Prime Ideal is Prime Element

Theorem
Let $L = \left({S, \vee, \wedge, \preceq}\right)$ be a lattice.

Let $I$ be an ideal in $L$.

Then
 * $I$ is a prime ideal


 * $I$ is a prime element in $\left({ \mathit{Ids}\left({L}\right), \precsim}\right)$

where
 * $\mathit{Ids}\left({L}\right)$ denotes the set of all ideals in $L$,
 * $\mathord\precsim = \mathord\subseteq\restriction_{\mathit{Ids}\left({L}\right) \times \mathit{Ids}\left({L}\right) }$

Sufficient Condition
Let $I$ be a prime ideal.

Let $x, y \in \mathit{Ids}\left({L}\right)$ such that
 * $x \wedge y \precsim I$

By definition of $\precsim$:
 * $x \wedge y \subseteq I$

By Meet is Intersection in Set of Ideals:
 * $x \cap y \subseteq I$

Aiming for a contraindication suppose
 * $x \not\precsim I$ and $y \not\precsim I$

By definition of $\precsim$: $x \nsubseteq I$ and $y \nsubseteq I$

By definition of subset:
 * $\exists a \in x: a \notin I$

and
 * $\exists b \in y: b \notin I$

By Meet Precedes Operands:
 * $a \wedge b \preceq a$ and $a \wedge b \preceq b$

By definition lower set:
 * $a \wedge b \in x$ and $a \wedge b \in y$

By definitions subset and intersection:
 * $a \wedge b \in I$

By Characterization of Prime Ideal:
 * $a \in I$ or $b \in I$

This contradicts $a \notin I$ and $b \notin I$

Necessary Condition
Let $I$ be prime element in $\left({\mathit{Ids}\left({L}\right), \precsim}\right)$

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

By Lower Closure of Element is Ideal:
 * $X := x^\preceq$ and $Y := y^\preceq$ are ideals in $L$

By Meet is Intersection in Set of Ideals:
 * $X \cap Y = X \wedge Y$

We will prove that
 * $X \wedge Y \subseteq I$

Let $a \in X \wedge Y$.

By definition of intersection:
 * $a \in X$ and $a \in Y$

By definition of lower closure of element:
 * $a \preceq x$ and $a \preceq y$

By definition of infimum:
 * $a \preceq x \wedge y$

Thus by definition of lower set:
 * $a \in I$

By definition of $\precsim$:
 * $X \wedge Y \precsim I$

By definition of prime element:
 * $X \precsim I$ or $Y \precsim I$

By definition of $\precsim$:
 * $X \subseteq I$ or $Y \subseteq I$

By definition of reflexivity:
 * $x \preceq x$ and $y \preceq y$

By definition of lower closure of element:
 * $x \in X$ and $y \in Y$

Thus by definition of subset:
 * $x \in I$ or $y \in I$

Hence by Characterization of Prime Ideal:
 * $I$ is prime ideal.