4394 lines
236 KiB
TeX
4394 lines
236 KiB
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\documentclass[envcountsect,runningheads,draft]{llncs} % fails
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%
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\title{Coalgebraic Simulation.}
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% an abbreviated paper title here
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\author{Sergey Goncharov\inst{1}\orcidID{0000-0001-6924-8766} \and
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Pouya Partow\inst{1}\orcidID{0009-0003-9652-9469}}
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%
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\authorrunning{S.~Goncharov, P.~Partow}
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% First names are abbreviated in the running head.
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% If there are more than two authors, 'et al.' is used.
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%
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\institute{University of Birmingham, UK\\
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\email{\{s.goncharov,p.partow\}@bham.ac.uk}}
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%
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\maketitle % typeset the header of the contribution
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%
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\begin{abstract}
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Hello simulation!
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\end{abstract}
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%
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%
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\section{Liftable Orders}
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\begin{definition}[Natural Order Structure]\label{def:nat-ord}
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A \emph{natural order structure} on a functor $F$ is a poset $\appr$ on each Hom-set of the form $\Hom(X,FY)$ such that if $\alpha\appr\beta$ in $\Hom(X,FY)$, $f\c X'\to X$, $g\c Y\to Y'$, then:
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\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
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\item $\alpha\comp f\appr\beta\comp f$ in $\Hom(X',FY)$. \label{item:nat-ord:I}
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\item $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,FY')$. \label{item:nat-ord:II}
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\end{enumerate}
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\end{definition}
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We want to say that a natural order structure entails an order over a functor defined by Jacobs and Hughes that is a functor of the type $\Set\to\poset$.
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\begin{prop}
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Assuming that $F\c\Set\to\Set$ is a functor, and we have a natural order structure on $F$, then we have a functor $\tilde{F}\c\Set\to\poset$, such that $U\comp \tilde{F}=F$, where $U$ is a forgetful functor.
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\end{prop}
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\begin{proof}
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We take $\tilde{F}X=\Hom(1,FX)$. Assuming that $U$ takes every poset to its carrier set, then we have $U\comp\tilde{F}X=FX$ for every object $X$.\qed
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\end{proof}
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\begin{definition}[Liftable Order Structure]\label{def:liftable-ord}
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A \emph{liftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $h\appr Fg\comp k$ in $\Hom(X,FZ)$, and $g$ is epic, then there is $k'\c X\to FY$ such that $k'\appr k$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.\ppnote{I added the surjection condition on $g$.}
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\end{definition}
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\begin{definition}[Coliftable Order Structure]\label{def:coliftable-ord}
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A \emph{coliftable order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $Fg\comp k\appr h $ in $\Hom(X,FZ)$, and $g$ is epic, then there is $k'\c X\to FY$ such that $k\appr k'$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
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\end{definition}
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\begin{lemma}\label{lem:set-ord-str}
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In $\Set$, assuming that for every set $Y$ we have an order $\leq$ on $\Hom(1,FY)$ that satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, and if for $A\in\Hom(1,FZ)$, $B\in\Hom(1,FY)$, and $g\c Y\to Z$, $h\leq Fg(B)$ in $\Hom(1,FZ)$ there exists $B'\in\Hom(1,FY)$ such that $B'\leq B$ and $B\in\Hom(1,FY)$ and $A=Fg(B')$, then there exists an order structure $\appr$ on $F$ that is a liftable order structure.
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\end{lemma}
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\begin{proof}
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For every $f,g\in\Hom(X,FY)$, we define $f\appr g$ whenever for all $x\in X$, $f(x)\leq g(x)$.
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Now, assuming $f\c X'\to X$, $\alpha,\beta\c X\to FY$, and $\alpha\appr\beta$, then for every $x'\in X'$, we have $\alpha(f(x'))\leq\beta(f(x'))$, so we have $\alpha\comp f\appr \beta\comp f$. Additionally, if we assume $g\c Y\to Y'$, then since $\leq$ satisfies \autoref{def:nat-ord}.\eqref{item:nat-ord:II}, for every $x\in X$ we have $Fg\comp\alpha(x)\leq Fg\comp\beta(x)$.
