Strict_Lower_Closure_of_Sum_with_One

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 $n$.