Bottom not in Proper Filter

Theorem
Let $L = \left({S, \preceq}\right)$ be a bounded below preordered set.

Let $F$ be a filter on $L$.

Then $F$ is proper filter $\bot \notin F$

where $\bot$ denotes the smallest element of $S$.

Sufficient Condition
Suppose
 * $F$ is proper.

By definition of proper subset:
 * $F \subseteq S$ and $F \ne S$

By definitions of set equality and subset:
 * $\exists x: x \in S \land x \notin F$

By definition of smallest element:
 * $\bot \preceq x$

Thus by definition of upper set:
 * $\bot \notin F$

Necessary Condition
Suppose $\bot \notin F$

By definition of set equality:
 * $F \ne S$

Hence $F$ is proper filter.