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Furthermore, if we have $g\c Y\to Z$, $k\c X\to FY$, $h\c X\to FZ$, and $Fg\comp k\appr h$, then for every $x\in X$ we have $Fg\comp k(x)\leq h(x)$, thus by the assumption of the lemma, for every $x$, there exists a set that we call $k'(x)$, for which we have $Fg(k'(x))\leq h(x)$. So, we have a function $k'\c X\to FY$, such that for every $x$ we have $Fg\comp k'(x)\leq h(x)$ that means $Fg\comp k'\appr h$.
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\qed
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\end{proof}
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\begin{remark}\label{rem:set-ord-str-co}
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The exact same argument in~\autoref{lem:set-ord-str} is true for coliftable order structures.
|
|
\end{remark}
|
|
|
|
\begin{lemma}\label{lem:liftable}
|
|
Assuming that a a functor $F$ has an order structure $\appr$ that is liftable, then for every surjective $f\in\Hom(X,FY)$ we have:
|
|
\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
\item $Ff\comp\sappr\quad=\quad\sappr\comp Ff$
|
|
\item $(Ff)^\op\comp\appr\quad=\quad\appr\comp (Ff)^\op$
|
|
\end{enumerate}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
$(I)$ Assuming $t\mathrel{(Ff\comp\sappr)} x$, there exists $s$ such that $t\sappr s$ and $Ff(s)=x$. Since $\appr$ is liftable, and thus natural, by~\autoref{def:nat-ord}, from $s\appr t$ we get $x\appr Ff(t)$ that is $t\mathrel{(\sappr\comp Ff)} x$.
|
|
|
|
Assuming $t\mathrel{(\sappr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\sappr x$. By~\autoref{def:liftable-ord} since $Ff(t)\sappr x$ there exists $s$ that $t\sappr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\sappr)}s$.
|
|
|
|
$(II)$ Basically, by definition of $\op$ and relation composition we have
|
|
\begin{gather*}
|
|
(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
|
|
(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
|
|
\end{gather*}
|
|
So it follows directly from applying $\op$ on both sides of $(I)$.\qed
|
|
\end{proof}
|
|
\begin{lemma}\label{lem:coliftable}
|
|
Assuming that a a functor $F$ has an order structure $\appr$ that is coliftable, then for every surjective $f\in\Hom(X,FY)$ we have:
|
|
\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
\item $Ff\comp\appr\quad=\quad\appr\comp Ff$
|
|
\item $(Ff)^\op\comp\sappr\quad=\quad\sappr\comp (Ff)^\op$
|
|
\end{enumerate}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
$(I)$ Assuming $t\mathrel{(Ff\comp\appr)} x$, there exists $s$ such that $t\appr s$ and $Ff(s)=x$. Since $\appr$ is coliftable, and thus natural, by~\autoref{def:nat-ord}, from $t\appr s$ we get $Ff(t)\appr x$ that is $t\mathrel{(\appr\comp Ff)} x$.
|
|
|
|
Assuming $t\mathrel{(\appr\comp Ff)} x$, there exists $y$ such that $Ff(t)=y$ and $y\appr x$. By~\autoref{def:coliftable-ord} since $Ff(t)\appr x$ there exists $s$ that $t\appr s$ and $Ff(s)=x$ that is $t\mathrel{(Ff\comp\appr)}s$.
|
|
|
|
$(II)$ Basically, by definition of $\op$ and relation composition we have
|
|
\begin{gather*}
|
|
(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
|
|
(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
|
|
\end{gather*}
|
|
So it follows directly from applying $\op$ on both sides of $(I)$.\qed
|
|
\end{proof}
|
|
|
|
\begin{prop}
|
|
Assuming that $F$ and $G$ are endofunctors on a category $\BC$, and $\appr$ is a liftable order structure on $F$, if $G$ preserves epics, then $\appr$ is a liftable order structure on $FG$ as well.
