=== 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: :$x \in p^\succeq$ {{qed|lemma}} 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: :$x \in p^\succeq$ {{qed|lemma}} 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: :$q \preceq s$ {{qed|lemma}} 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: :$x \in \set p \cup q^\succeq$ {{qed|lemma}} Thus by definition of set equality: :$p^\succeq = \set p \cup q^\succeq$ {{qed|lemma}} === 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: :$q \preceq a$ {{qed|lemma}} 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: :$b \preceq q$ {{qed|lemma}} 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. {{qed}}