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, 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 $n \prec m$.