=== Sufficient Condition === Assume that :$\mathscr S$ is meet-continuous. We will prove that :for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$ Let $D_1, D_2$ be directed subsets of $S$. we will prove as sublemma that :for every an element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x = \sup \set {x \wedge d: d \in D}$ Let $x \in S$, $D$ be a directed subset of $S$ such that: :$x \preceq \sup D$ Thus {{begin-eqn}} {{eqn | l = x       | r = x \wedge \sup D       | c = Preceding iff Meet equals Less Operand }} {{eqn | r = \sup \set {x \wedge d: d \in D}       | c = {{Defof|Meet-Continuous Lattice}} }} {{end-eqn}} By definition of reflexivity: :for every element $x$ of $S$, a directed subset $D$ of $S$ if $x \preceq \sup D$, then $x \preceq \sup \set {x \wedge d: d \in D}$ Define a mapping $g: S \times S \to S$: :$\forall \tuple {s, t} \in S \times S: \map g {s, t} = s \wedge t$ By Meet is Directed Suprema Preserving: :$g$ preserves directed suprema as a mapping from Cartesian product $\struct {S \times S, \precsim}$ of $\mathscr S$ and $\mathscr S$ into $\mathscr S$. Thus by Meet is Directed Suprema Preserving implies Meet of Suprema equals Supremum of Meet of Directed Subsets:  :$\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$ By exemplification: :for every ideals $I_1, I_2$ of $S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {d_1 \wedge d_2: d_1 \in I_1, d_2 \in I_2}$ Thus by Meet of Suprema equals Supremum of Meet of Ideals implies Ideal Supremum is Meet Preserving: :$f$ preserves meet. {{qed|lemma}} === Necessary Condition === Assume that :$f$ preserves meet Thus :$\mathscr S$ is up-complete We will prove that :for every ideals $I_1, I_2$ in $\mathscr S$: $\paren {\sup I_1} \wedge \paren {\sup I_2} = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}$ Let $I_1, I_2$ be ideals in $\mathscr S$. Thus {{begin-eqn}} {{eqn | l = \paren {\sup I_1} \wedge \paren {\sup I_2}       | r = \map f {I_1} \wedge \map f {I_2}       | c = Definition of $f$ }} {{eqn | r = \map f {I_1 \wedge I_2}       | c = {{Defof|Mapping Preserves Infimum|subdef = Meet}} }} {{eqn | r = \map \sup {I_1 \wedge I_2}       | c = Definition of $f$ }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in I_1, i_2 \in I_2}       | c = Meet in Set of Ideals }} {{end-eqn}} We will prove that :for every directed subsets $D_1, D_2$ of $S$: $\paren {\sup D_1} \wedge \paren {\sup D_2} = \sup \set {d_1 \wedge d_2: d_1 \in D_1, d_2 \in D_2}$ Let $D_1, D_2$ be directed subsets of $S$. By definition of up-complete: :$D_1$ and $D_2$ admit suprema By Supremum of Lower Closure of Set: :$D_1^\preceq$ and $D_2^\preceq$ admit suprema Thus {{begin-eqn}} {{eqn | l = \paren {\sup D_1} \wedge \paren {\sup D_2}       | r = \paren {\sup D_1^\preceq} \wedge \paren {\sup D_2^\preceq}       | c = Supremum of Lower Closure of Set }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}       | c = $D_1^\preceq$ is ideal in $\mathscr S$ }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1^\preceq, i_2 \in D_2^\preceq}^\preceq       | c = Supremum of Lower Closure of Set }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}^\preceq       | c = Lower Closure of Meet of Lower Closures }} {{eqn | r = \sup \set {i_1 \wedge i_2: i_1 \in D_1, i_2 \in D_2}       | c = Supremum of Lower Closure of Set }} {{end-eqn}} It remains to prove (MC) of definition of meet-continuous. Let $x$ be an element of $S$. Let $D$ be a directed subset of $S$. By Singleton is Directed and Filtered Subset: :$\set x$ is directed. Thus {{begin-eqn}} {{eqn | l = x \wedge \sup D       | r = \paren {\sup \set x} \wedge \sup D       | c = Supremum of Singleton }} {{eqn | r = \sup \set {d_1 \wedge d_2: d_1 \in \set x, d_2 \in D} }} {{eqn | r = \sup \set {x \wedge d: d \in D}       | c = {{Defof|Singleton}} }} {{end-eqn}} {{qed}}