|
|
\end{prop}
|
|
\begin{proof}
|
|
$\appr$ being a liftable order structure on $F$ means that for morphisms $g\c Y\to Z$, $k\in\Hom(X,FY)$, and $h\in\Hom(X,FZ)$ that $g$ is epic, if $h\appr Fg\comp k$ then exists $k'\in\Hom(X,FY)$, such that $k\appr k'$ and $h=Fg\comp k'$. %Now, assuming $\alpha\c Y\to Z$, $\mu\in\Hom(X,FGY)$, and $\nu\in\Hom(X,FGZ)$ that $\alpha$ is epic, we need to prove that exists some $\mu'\in\Hom(X,FGY)$ such that $\mu'\appr\mu$ and $\nu=FG\alpha\comp\mu'$. Since $G$ preserves epimorphisms and $\alpha$ is epic, then $G\alpha$ is also epic. Furthermore, since $\appr$ is a liftable order structure on $F$, then the mentioned $\mu'$ exists.\qed
|
|
Now, assuming $\alpha\c Y\to Z$, $\mu\in\Hom(X,FGY)$, and $\nu\in\Hom(X,FGZ)$ that $\alpha$ is epic. Since $G$ is assumed to preserve epimorphisms, $G\alpha$ is epic. Since $\appr$ is a liftable order structure on $F$, then there exists some $\mu'\in\Hom(X,FGY)$ such that $\mu'\appr\mu$ and $\nu=FG\alpha\comp\mu'$. So, $\appr$ is a liftable order structure for $FG$ as well.\qed
|
|
\end{proof}
|
|
\begin{rem}
|
|
For every regular category $\BC$ with the axiom of choice, every functor $G\c\BC\to\BC$ preserves epimorphisms. Assuming $e\c X\to Y$ is epic, then it has a section $s\c Y\to X$. Applying $G$ on both of them we have $Ge\c GX\to GY$ and $Gs\c GY\to GX$ that
|
|
\begin{align*}
|
|
Ge\comp Gs&\\
|
|
=&G(e\comp s)\\
|
|
=&G(\id_Y)\\
|
|
=&\id_{GY}
|
|
\end{align*}
|
|
that means that $Gs$ is a section for $Ge$ that entails that $Ge$ is epic.
|
|
\end{rem}
|
|
\todo{It seems too much, but perhaps you can study if you can derive an order structure from $F$ to $G$ and consequently $GF$, by the following rule:
|
|
\begin{gather*}
|
|
\infer{Gh\appr GFg\comp Gk}{h\appr Fg\comp k}
|
|
\end{gather*}}
|
|
\subsection{Powerset Functor}
|
|
In this section we discuss set inclusion as an ordering over the powerset functor.
|
|
\begin{prop}\label{prop:lift-gen-func}
|
|
For a functor $F\c\Set\to\Set$, a functor of the form $\powf F$, where $\powf$ is the powerset functor, the set inclusion is a liftable order structure.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Using~\autoref{lem:set-ord-str} we only prove the case for every $h\in\Hom(1,\powf FZ)$, $k\in\Hom(1,\powf FY)$. Additionally, for every $g\c Y\to Z$, such that $h\subseteq\powf Fg(k)$, we define $k'\in\powf FY$ that that $k'=\{y\mid Fg(y)\in h\}$. We show that $k'\subseteq k$ and $\powf Fg(k')=h$.
|
|
|
|
Assuming $y\in k'$ we have $Fg(y)\in h$. Since $h\subseteq \powf Fg(k)$ we have $Fg(y)\in\powf Fg(k)$, so we have $y\in k$ that means $k'\subseteq k$.
|
|
|
|
Assuming $z\in\powf Fg(k')$, then there exists $y'\in k'$ that $z=Fg(y')$, so $z\in h$ and $\powf Fg(k')\subseteq h$.
