Let $R\colon A\mathrel {\rightarrow \kern -9.5pt\mathrlap {|}\kern 6pt}B$ be a relation.
Proof of Proposition 6.1.5.1.2.
Item 1: Characterisations
We claim that Item (a) and Item (b) are indeed equivalent: - Item (a)$\implies $Item (b): We have to show that, for each $\webleft (a,a'\webright )\in A$, we have
\[ \chi _{A}\webleft (a,a'\webright )\preceq _{\{ \mathsf{t},\mathsf{f}\} }\webleft [R^{\dagger }\mathbin {\diamond }R\webright ]\webleft (a,a'\webright ), \]
i.e. that if $a=a'$, then there exists some $b\in B$ such that $a\sim _{R}b$ and $b\sim _{R^{\dagger }}a'$ (i.e. $a\sim _{R}b$ again), which follows from the totality of $R$.
- Item (b)$\implies $Item (a): Given $a\in A$, since $\chi _{A}\subset R^{\dagger }\mathbin {\diamond }R$, we must have
\[ \webleft\{ a\webright\} \subset \webleft [R^{\dagger }\mathbin {\diamond }R\webright ]\webleft (a\webright ), \]
implying that there must exist some $b\in B$ such that $a\sim _{R}b$ and $b\sim _{R^{\dagger }}a$ (i.e. $a\sim _{R}b$) and thus $R\webleft (a\webright )\neq \text{Ø}$, as $b\in R\webleft (a\webright )$.
