Strict Lower Closure of Sum with One

Theorem
Let $\struct {S, \circ, \preceq}$ be a naturally ordered semigroup.

Then:
 * $\forall n \in \struct {S, \circ, \preceq}: \paren {n \circ 1}^\prec = n^\prec \cup \set n$

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 $\struct {S, \circ, \preceq}$ is well-ordered and hence totally ordered, the Trichotomy Law applies.

Thus:

So:

Similarly:

So:
 * $p \notin n^\prec \cup \set n \iff p \notin \paren {n \circ 1}^\prec$

Thus:
 * $\relcomp S {\paren {n \circ 1}^\prec} = \relcomp S {n^\prec \cup \set n}$

from the definition of relative complement.

So: