Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation

Theorem
Let $R = \left({S, \preceq}\right)$ be a bounded below join semilattice.

Let $\mathit{Ids}\left({R}\right)$ be the set of all ideals in $R$.

Let $L = \left({ \mathit{Ids}\left({R}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\mathit{Ids}\left({R}\right) \times \mathit{Ids}\left({R}\right)}$.

Let $M = \left({F, \preccurlyeq}\right)$ be the ordered set of increasing mappings $g:S \to \mathit{Ids}\left({R}\right)$ satisfying $\forall x \in S: g\left({x}\right) \subseteq x^\preceq$.

Let $f \in F$.

Then
 * there exists an auxiliary relation $\mathcal R$ on $S$ such that
 * $\forall x \in S:f\left({x}\right) = x^{\mathcal R}$

Proof
Define relation $\mathcal R$ on $S$:
 * $\forall x, y \in S: \left({x, y}\right) \in \mathcal R \iff x \in f\left({y}\right)$

We will prove that
 * $(i): \quad \forall x, y \in S: \left({x, y}\right) \in \mathcal R \implies x \preceq y$

Let $x, y \in S$ such that
 * $\left({x, y}\right) \in \mathcal R$

By definitions of $\mathcal R$ and $F$:
 * $x \in f\left({y}\right) \subseteq y^\preceq$

By definition of subset;
 * $x \in y^\preceq$

Thus be definition of lower closure of element:
 * $x \preceq y$

We will prove that
 * $(ii): \quad \forall x, y, z, u \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u \implies \left({x, u}\right) \in \mathcal R$

Let $x, y, z, u \in S$ such that
 * $x \preceq y \land \left({y, z}\right) \in \mathcal R \land z \preceq u$

By definition of $\mathcal R$:
 * $y \in f\left({z}\right)$

By definition of increasing mapping:
 * $f\left({z}\right) \precsim f\left({u}\right)$

By definition of $\precsim$:
 * $f\left({z}\right) \subseteq f\left({u}\right)$

By definition of subset:
 * $y \in f\left({u}\right)$

By definition of $F$:
 * $f\left({u}\right)$ is an ideal in $R$.

By definition of ideal:
 * $f\left({u}\right)$ is a lower set.

By definition of lower set:
 * $x \in f\left({u}\right)$

Thus by definition of $\mathcal R$:
 * $\left({x, u}\right) \in \mathcal R$

We will prove that
 * $(iii): \quad \forall x, y, z \in S: \left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \implies \left({x \vee y, z}\right) \in \mathcal R$

Let $x, y, z \in S$ such that
 * $\left({x, z}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R$

By definition of $\mathcal R$:
 * $x \in f\left({z}\right)$ and $y \in f\left({z}\right)$

By definition of $F$:
 * $f\left({z}\right)$ is an ideal in $R$.

By definition of ideal:
 * $f\left({z}\right)$ is a directed lower set.

By definition of directed subset:
 * $\exists d \in f\left({z}\right): x \preceq d \land y \preceq d$

By definition of supremum:
 * $x \vee y \preceq d$

By definition of lower set:
 * $x \vee y \in f\left({z}\right)$

Thus by definition of $\mathcal R$:
 * $\left({x \vee y, z}\right) \in \mathcal R$

We will prove that
 * $(iv): \quad \forall x \in S: \left({\bot, x}\right) \in \mathcal R$

where $\bot$ denotes the smallest element in $R$.

Let $x \in S$.

By definition of $F$:
 * $f\left({x}\right)$ is an ideal in $R$.

By definition of ideal:
 * $f\left({z}\right)$ is a non-empty lower set.

By definition of non-empty set:
 * $\exists z: z \in f\left({x}\right)$

By definition of smallest element:
 * $\bot \preceq z$

By definition of lower set:
 * $\bot \in f\left({x}\right)$

Thus by definition of $\mathcal R$:
 * $\left({\bot, x}\right) \in \mathcal R$

By definition:
 * $\mathcal R$ is an auxiliary relation on $S$.

Thus by definitions of $\mathcal R$ and $\mathcal R$-segment:
 * $\forall x \in S: f\left({x}\right) = x^{\mathcal R}$