Assume that :$(1): \quad f$ preserves finite infima and :$(2): \quad f$ preserves filtered infima. Let $X$ be a subset of $S_1$. Let $X$ admits an infimum in $\left({S_1, \preceq_1}\right)$ Define $Z := \left\{ {\inf A: A \in {\it Fin}\left({X}\right) \land A \ne \varnothing}\right\}$ where :$\inf A$ denotes the infimum of $A$ in $\left({S_1, \preceq_1}\right)$ :${\it Fin}\left({X}\right)$ denotes the set of all finite subsets of $X$ By Existence of Non-Empty Finite Infima in Meet Semilattice: :for every non-empty finite subset $A$ of $X$, $A$ admits an infimum in $\left({S_1, \preceq_1}\right)$ By Infimum of Infima: :$\inf Z = \inf\bigcup \left({ {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\} }\right) = \inf X$ :$Z$ admits an infimum in $\left({S_1, \preceq_1}\right)$ We will prove that :$Z$ is filtered Let $x, y \in Z$. By definition of $Z$: :$\exists A \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: x = \inf A$ and :$\exists B \in {\it Fin}\left({X}\right) \setminus \left\{ {\varnothing}\right\}: y = \inf B$ By Finite Union of Finite Sets is Finite: :$A \cup B$ is finite :$A \cup B \ne \varnothing$ By definition of $Z$: :$\inf\left({A \cup B}\right) \in Z$ By Infimum of Infima: :$\left({\inf A}\right) \wedge \left({\inf B}\right) = \inf\left({A \cup B}\right)$ Thus by definition of infimum: :$\inf\left({A \cup B}\right) \preceq_1 x \land \inf\left({A \cup B}\right) \preceq_1 y$ Thus by definition :$Z$ is filtered By $(2)$ and definitions of mapping preserves filtered infima and mapping preserves infimum on subset: :$f^\to\left({Z}\right)$ admits an infimum in $\left({S_2, \preceq_2}\right)$ and $\inf\left({f^\to\left({Z}\right)}\right) = f\left({\inf Z}\right)$ We will prove that :$X \subseteq Z$ Let $x \in X$. By definition of ${\it Fin}\left({X}\right)$: :$\left\{ {x}\right\} \in {\it Fin}\left({X}\right)$ and $\left\{ {x}\right\} \ne \varnothing$ By definition of $Z$: :$\inf \left\{ {x}\right\} \in Z$ Thus by Infimum of Singleton: :$x \in Z$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$f^\to\left({X}\right) \subseteq f^\to\left({Z}\right)$ By definition of infimum: :$\inf\left({f^\to\left({Z}\right)}\right)$ is a lower bound for $f^\to\left({Z}\right)$ By Lower Bound is Lower Bound for Subset :$\inf\left({f^\to\left({Z}\right)}\right)$ is a lower bound for $f^\to\left({X}\right)$ We will prove that :for every lower bound $a$ for $f^\to\left({X}\right)$, $a$ is lower bound for $f^\to\left({Z}\right)$ Assume that :$a$ is lower bound for $f^\to\left({X}\right)$ Let $x \in f^\to\left({Z}\right)$. By definition of image of set: :$\exists y \in S_1: y \in Z \land f\left({y}\right) = x$ By definition of $Z$: :$\exists A \in {\it Fin}\left({X}\right): y = \inf A \land A \ne \varnothing$ By $(1)$ and definition of mapping preserves finite infima: :$f$ preserves the infimum of $A$ By Existence of Non-Empty Finite Infima in Meet Semilattice :$A$ admits an infimum in $\left({S_1, \preceq_1}\right)$ and $\inf\left({f^\to\left({A}\right)}\right) = f\left({\inf A}\right) = x$ By Image of Subset under Relation is Subset of Image/Corollary 2: :$f^\to\left({A}\right) \subseteq f^\to\left({X}\right)$ By Lower Bound is Lower Bound for Subset :$a$ is lower bound for $f^\to\left({A}\right)$ Thus by definition of infimum: :$a \preceq_2 x$ Thus by definition :$a$ is lower bound for $f^\to\left({Z}\right)$ By definition of infimum: :for every lower bound $a$ for $f^\to\left({X}\right)$, $a \preceq_2 f\left({\inf X}\right)$ Again by definition of infimum: :$f^\to\left({X}\right)$ admits an infimum in $\left({S_2, \preceq_2}\right)$ :$\inf\left({f^\to\left({X}\right)}\right) = f\left({\inf X}\right)$ Thus the result follows by definition of mapping preserves all infima. {{qed}}