natural number such that:
:$s \left({k}\right) + c = b$
Then:
which is a contradiction of $(P4)$: $0$ is not in the image of $s$.
== Sources ==
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
Category:Natural Numbers