|
|
|
|
Assuming $z\in h$, since $h\subseteq \powf Fg(k)$, then $z\in\powf Fg(k)$. So, there exists $y'\in k$ such that $Fg(y')=z$. So, by the definition of $k'$ we have $y'\in k'$ that means $z\in\powf Fg(k')$.\qed
|
|
\end{proof}
|
|
|
|
\begin{example}
|
|
In the category of sets, subset over the powerset functor is an example of a coliftable order structure if the $g$ in~\autoref{def:coliftable-ord} is a surjective. Using~\autoref{rem:set-ord-str-co} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, $g\c Y\to Z$, such that $\powf g(k)\subseteq h$. We define $k'=k\cup\{y\mid g(y)\in h\}$, and we show that $\powf g(k')=h$.
|
|
|
|
Obviously, $\powf g(k')\subseteq h$. Now, assuming $z\in h$ we prove that $z\in \powf g(k')$. Since $g$ is surjective, then exists $A\subseteq Y$ such that $\powf g(A)=h$. By the definition of $k'$, $A\subseteq k'$. So from $z\in h$ we have $z\in \powf g(A)$ that means that exists $a\in A$, such that $g(a)=z$. Now, since $A\subseteq k'$, then $a\in k'$. So, we have $z\in \powf g(k')$.\qed
|
|
\end{example}
|
|
|
|
\begin{remark}
|
|
We give an example that justifies why the $g$ in~\autoref{def:coliftable-ord} should be surjective.
|
|
If we take the surjectivity of $g$ out of~\autoref{def:coliftable-ord}, subset relation over the powerset functor is not an example of a coliftable order! For some set $X$ we take $h\c X\to\powf\mathbb{Z}$, for every $x\in X$, $h(x)=\mathbb{Z}$, $g\c\mathbb{Z}\to\mathbb{Z}$, and for every $z\in\mathbb{Z}$,
|
|
\begin{gather*}
|
|
g(z)=
|
|
\begin{cases}
|
|
0 & z\in\mathbb{Z}^+ \\
|
|
1 & \mathsf{otherwise}
|
|
\end{cases}
|
|
\end{gather*}
|
|
then
|
|
\begin{gather*}
|
|
\powf g(A)=
|
|
\begin{cases}
|
|
\emptyset & A=\emptyset \\
|
|
\{0\} & A\subseteq \mathbb{Z}^+,A\neq\emptyset\\
|
|
\{1\} & A\subseteq \mathbb{Z}^-\cup\{0\},A\neq\emptyset\\
|
|
\{0,1\} & \mathsf{otherwise}
|
|
\end{cases}
|
|
\end{gather*}
|
|
then no matter what $k\c X\to\powf\mathbb{Z}$ is there will be no $k'\c X\to\powf\mathbb{Z}$, for every $x\in X$, $\powf g(k'(x))=h(x)$, as $\powf g(k'(x))\subseteq\{0,1\}$, while $h(x)=\mathbb{Z}$, so $\powf g(k'(x))\subset h(x)$.
|
|
\end{remark}
|
|
|
|
\begin{remark}
|
|
We could not prove a general statement like~\autoref{prop:lift-gen-func} for coliftability even if the surgectivity of $g$ remains in the definition. Perhaps more conditions on $F$ in~\autoref{prop:lift-gen-func} are needed. Conditions like preserving surjections and unions that make the statement extremely limited, and meaningless.
|
|
\end{remark}
|
|
|
|
\subsection{Maybe Functor}
|
|
The order structure that we can define for this functor is that for sets $X$ and $Y$, and functions $f,g\c X\to Y+1$ we have $f\appr g$ whenever $\Dom(f)\subseteq\Dom(g)$, and for every $x\in\Dom(f)$ we have $f(x)=g(x)$ ($\Dom(f)$ is the domain of a function $f$).
|
|
\begin{prop}\label{prop:maybe-lif}
|
|
The order structure on the set-functor $FX=X+1$ is a liftable order.
|
|
\end{prop}
|
|
% By~\autoref{lem:set-ord-str}, assuming $h\in\Hom(1,Y+1)$, $k\in\Hom(1,X+1)$, $g\c X\to Y$, and $h\appr Fg(k)$, we need to prove that exists $k'\in\Hom(1,X+1)$ such that $k'\appr k$ and $Fg(k')=h$. Since $h\in\Hom(1,Y+1)$ we have two cases:
|
|
% \begin{itemize}
|
|
% \item $h=\bot$: In this case we take $k'=\bot$, so we have $k'\appr k$, and then we have $Fg(k')=\bot=h$.
|
|
% \item $h\in Y$: Since $h\appr Fg(k)$ and $h\neq\bot$, we have $h=Fg(k)$. In this case we take $k'=k$, so $k'\appr k$ and $Fg(k')=h$.\qed
|
|
% \end{itemize}
|
|
\begin{proof}
|
|
Assuming that $h\in\Hom(X,Z+1)$, $g\c Y\to Z$, $k\in\Hom(X,Y+1)$, such that $h\appr (g+1)\comp k$. We define $k'\in\Hom(X,Y+1)$ as follows:
|
|
\begin{gather*}
|
|
k'(x)=
|
|
\begin{cases}
|
|
\bot&x\notin\Dom(h)\\
|
|
k(x)&x\in\Dom(h)
|
|
\end{cases}
|
|
\end{gather*}
|
|
We have $\Dom((g+1)\comp k')=\Dom(k')$ and $\Dom(k')=\Dom(h)$, so we have $\Dom((g+1)\comp k')=\Dom(h)$. If $x\in\Dom(h)$, then $h(x)=(g+1)\comp k(x)$ and $(g+1)\comp k(x)=(g+1)\comp k'(x)$, so we have $h(x)=(g+1)\comp k'(x)$. So, we have $h=(g+1)\comp k'$. Additionally, $\Dom(k')\subseteq\Dom(k)$, and for every $x\in \Dom(k')$, we have $k'(x)=k(x)$, so we have $k'\appr k$. \qed
|
|
\end{proof}
|
|
\begin{prop}\label{prop:maybe-colif}
|
|
The order structure on the set-functor $FX=X+1$ is a coliftable order.
|
|
\end{prop}
|
|
\begin{proof}
|
|
By~\autoref{lem:set-ord-str}, assuming $h\in\Hom(1,Y+1)$, $k\in\Hom(1,X+1)$, $g\c X\to Y$, and $Fg(k)\appr h$, we need to prove that exists $k'\in\Hom(1,X+1)$ such that $k\appr k'$ and $Fg(k')=h$. Since $h\in\Hom(1,Y+1)$ we have two cases:
|
|
\begin{itemize}
|
|
\item $Fg(k)=h$: In this case we take $k'=k$, and then we have $Fg(k')=h$.
|
|
|
|
\item $Fg(k)=\bot$: It entails that $k=\bot$. We either have $h=\bot$ or $h\in Y$. If $h=\bot$ then we take $k'=k$, and we are done. If $h\in Y$, then by the surjectivity of $g$, there exists $k'$ such that $g(k')=h$ that entails $Fg(k')=h$ as well.\qed
|
|
\end{itemize}
|
|
\end{proof}
|
|
|
|
\subsection{Subdistribution Functor}
|
|
The subdistribution functor $\sub\c\Set\to\Set$ is defined as $\sub X=\{\mu\c X\to[0,1]\mid \sum_{x\in X}\mu(x)\leq 1\}$ on objects, and for $\sub f\c \sub X\to\sub Y$, we have
|
|
\begin{gather*}
|
|
\sub f(\mu)=y\mapsto\sum_{x\in f^{\mone}(y)}\mu(x),
|
|
\end{gather*}
|
|
on morphisms. We define the ordering on $\sub X$ for $\mu_1,\mu_2\in \sub X$ as
|
|
\begin{gather*}
|
|
\mu_1\appr \mu_2\iff \forall x\in X, \mu_1(x)\leq\mu_2(x).
|
|
\end{gather*}
|
|
The definition derives the ordering on morphisms of each hom-set $\Hom(X,\sub Y)$ as usual in $\Set$.
|
|
\begin{prop}
|
|
The pointwise ordering on $\sub$ is a liftable ordering.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming that $\mu\in SX$, $g\c X\to Y$ is surjective, $\nu\in SY$, and $\nu\appr \sub g(\mu)$, then there exists $\mu'\in \sub X$ such that $\mu'\appr \mu$ and $\sub g(\mu')=\nu$.\\
|
|
The assumption $\nu \appr Sg(\mu)$ means that for every $y \in Y$,
|
|
\begin{gather*}
|
|
\nu(y) \leq Sg(\mu)(y) = \sum_{x \in g^{\mone}(y)} \mu(x).
|
|
\end{gather*}
|
|
We construct $\mu'$ as follows:
|
|
\begin{gather*}
|
|
\mu'(x)=
|
|
\begin{cases}
|
|
\frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)&Sg(\mu)(g(x))\neq 0\\
|
|
0&Sg(k)(g(x))= 0
|
|
\end{cases}
|
|
\end{gather*}
|
|
Now, we need to prove that $\mu'$ is a subdistribution, but we do not need to prove it directly. Proving that $\mu'\appr\mu$ entails that $\mu'$ is a subdistribution. So, we prove that $\mu'\appr\mu$.
|
|
For every $x\in X$, have the following cases:
|
|
\begin{itemize}
|
|
\item $Sg(\mu)(g(x))= 0$: We have $\mu'(x)=0$, and obviously $\mu'\appr\mu$.
|
|
\item $Sg(\mu)(g(x))\neq 0$: We have $\frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)$, and since $\nu\appr Sg(\mu)$ then we have:
|
|
\begin{align*}
|
|
&\quad\frac{\nu(g(x))}{Sg(\mu)(g(x))}\leq 1\\
|
|
\Rightarrow&\quad\frac{\nu(g(x))}{Sg(\mu)(g(x))}\comp \mu(x)\leq \mu(x)
|
|
\end{align*}
|
|
\end{itemize}
|
|
% First, we need to prove that $\mu'$ is a subdistribution.
|
|
% For every $x \in X$, $\mu'(x) \ge 0$. We have:
|
|
% \begin{align*}
|
|
% \sum_{x \in X} \mu'(x)&\\
|
|
% =& \sum_{x \in X} \frac{\nu(g(x))}{Sg(\mu)(g(x))} \comp \mu(x)&(Sg(\mu)(g(x))\neq 0)\\
|
|
% =& \sum_{y \in Y} \sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)&(Sg(\mu)(y)\neq 0)\\
|
|
% =& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)&(Sg(\mu)(y)\neq 0)\\
|
|
% =& \sum_{y \in Y} \frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)&(Sg(\mu)(y)\neq 0)\\
|
|
% =& \sum_{y \in Y} \nu(y)\\
|
|
% \leq& 1
|
|
% \end{align*}
|
|
% As for every $x\in X$ that $Sg(\mu)(g(x))=0$, $\mu'(x)=0$, it does not effect the inequality. Thus $\mu'$ is a subdistribution.
|
|
Now, we prove $Sg(\mu') = \nu$.
|
|
For any $y \in Y$, we have:
|
|
\begin{itemize}
|
|
\item $Sg(\mu)(y) = 0$: We have $\nu(y) = 0$, and the sum is $0$ as well, so the equality holds.\\
|
|
\item $Sg(\mu)(y)\neq 0$:
|
|
\begin{align*}
|
|
Sg(\mu')(y)&\\
|
|
= &\sum_{x \in g^{\mone}(y)} \mu'(x)\\
|
|
= &\sum_{x \in g^{\mone}(y)} \frac{\nu(y)}{Sg(\mu)(y)} \comp \mu(x)\\
|
|
= &\frac{\nu(y)}{Sg(\mu)(y)} \comp \sum_{x \in g^{\mone}(y)} \mu(x)\\
|
|
= &\frac{\nu(y)}{Sg(\mu)(y)} \comp Sg(\mu)(y)\\
|
|
= &\nu(y)
|
|
\end{align*}
|
|
\end{itemize}\qed
|
|
\end{proof}
|
|
|
|
\section{Spans and Relations}
|
|
%In this section, by $\spa(\BC)$ we refer to spans in a category $\BC$ that has
|
|
%products, and by $\rel(\BC)$ we refer to the category of relations in $\BC$, i.e.\
|
|
%such spans $(X \stackrel{p_1}{\leftarrow} R
|
|
%\stackrel{p_2}{\to}Y)$ that the morphism $\brks{p_1,p_2}$ is a mono.
|
|
%We denote morphisms in $\spa(\BC)$ by $\spto$, and in $\rel(\BC)$ by $\rto$.
|
|
%A morphism $R\rto S$ ($R\spto S$) in $\rel(\BC)$
|
|
%($\spa(\BC)$) is such a triple of morphisms $(f\c R\to S, g_1\c X_1\to Y_1, g_2\c X_2\to Y_2)$
|
|
%in $\BC$ that the following diagram commutes:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_1} \& R \& {X_2} \\
|
|
% {Y_1} \& S \& {Y_2}
|
|
% \arrow["{g_1}"', from=1-1, to=2-1]
|
|
% \arrow["{{p_1}}"', from=1-2, to=1-1]
|
|
% \arrow["{{p_2}}", from=1-2, to=1-3]
|
|
% \arrow["f", from=1-2, to=2-2]
|
|
% \arrow["{g_2}", from=1-3, to=2-3]
|
|
% \arrow["{q_1}", from=2-2, to=2-1]
|
|
% \arrow["{q_2}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%
|
|
%There exists a well-known notion named relator, instead of relation lifting. A relator does not need to be a functor, but it should be a map of a specific type format that is monotone with respect to inclusion. Also, relators are defined only on $\Set$ unlike relation liftings. %We discuss relators more in depth in the later chapters.
|
|
%For the time being, we limit the discussion to the case $\BC=\Set$. For simplicity, by $\rel$ and $\spa$ we mean $\rel(\Set)$ and $\spa(\Set)$, accordingly.
|
|
%
|
|
%We can define morphisms in $\rel$ differently by only requesting such functions $g_1$ and $g_2$ that $x\mathrel{R}y$ entails $g_1(x)\mathrel{S}g_2(y)$. Let us
|
|
%call such morphisms \emph{anonymous} (because they omit the witnessing part $f$, which is unique for relations but not for general spans). This however yields
|
|
%an equivalent definition. \todo{Add a proof.} \todo{Do the same for spans; prove equivalence under the axiom of choice.}
|
|
%We define the category of spans and relations.
|
|
\begin{definition}[Category of Spans]\ppnote{Paul Levy said he doesn't like this name for this category as "category of spans" is used for a different category that is a bicategory.}
|
|
For an arbitrary category $\BC$, the category of spans, denoted by $\spa(\BC)$ is the category that for every objects $R$, $X_1$ and $X_2$ in $\BC$, and every morphisms $p_1\c R\to X_1$, and $p_2\c R\to X_2$ in $\BC$, has an object $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$, and for every morphisms $g_1\c X_1\to Y_1$, $g_2\c X_2\to Y_2$, and $w\c R\to S$ in $\BC$, has a morphism $(g_1,g_2,w)$ of type $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to (Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, whenever the following diagram commutes:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{X_1} \& R \& {X_2} \\
|
|
{Y_1} \& S \& {Y_2}
|
|
\arrow["{g_1}"', from=1-1, to=2-1]
|
|
\arrow["{{p_1}}"', from=1-2, to=1-1]
|
|
\arrow["{{p_2}}", from=1-2, to=1-3]
|
|
\arrow["f", from=1-2, to=2-2]
|
|
\arrow["{g_2}", from=1-3, to=2-3]
|
|
\arrow["{q_1}", from=2-2, to=2-1]
|
|
\arrow["{q_2}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{definition}
|
|
|
|
\begin{definition}[Category of Relations]
|
|
For an arbitrary category $\BC$\sgnote{In this definition $\BC$ is not arbitrary -- it must have products. But this can be fixed by reformulating in terms of joint monics.}, the category of relations, denoted by $\rel(\BC)$ has as objects such spans $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ that $\brks{p_1,p_2}$ is a monomorphism. A morphism $(g_1,g_2,w)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to (Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa(\BC)$ is a morphism in $\rel(\BC)$ as well, whenever both $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ are objects in $\rel(\BC)$ as well, and $(g_1,g_2,w)$ is a morphism in $\spa(\BC)$.
