Maximal implies Completely Irreducible

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

Let $p \in S$ such that
 * $\exists k \in S: p$ is maximal in $S \setminus k^\succeq$

Then $p$ is completely irreducible.

Proof
By Join Succeeds Operands:
 * $k \preceq p \vee k$ and $p \preceq p \vee k$

By definition of upper closure of element:
 * $p \vee k \in k^\succeq$ and $p \vee k \in p^\succeq$

By definition of intersection:
 * $p \vee k \in p^\succeq \cap k^\succeq$

By Maximal implies Difference equals Intersection:
 * $p^\succeq \setminus \left\{ {p}\right\} = p^\succeq \cap k^\succeq$

By Infimum of Intersection of Upper Closures equals Join Operands
 * $\inf\left({p^\succeq \cap k^\succeq}\right) = p \vee k$

By definition:
 * $p^\succeq \setminus \left\{ {p}\right\}$ admits a minimum.

Thus by definition:
 * $p$ is completely irreducible.