Increasing Mappings Satisfying Inclusion in Lower Closure is Isomorphic to Auxiliary Relations

Theorem
Let $R = \struct {S, \preceq}$ be a bounded below join semilattice.

Let $\map {\operatorname{Ids} } R$ be the set of all ideals in $R$.

Let $L = \struct {\map {\operatorname{Ids} } R, \precsim}$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{\map {\operatorname{Ids} } R \times \map {\operatorname{Ids} } R}$.

Let $M = \struct {F, \preccurlyeq}$ be the ordered set of increasing mappings $g:S \to \map {\operatorname{Ids} } R$ satisfying $\forall x \in S: \map g x \subseteq x^\preceq$.

Let $\map {\operatorname{Aux} } R$ be the set of all auxiliary relations on $S$.

Let $P = \struct {\map {\operatorname{Aux} } R, \precsim'}$ be an ordered set where $\precsim' \mathop = \subseteq\restriction_{\map {\operatorname{Aux} } R \times \map {\operatorname{Aux} } R}$.

Then
 * there exists an order isomorphism between $P$ and $M$

Proof
By Segment of Auxiliary Relation Mapping is Element of Increasing Mappings Satisfying Inclusion in Lower Closure define $G: \map {\operatorname{Aux} } R \to F$:
 * $\forall \RR \in \map {\operatorname{Aux} } R: \map G \RR = \paren {S \ni x \mapsto x^\RR}$

We will prove by Order Isomorphism is Surjective Order Embedding that:
 * $G$ is an order isomorphism.

Surjection
By Element of Increasing Mappings Satisfying Inclusion in Lower Closure is Generated by Auxiliary Relation:
 * $\forall f \in F: \exists \RR \in \map {\operatorname{Aux} } R: f = \paren {S \ni x \mapsto x^\RR}$

By definition of image of mapping:
 * $\forall f \in F: f \in \Img G$

Thus by definition:
 * $G$ is a surjection.

Order Embedding
Let $\RR, \QQ \in \map {\operatorname{Aux} } R$.

We will prove that:
 * $\RR \precsim' \QQ \implies \map G \RR \preccurlyeq \map G \QQ$

Let:
 * $\RR \precsim' \QQ$

By definition of $\precsim'$:
 * $\RR \subseteq \QQ$

Let $x \in S$.

By Relation Segment is Increasing:
 * $x^\RR \subseteq x^\QQ$

By Relation Segment of Auxiliary Relation is Ideal:
 * $x^\RR \in \map {\operatorname{Ids} } R$ and $x^\QQ \in \map {\operatorname{Ids} } R$

By definition of $\precsim$:
 * $x^\RR \precsim x^\QQ$

By definition of ordering on mappings:
 * $\paren {S \ni x \mapsto x^\RR} \precsim \paren {S \ni x \mapsto x^\QQ}$

Thus by definitions of $G$ and $\preccurlyeq$:
 * $\map G \RR \preccurlyeq \map G \QQ$

We will prove that:
 * $\map G \RR \preccurlyeq \map G \QQ \implies \RR \precsim' \QQ$

Let:
 * $\map G \RR \preccurlyeq \map G \QQ$

By definition of $G$ and $\preccurlyeq$:
 * $\paren {S \ni x \mapsto x^\RR} \precsim \paren {S \ni x \mapsto x^\QQ}$

Let $x, y \in S$ such that:
 * $\tuple {x, y} \in \RR$

By definition of $\RR$-segment:
 * $x \in y^\RR$

By definition of ordering on mappings:
 * $y^\RR \precsim y^\QQ$

By definition of $\precsim$:
 * $y^\RR \subseteq y^\QQ$

By definition of subset:
 * $x \in y^\QQ$

Thus by definition of $\QQ$-segment:
 * $\tuple {x, y} \in \QQ$

By definition of subset:
 * $\RR \subseteq \QQ$

Thus by definition of $\precsim'$:
 * $\RR \precsim' \QQ$