Strict Lower Closure of Sum with One

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

Then $$\forall n \in \left({S, \circ; \preceq}\right): S_{n \circ 1} = S_n \cup \left\{{n}\right\}$$

where $$S_n$$ is defined as the set of preceding elements of $$n$$.

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 $$p \notin S_n \cup \left\{{n}\right\} \iff p \notin S_{n \circ 1}$$.

Thus $$\mathcal{C}_S \left({S_{n \circ 1}}\right) = \mathcal{C}_S \left({S_n \cup \left\{{n}\right\}}\right)$$ from the definition of relative complement.

So: