Meet-Continuous and Distributive implies Shift Mapping Preserves Finite Suprema

Theorem
Let $\left({S, \vee, \wedge, \preceq}\right)$ be a meet-continuous distributive complete lattice.

Let $x \in S$.

Let $f: S \to S$ be a mapping such that
 * $\forall y \in S: f\left({y}\right) = x \wedge y$

Then
 * $f$ preserves finite suprema

Proof
Let $X$ be finite subset of $S$ such that
 * $X$ admits a supremum.

By definition of complete lattice:
 * $f^\to\left({X}\right)$ admits a supremum.

We will prove by induction of cardinality of $X$.

Basis Case

 * $\forall X \subseteq S: \left\vert{X}\right\vert = 0 \implies \sup \left({f^\to\left({X}\right)}\right) = f\left({\sup X}\right)$

where
 * $\left\vert{X}\right\vert$ denotes the cardinality of $X$

Let $X \subseteq S$ such that
 * $\left\vert{X}\right\vert = 0$

By Cardinality of Empty Set:
 * $X = \varnothing$

By definition of image of set:
 * $f^\to\left({X}\right) = \varnothing$

By definitions of bottom and smallest element:
 * $\bot \preceq x$

Thus

Induction Hypothesis

 * $\forall X \subseteq S: \left\vert{X}\right\vert = n \implies \sup \left({f^\to\left({X}\right)}\right) = f\left({\sup X}\right)$

Induction Step

 * $\forall X \subseteq S: \left\vert{X}\right\vert = n+1 \implies \sup \left({f^\to\left({X}\right)}\right) = f\left({\sup X}\right)$

Let $X \subseteq S$ such that
 * $\left\vert{X}\right\vert = n+1$

Then
 * $\exists x_1, \dots, x_n, x_{n+1} \in S: X = \left\{ {x_1, \dots, x_n, x_{n+1} }\right\}$

Thus