Prime Element iff Element Greater is Top

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

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

Then
 * $p$ is prime element


 * $\forall x \in S: \left({ p \prec x \implies x = \top }\right)$

Sufficient Condition
Suppose that
 * $p$ is prime element

Let $x \in S$ such that
 * $p \prec x$

By definition of Boolean lattice:
 * $L$ is complemented distributive lattice.

By definition of complemented:
 * $\exists y \in S: y$ is complement of $x$.

By definition of complement
 * $x \wedge y = \bot$

By definition of smallest element:
 * $x \wedge y \preceq p$

By definition of prime element:
 * $x \preceq p$ or $y \preceq p$

Then by definition of antisymmetry:
 * $y \prec x$

By definition of $\prec$:
 * $y \preceq x$

By definition of complement:
 * $x \vee y = \top$

Thus by Preceding iff Join equals Larger Operand:
 * $x = \top$

Necessary Condition
Suppose that
 * $\forall x \in S: \left({ p \prec x \implies x = \top }\right)$

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 Preceding iff Join equals Larger Operand:
 * $p \ne p \vee y$ and $p \ne p \vee x$

By Join Succeeds Operands:
 * $p \preceq p \vee y$ and $p \preceq p \vee x$

By definition of $\prec$:
 * $p \prec p \vee y$ and $p \prec p \vee x$

By assumption:
 * $p \vee y = \top$ and $p \vee x = \top$

By Meet is Idempotent:
 * $p = \top$

This contradicts $p \ne \top$