Characterization of Prime Element in Meet Semilattice

Theorem
Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.

Let $p \in S$,

Then:
 * $p$ is prime element


 * for all non-empty finite subsets $A$ of $S$:
 * if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.
 * if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.

Sufficient Condition
Let $p$ be prime element.

Let $A$ be non-empty finite subsets of $S$.

Define
 * $\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$

where $X \subseteq S$.

We will prove that
 * $\forall x \in A, B \subseteq A: \map P B \implies \map P {B \cup \set x}$

Let $x \in A, B \subseteq A$ such that
 * $\map P B$

and
 * $B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \preceq p$

Case $B = \O$.

By Union with Empty Set:
 * $B \cup \set x = \set x$

By Infimum of Singleton:
 * $\inf \set x = x$

By definition of singleton:
 * $x \in \set x$

Thus
 * $\exists z \in B \cup \set x: z \preceq p$

Case $B \ne \O$.

By Subset of Finite Set is Finite:
 * $B$ is finite.

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $B, \set x$ admit infima.

By Infimum of Infima:
 * $\map \inf {B \cup \set x} = \inf B \wedge \inf \set x$

By definition of prime element:
 * $\inf B \preceq p$ or $x \preceq p$

Case $\inf B \preceq p$.

By assumption:
 * $\exists z \in B: z \preceq p$

By definition of union:
 * $z \in B \cup \set x$

Thus
 * $\exists z \in B \cup \set x: z \preceq p$

Case $x \preceq p$.

By definitions of union and singleton:
 * $x \in B \cup \set x$

Thus
 * $\exists z \in B \cup \set x: z \preceq p$

By definition of empty set:
 * $\map P \O$

By Induction of Finite Set:
 * $\map P A$

Thus
 * if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.

Necessary Condition
Suppose that
 * for all non-empty finite subsets $A$ of $S$:
 * if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.

Let $x, y \in S$ such that
 * $x \wedge y \preceq p$

Define $A := \set {x, y}$

By definitions of non-empty set and finite set:
 * $A$ is non-empty finite subset of $S$.

By definition of meet:
 * $\inf A = x \wedge y$

By assumption:
 * $\exists z \in A: z \preceq p$

Thus by definition of unordered tuple:
 * $x \preceq p$ or $y \preceq p$