Strict Lower Closure of Sum with One

Theorem
Let $\left({S, \circ, \preceq}\right)$ be a naturally ordered semigroup.

Then:
 * $\forall n \in \left({S, \circ, \preceq}\right): \left({n \circ 1}\right)^\prec = n^\prec \cup \left\{{n}\right\}$

where $n^\prec$ is defined as the strict lower closure of $n$, that is, the set of elements strictly preceding $n$.

Proof
First note that as $\left({S, \circ, \preceq}\right)$ is well-ordered and hence totally ordered, the Trichotomy Law applies.

Thus:

So:

Similarly:

So:
 * $p \notin n^\prec \cup \left\{{n}\right\} \iff p \notin \left({n \circ 1}\right)^\prec$

Thus:
 * $\complement_S \left({\left({n \circ 1}\right)^\prec}\right) = \complement_S \left({n^\prec \cup \left\{{n}\right\}}\right)$

from the definition of relative complement.

So: