Sum with One is Immediate Successor in Naturally Ordered Semigroup
Theorem
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.
Let $1$ be the one of $S$.
Let $n \in S$.
Then $n \circ 1$ is the immediate successor of $n$.
That is, for all $m \in S$:
- $n \prec m \iff n \circ 1 \preceq m$
Proof
By Zero Strictly Precedes One, $0 \prec 1$, where $0$ is the zero of $S$.
Hence from Strict Ordering of Naturally Ordered Semigroup is Strongly Compatible:
- $n \circ 0 \prec n \circ 1$
and by Zero is Identity in Naturally Ordered Semigroup, $n \circ 0 = n$.
Now suppose that $n \prec m$.
Then by Naturally Ordered Semigroup Axiom $\text {NO} 3$: Existence of Product, there exists $p \in S$ such that:
- $n \circ p = m$
Moreover, since $n \ne m$, it follows that $p \ne 0$.
Hence $0 \prec p$ by definition of zero.
Therefore, by definition of one:
- $1 \preceq p$
Now by compatibility of $\preceq$ with $\circ$:
- $n \circ 1 \preceq n \circ p = m$
as desired.
Conversely, if $n \circ 1 \preceq m$, it is immediate from:
- $n \prec n \circ 1$
that $n \prec m$.
$\blacksquare$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.4$