Prime Element iff Complement of Lower Closure is Filter

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a bounded above lattice.

Let $p \in S$ such that
 * $p \ne \top$

where $\top$ denotes the top of $L$.

Then
 * $p$ is prime element


 * $\complement_S\left({p^\preceq}\right)$ is filter in $L$

where
 * $p^\preceq$ denotes the lower closure of $p$,
 * $\complement_S\left({p^\preceq}\right)$ denotes the relative complement of $p^\preceq$ relative to $S$.

Sufficient Condition
Assume that
 * $p$ is prime element.

Ny definition of the greatest element:
 * $p \preceq \top$

By definition of $\prec$:
 * $p \prec \top$

By definition of antisymmetry:
 * $\top \npreceq p$

By definition of lower closure of element:
 * $\top \notin p^\preceq$

By definition of relative complement:
 * $\top \in \complement_S\left({p^\preceq}\right)$

By definition
 * $X := \complement_S\left({p^\preceq}\right)$ is a non-empty set

We will prove that
 * $X$ is filtered.

Let $x, y \in X$.

By definition of relative complement:
 * $x \notin p^\preceq$ and $y \notin p^\preceq$

By definition of lower closure of element:
 * $x \npreceq p$ and $y \npreceq p$

Then by definition of prime element:
 * $x \wedge y \npreceq p$

By definition of lower closure of element:
 * $x \wedge y \notin p^\preceq$

Thus by definition of relative complement:
 * $x \wedge y \in X$

Thus by Meet Precedes Operands:
 * $x \wedge y \preceq x$ and $x \wedge y \preceq y$

Thus
 * $\exists z \in X: z \preceq x \land z \preceq y$

We will prove that
 * $X$ is an upper set.

Let $x \in X, y \in S$ such that
 * $x \preceq y$

By definition of relative complement:
 * $x \notin p^\preceq$

By definition of lower closure of element:
 * $x \npreceq p$

By definition of transitivity:
 * $y \npreceq p$

By definition of lower closure of element:
 * $y \notin p^\preceq$

Thud by definition of relative complement:
 * $y \in X$

Hence by definition
 * $X$ is filter in $L$.

Necessary Condition
Assume that
 * $\complement_S\left({p^\preceq}\right)$ is filter in $L$.

Let $x, y \in S$ such that
 * $x \wedge y \preceq p$

Aiming for a contradiction suppose that
 * $x \npreceq p$ and $y \npreceq p$

By definition of lower closure of element: $x \notin p^\preceq$ and $y \notin p^\preceq$

By definition of relative complement:
 * $x \in \complement_S\left({p^\preceq}\right)$ and $y \in \complement_S\left({p^\preceq}\right)$

By definition of filtered:
 * $\exists z \in \complement_S\left({p^\preceq}\right): z\preceq x \land z \preceq y$

Then $z \preceq x\wedge y$

By definition of transitivity:
 * $z \preceq p$

By definition of upper set:
 * $p \in \complement_S\left({p^\preceq}\right)$

By definition of reflexivity:
 * $p \preceq p$

By definition of lower closure of element:
 * $p \in p^\preceq$

This contradicts $p \in \complement_S\left({p^\preceq}\right)$