Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element

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

Let $p \in S$.

Then $p$ is completely irreducible
 * $\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$

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

Sufficient Condition
Assume that
 * $p$ is completely irreducible.

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

By definitions of minimum and infimum:
 * $p^\succeq \setminus \left\{ {p}\right\}$ admits an infimum.

Define $q = \inf\left({p^\succeq \setminus \left\{ {p}\right\} }\right)$

By definition of minimum:
 * $q \in p^\succeq \setminus \left\{ {p}\right\}$

By definition of difference:
 * $q \in p^\succeq$

By definition of upper closure of element:
 * $p \preceq q$

By Completely Irreducible implies Infimum differs from Element:
 * $q \ne p$

Thus by definition of strictly precede:
 * $p \prec q$

We will prove that
 * $\left\{ {p}\right\} \cup q^\succeq \subseteq p^\succeq$

Let $x \in \left\{ {p}\right\} \cup q^\succeq$

By definition of union:
 * $x \in \left\{ {p}\right\}$ or $x \in q^\succeq$

In the first case:
 * $x \in \left\{ {p}\right\}$

By definition of singleton:
 * $x = p$

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

Thus by definition of upper closure of element:
 * $x \in p^\succeq$

In the second case:
 * $x \in q^\succeq$

By definition of upper closure of element:
 * $q \preceq x$

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

Thus by definition of upper closure of element:
 * $x \in p^\succeq$

By definition of infimum:
 * $q$ is lower bound for $p^\succeq \setminus \left\{ {p}\right\}$

We will prove that
 * $\forall s \in S: p \prec s \implies q \preceq s$

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

By definition of strictly precede
 * $p \preceq s$ and $p \ne s$

By definition of upper closure of element:
 * $s \in p^\succeq$

By definition of singleton:
 * $s \notin \left\{ {p}\right\}$

By definition of difference:
 * $s \in p^\succeq \setminus \left\{ {p}\right\}$

Thus by definition of lower bound:
 * $q \preceq s$

We will prove that
 * $p^\succeq \subseteq \left\{ {p}\right\} \cup q^\succeq$

Let $x \in p^\succeq$.

By definition of singleton:
 * $x = p$ or $x \in p^\succeq$ and $x \notin \left\{ {p}\right\}$

By definition of difference:
 * $x = p$ or $x \in p^\succeq \setminus \left\{ {p}\right\}$

By definitions of infimum and lower bound:
 * $x = p$ or $q \preceq x$

By definitions of singleton and upper closure of element:
 * $x \in \left\{ {p}\right\}$ or $x \in q^\succeq$

Thus by definition of union:
 * $x \in \left\{ {p}\right\} \cup q^\succeq$

Thus by definition of set equality:
 * $p^\succeq = \left\{ {p}\right\} \cup q^\succeq$

Necessary Condition
Assume that
 * $\exists q \in S: p \prec q \land \left({\forall s \in S: p \prec s \implies q \preceq s}\right) \land p^\succeq = \left\{ {p}\right\} \cup q^\succeq$

We will prove that
 * $q$ is lower bound for $p^\succeq \setminus \left\{ {p}\right\}$

Let $a \in p^\succeq \setminus \left\{ {p}\right\}$

By definition of difference:
 * $a \in p^\succeq$ and $a \notin \left\{ {p}\right\}$

By definitions of upper closure of element and singleton:
 * $p \preceq a$ and $a \ne p$

By definition of strictly precede:
 * $p \prec a$

Thus by assumption:
 * $q \preceq a$