Ordering on 1-Based Natural Numbers is Transitive

Theorem

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

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

Then $<$ is a transitive relation.

Proof

Let $a < b$ and $b < c$.

Then:

\(\ds \exists x \in \N_{> 0}: \, \)

\(\ds a + x\)

\(=\)

\(\ds b\)

\(\ds \land \ \ \)

\(\ds \exists y \in \N_{> 0}: \, \)

\(\ds b + y\)

\(=\)

\(\ds c\)

\(\ds \leadsto \ \ \)

\(\ds \paren {a + x} + y\)

\(=\)

\(\ds c\)

substituting $a + x$ for $b$

\(\ds \leadsto \ \ \)

\(\ds a + \paren {x + y}\)

\(=\)

\(\ds c\)

Natural Number Addition is Associative

\(\ds \leadsto \ \ \)

\(\ds a\)

\(<\)

\(\ds c\)

as $x + y \in \N_{> 0}$

$\blacksquare$

Sources

• 1964: W.E. Deskins: Abstract Algebra ... (previous) ... (next): $2.2$: Theorem $2.9$