Ordering on 1-Based Natural Numbers is Trichotomy

Theorem

Let $\N_{> 0}$ be the $1$-based natural numbers.

Let $<$ be the strict ordering on $\N_{>0}$.

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 $\N_{> 0}$.

Proof

Using the following axioms:

\((\text A)\)	$:$			\(\ds \exists_1 1 \in \N_{> 0}:\)	\(\ds a \times 1 = a = 1 \times a \)
\((\text B)\)	$:$			\(\ds \forall a, b \in \N_{> 0}:\)	\(\ds a \times \paren {b + 1} = \paren {a \times b} + a \)
\((\text C)\)	$:$			\(\ds \forall a, b \in \N_{> 0}:\)	\(\ds a + \paren {b + 1} = \paren {a + b} + 1 \)
\((\text D)\)	$:$			\(\ds \forall a \in \N_{> 0}, a \ne 1:\)	\(\ds \exists_1 b \in \N_{> 0}: a = b + 1 \)
\((\text E)\)	$:$			\(\ds \forall a, b \in \N_{> 0}:\)	\(\ds \)Exactly one of these three holds:\( \)
					\(\ds a = b \lor \paren {\exists x \in \N_{> 0}: a + x = b} \lor \paren {\exists y \in \N_{> 0}: a = b + y} \)
\((\text F)\)	$:$			\(\ds \forall A \subseteq \N_{> 0}:\)	\(\ds \paren {1 \in A \land \paren {z \in A \implies z + 1 \in A} } \implies A = \N_{> 0} \)

Axiom $E$ states:

$\forall a, b \in \N_{>0}$, either:

$a = b$, in which case $(1)$ holds

$\exists x \in \N_{> 0}: a = b + x$, in which case, by definition of the ordering defined, $a > b$, in which case $(2)$ holds

$\exists x \in \N_{> 0}: a + x = b$, in which case, by definition of the ordering defined, $a < b$, in which case $(3)$ holds.

Hence the result.

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.8$