Completely Irreducible Element iff Exists Element that Strictly Succeeds First Element

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

Let $p \in S$.

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

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

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

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

By definitions of minimum and infimum:
 * $p^\succeq \setminus \set p$ admits an infimum.

Define $q = \map \inf {p^\succeq \setminus \set p}$

By definition of minimum:
 * $q \in p^\succeq \setminus \set p$

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
 * $\set p \cup q^\succeq \subseteq p^\succeq$

Let $x \in \set p \cup q^\succeq$

By definition of union:
 * $x \in \set p$ or $x \in q^\succeq$

In the first case:
 * $x \in \set p$

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 $p^\succeq \setminus \set p$

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:
 * $s \notin \set p$

By definition of difference:
 * $s \in p^\succeq \setminus \set p$

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

We will prove that:
 * $p^\succeq \subseteq \set p \cup q^\succeq$

Let $x \in p^\succeq$.

By definition of singleton:
 * $x = p$ or $x \in p^\succeq$ and $x \notin \set p$

By definition of difference:
 * $x = p$ or $x \in p^\succeq \setminus \set p$

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

By definitions of singleton and upper closure of element:
 * $x \in \set p$ or $x \in q^\succeq$

Thus by definition of union:
 * $x \in \set p \cup q^\succeq$

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

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

We will prove that:
 * $q$ is lower bound for $p^\succeq \setminus \set p$

Let $a \in p^\succeq \setminus \set p$.

By definition of difference:
 * $a \in p^\succeq$ and $a \notin \set p$

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:
 * $q \preceq a$

By definition of reflexivity:
 * $q \preceq q$

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

By definition of union:
 * $q \in \set p \cup q^\succeq$

By definition of strictly precede:
 * $p \ne q$

By definition of singleton:
 * $q \notin \set p$

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

We will prove that:
 * $\forall b \in S: b$ is lower bound for $p^\succeq \setminus \set p \implies b \preceq q$

Let $b \in S$ such that:
 * $b$ is lower bound for $p^\succeq \setminus \set p$

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

By definition of infimum:
 * $q = \map \inf {p^\succeq \setminus \set p}$

By definition:
 * $p^\succeq \setminus \set p$ admits a minimum.

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