Completely Irreducible implies Infimum differs from Element

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $p \in S$ such that
 * $p$ is completely irreducible.

Then $\map \inf {p^\succeq \setminus \set p} \ne p$

where $p^\succeq$ donotes the upper closure of $p$.

Proof
By definition of completely irreducible:
 * $p^\succeq \setminus \set p$ admits a minimum.

Then:
 * $p^\succeq \setminus \set p$ admits a infimum and $\map \inf {p^\succeq \setminus \set p} \in p^\succeq \setminus \set p$

By definition of difference:
 * $\map \inf {p^\succeq \setminus \set p} \notin \set p$

Thus by definition of singleton:
 * $\map \inf {p^\succeq \setminus \set p} \ne p$