Preceding is Top in Ordered Set of Auxiliary Relations

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

Let ${\it Aux}\left({L}\right)$ be the set of all auxiliary relations on $S$.

Let $P = \left({ {\it Aux}\left({L}\right), \precsim}\right)$ be an ordered set where $\precsim \mathop = \subseteq\restriction_{{\it Aux}\left({L}\right) \times {\it Aux}\left({L}\right)}$

Then
 * $\preceq \mathop = \top_P$

where $\top_P$ denotes the greatest element in $P$.

Proof
By Preceding is Auxiliary Relation:
 * $\preceq \mathop \in {\it Aux}\left({L}\right)$

By definition
 * $\preceq$ is lower bound for $\varnothing$ in $P$

We will prove that
 * $\forall R \in {\it Aux}\left({L}\right): R$ is lower bound for $\varnothing \implies R \mathop \precsim \preceq$

Let $R \in {\it Aux}\left({L}\right)$

By condition $(i)$ of definition of auxiliary relation:
 * $R \mathop \subseteq \preceq$

Thus by definition of $\precsim$:
 * $R \mathop \precsim \preceq$

By definition of infimum:
 * $\preceq \mathop = \inf_P \varnothing$

Thus by Infimum of Empty Set is Greatest Element:
 * $\preceq \mathop = \top_P$