Define $I := \left\{ {u \in S: \exists y \in S: \left({u, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R}\right\}$ By definition of auxiliary relation: :$\left({\bot, \bot}\right) \in \mathcal R$ and $\left({\bot, z}\right) \in \mathcal R$ where $\bot$ denotes the smallest element in $L$. Then :$\bot \in I$ By definition: :$I$ is a non-empty set. We will prove that :$I$ is a lower set. Let $a \in I, b \in S$ such that :$b \preceq a$ By definition of $I$: :$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$ By definitions of auxiliary relation and reflexivity: :$\left({b, s}\right) \in \mathcal R$ Thus :$b \in I$ {{qed|lemma}} We will prove that :$I$ is directed. Let $a, b \in I$. By definition of $I$: :$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$ and :$\exists t \in S: \left({b, t}\right) \in \mathcal R \land \left({t, z}\right) \in \mathcal R$ By Auxiliary Relation is Congruent: :$\left({a \vee b, s \vee t}\right) \in \mathcal R$ By definition of auxiliary relation: :$\left({s \vee t, z}\right) \in \mathcal R$ Thus by definition of $I$: :$a \vee b \in I$ Thus by Join Succeeds Operands: :$a \preceq a \vee b$ and $b \preceq a \vee b$ {{qed|lemma}} By definition: :$I$ is ideal in $L$. We will prove that :$I \subseteq z^{\mathcal R}$ Let $a \in I$. By definition of $I$: :$\exists s \in S: \left({a, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$ By definition of auxiliary relation: :$a \preceq s$ Again by definition of auxiliary relation and reflexivity: :$\left({a, z}\right) \in \mathcal R$ Thus by definition of $\mathcal R$-segment: :$a \in z^{\mathcal R}$ {{qed|lemma}} By Supremum of Subset and definition of approximating relation: :$\sup I \preceq \sup \left({z^{\mathcal R} }\right) = z$ We will prove that :$\sup I = z$ Aiming for a contradiction suppose that :$\sup I \ne z$ By definition of $\prec$: :$\sup I \prec z$ Then by definition of antisymmetry: :$z \npreceq \sup I$ By Not Preceding implies Approximating Relation and not Preceding: :$\exists y \in S: \left({y, z}\right) \in \mathcal R \land y \npreceq \sup I$ Again by Not Preceding implies Approximating Relation and not Preceding: :$\exists u \in S: \left({u, y}\right) \in \mathcal R \land u \npreceq \sup I$ By definition of $I$: :$u \in I$ This contradicts $u \preceq \sup I$ by definition of supremum. {{qed|lemma}} By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal: :$x \in I$ By definition of $I$: :$\exists s \in S: \left({x, s}\right) \in \mathcal R \land \left({s, z}\right) \in \mathcal R$ By definition of auxiliary relation: :$x \preceq s$ By definition of auxiliary relation and reflexivity: :$\left({x, z}\right) \in \mathcal R$ By Auxiliary Approximating Relation has Quasi Interpolation Property: :$\exists y \in S: x \preceq y \land \left({y, z}\right) \in \mathcal R \land x \ne y$ By definition of $\prec$: :$ x \prec y$ Define $Y := s \vee y$ By Join Succeeds Operands: :$s \preceq Y$ and $y \preceq Y$ Then  :$x \ne Y$ By definition of reflexivity: :$x \preceq x$ By definition of auxiliary relation: :$\left({x, Y}\right) \in \mathcal R$ Again by definition of auxiliary relation: :$\left({Y, z}\right) \in \mathcal R$ Hence :$\exists y \in S: \left({x, y}\right) \in \mathcal R \land \left({y, z}\right) \in \mathcal R \land x \ne y$ {{qed}}