Strict Lower Closure is Lower Section/Proof 1

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

Let $p \in S$.

Then $p^\prec$, the strict lower closure of $p$, is a lower set.

 Let $l \in p^\prec$.

Let $s \in S$ with $s \preceq l$.

Then by the definition of strict lower closure:
 * $l \prec p$

Thus by Extended Transitivity:
 * $s \prec p$

So by the definition of strict lower closure:
 * $s \in p^\prec$

Since this holds for all such $l$ and $s$, $p^\prec$ is a lower set.