Characterization of Prime Element in Meet Semilattice

Theorem
Let $L = \left({S, \wedge, \preceq}\right)$ 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
 * $P\left({X}\right) :\equiv X \ne \varnothing \land \inf X \preceq p \implies \exists x \in X: x \preceq p$

where $X \subseteq S$.

We will prove that
 * $\forall x \in A, B \subseteq A: P\left({B}\right) \implies P\left({B \cup \left\{ {x}\right\} }\right)$

Let $x \in A, B \subseteq A$ such that
 * $P\left({B}\right)$

and
 * $B \cup \left\{ {x}\right\} \ne \varnothing$ and $\inf \left({B \cup \left\{ {x}\right\} }\right) \preceq p$

Case $B = \varnothing$.

By Union with Empty Set:
 * $B \cup \left\{ {x}\right\} = \left\{ {x}\right\}$

By Infimum of Singleton:
 * $\inf \left\{ {x}\right\} = x$

By definition of singleton:
 * $x \in \left\{ {x}\right\}$

Thus
 * $\exists z \in B \cup \left\{ {x}\right\}:z \preceq p$

Case $B \ne \varnothing$.

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

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

By Infimum of Infima:
 * $\inf \left({B \cup \left\{ {x}\right\} }\right) = \inf B \wedge \inf \left\{ {x}\right\}$

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:
 * $z \in B \cup \left\{ {x}\right\}$

Thus
 * $\exists z \in B \cup \left\{ {x}\right\}: z \preceq p$

Case $x \preceq p$.

By definitions of union and singleton:
 * $x \in B \cup \left\{ {x}\right\}$

Thus
 * $\exists z \in B \cup \left\{ {x}\right\}: z \preceq p$

By definition of empty set:
 * $P\left({\varnothing}\right)$

By Induction of Finite Set:
 * $P\left({A}\right)$

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