Define $X := \set {P \in \mathit{Ids}\left({L}\right): I \subseteq P \land P \cap F = \O}$ where $\mathit{Ids}\left({L}\right)$ denotes set of all ideals in $L$. By Set is Subset of Itself: :$I \in X$ We will prove that :$\forall Z: Z \ne \varnothing \land Z \subseteq X \land \left({\forall Y_1, Y_2 \in Z: Y_1 \subseteq Y_2 \lor Y_2 \subseteq Y_1}\right) \implies \bigcup Z \in X$ Let $Z$ such that :$Z \ne \varnothing$ and :$Z \subseteq X$ and :$\forall Y_1, Y_2 \in Z: Y_1 \subseteq Y_2 \lor Y_2 \subseteq Y_1$ By definition of $X$: :$Z$ is set of subsets of $S$. By definition of $X$: :$\forall Y \in Z: Y$ is lower set. Then by Every Element is Lower implies Union is Lower: :$J := \bigcup Z$ is lower set. By definition of non-empty set: :$\exists Y: Y \in Z$ By definition of subset: :$Y \in X$ By definition of $X$: :$I \subseteq Y$ By Set is Subset of Union/General Result: :$Y \subseteq J$ By Subset Relation is Transitive: :$I \subseteq J$ We will prove that :$\forall A, B \in Z: \exists C \in Z: A \cup B \subseteq C$ Let $A, B \in Z$. By assumption: :$A \subseteq B$ or $B \subseteq A$ By Union with Superset is Superset: :$A \cup B = B$ or $A \cup B = A$ Thus by Set is Subset of Itself: :$\exists C \in Z: A \cup B \subseteq C$ {{qed|lemma}} By definition of $X$: :$\forall Y \in Z: Y$ is directed. By Every Element is Directed and Every Two Elements are Included in Third Element implies Union is Directed: :$J$ is directed. By definition of ideal in ordered set: :$J$ is an ideal in $L$. We will prove that :$J \cap F = \varnothing$ Let $x \in J$. By definition of union: :$\exists A \in Z: x \in A$ By definition of subset: :$A \in X$ By definition of $X$: :$A \cap F = \varnothing$ Thus by definitions of intersection and empty set: :$x \notin F$ {{qed|lemma}} Thus by definition of $X$: :$\bigcup Z \in X$ {{qed|lemma}} By Zorn's Lemma: :there exists $Y \in X$: $Y$ is maximal set of $X$.  Then by definition of $X$: :$Y \in \mathit{Ids}\left({L}\right)$ and $I \subseteq Y$ and $Y \cap F = \varnothing$ We will prove that :$Y$ is a prime ideal. Let $x, y \in S$ such that :$x \wedge y \in Y$ Aiming for a contradiction suppose that :$x \notin Y$ and $y \notin Y$ Define $P_y = \operatorname{finsups}\left({Y \cap \left\{ {y}\right\} }\right)^\preceq$ By Finite Suprema Set and Lower Closure is Smallest Ideal: :$Y \cup \left\{ {y}\right\} \subseteq P_y$ By Set is Subset of Union: :$Y \subseteq Y \cup \left\{ {y}\right\}$ By Subset Relation is Transitive: :$Y \subseteq P_y$ By Subset Relation is Transitive: :$I \subseteq P_y$ By definition of singleton: :$y \in \left\{ {y}\right\}$ By definition of union: :$y \in Y \cup \left\{ {y}\right\}$ By definition of subset: :$y \in P_y$ We will prove that :$P_y \cap F \ne \varnothing$ Aiming for a contradiction suppose that  :$P_y \cap F = \varnothing$ By definition of $X$: :$P_y \in X$ By definition of minimal set: :$Y = P_y$ A contradiction between $y \notin Y$ and $y \in P_y$ {{qed|lemma}} By definitions of non-empty set and intersection: :$\exists v: v \in P_y \land v \in F$ Define $P_x = \operatorname{finsups}\left({Y \cap \left\{ {x}\right\} }\right)^\preceq$ Analogically: :$\exists u: u \in P_x \land u \in F$ By Finite Subset Bounds Element of Finite Suprema Set and Lower Closure: :$\exists u' \in Y: u \preceq u' \vee \sup \left\{ {x}\right\}$ By Finite Subset Bounds Element of Finite Suprema Set and Lower Closure: :$\exists v' \in Y: v \preceq v' \vee \sup \left\{ {y}\right\}$ By Join is Associative: :$\left({v' \vee u'}\right) \vee x = v' \vee \left({u' \vee x}\right)$ By Join Succeeds Operands:  :$\left({v' \vee u'}\right) \vee x \succeq u' \vee x$ By Supremum of Singleton: :$\sup \left\{ {x}\right\} = x$ By definition of transitivity: :$u \preceq v' \vee u' \vee x$ By definition of upper set: :$u' \vee v' \vee x \in F$ Analogically: :$u' \vee v' \vee y \in F$ By Filtered in Meet Semilattice: :$\left({u' \vee v' \vee x}\right) \wedge \left({u' \vee v' \vee y}\right) \in F$ By definition of distributive lattice: :$\left({u' \vee v'}\right) \vee \left({x \wedge y}\right) \in F$ By Directed in Join Semilattice: :$u' \vee v' \in Y$ By Directed in Join Semilattice: :$\left({u' \vee v'}\right) \vee \left({x \wedge y}\right) \in Y$ This contradicts $Y \cap F = \varnothing$ {{qed}}