Completely Irreducible implies Infimum differs from Element

Theorem
Let $\left({S, \preceq}\right)$ be an ordered set.

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

Then $\inf\left({p^\succeq \setminus \left\{ {p}\right\} }\right) \ne p$

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

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

Then
 * $p^\succeq \setminus \left\{ {p}\right\}$ admits a infimum and $\inf\left({p^\succeq \setminus \left\{ {p}\right\} }\right) \in p^\succeq \setminus \left\{ {p}\right\}$

By definition of difference:
 * $\inf\left({p^\succeq \setminus \left\{ {p}\right\} }\right) \notin \left\{ {p}\right\}$

Thus by definition of singleton:
 * $\inf\left({p^\succeq \setminus \left\{ {p}\right\} }\right) \ne p$