|
|
\end{definition}
|
|
%
|
|
%\begin{prop}
|
|
% Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects existing in both categories $\rel(\BC)$ and $\spa(\BC)$, there is a morphism $(g_1,g_2,w)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa(\BC)$ iff there is a morhpism $(g_1,g_2,w)$ of the same type in $\rel(\BC)$.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% Trivial by the definitions.\qed
|
|
%\end{proof}
|
|
%
|
|
\subsection{Spans and Relations in $\Set$}
|
|
From now on, instead of $\spa(\Set)$ and $\rel(\Set)$, we use $\spa$ and $\rel$ accordingly. There are two choices for defining morphisms in each of the categories $\spa$ and $\rel$, namely \emph{anonymous} and \emph{onymous}. The already introduced type of morphisms is onymous. For functions $g_1\c X_1\to Y_1$ and $g_2\c X_2\to Y_2$, $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2) \to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ is an anonymous morphism in $\spa$ if:
|
|
\begin{gather*}
|
|
u\in R\Rightarrow \exists v\in S, q_1(v)= g_1\comp p_1(u) \quad\&\quad q_2(v)=g_2\comp p_2(u)
|
|
\end{gather*}
|
|
In case we are defining anonymous morphisms in $\rel$ we have the stronger version of the above property that forces the elements of relations to be pairs and the existential quantifier refers to a unique element:
|
|
\begin{gather*}
|
|
(x_1,x_2)\in R\Rightarrow (g_1(x_1),g_2(x_2))\in S
|
|
\end{gather*}
|
|
From now on, we use $\spa_a$ and $\rel_a$ to denote the categories of spans and relations with anonymous morphisms, and $\spa$ and $\rel$ to denote the categories of spans and relations with onymous morphisms.
|
|
\begin{prop}\label{prop:rel-rela}
|
|
Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects existing in both categories $\rel$ and $\rel_a$, there is a morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\rel_a$ iff there is a morphism $(g_1,g_2,w)$ of the same type in $\rel$, with the witness\sgnote{Unique?} $w\c R\to S$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
$(\Rightarrow):$ For every $(x_1,x_2)\in R$ we define $w$ as $w(x_1,x_2)=(g_1(x_1),g_2(x_2))$. We have:
|
|
\begin{align*}
|
|
g_1\comp p_1(x_1,x_2)&\\
|
|
=&g_1(x_1)\\
|
|
=&q_1(g_1(x_1),g_2(x_2))\\
|
|
=&q_1\comp w(x_1,x_2)
|
|
\end{align*}
|
|
Similarly, we have $g_2\comp p_2(x_1,x_2)=q_2\comp w(x_1,x_2)$.
|
|
|
|
$(\Leftarrow):$ Assuming $x_1\mathrel{R}x_2$, then $w(x_1,x_2)\in S$ as well, and by the definition of onymous morphisms we have $w(x_1,x_2)=(g_1(x_1),g_2(x_2))$, so we have $g_1(x_1)\mathrel{S}g_2(x_2)$. \qed
|
|
\end{proof}
|
|
|
|
\begin{prop}\label{prop:spa-spaa}
|
|
Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects existing in both categories $\spa$ and $\spa_a$, there is a morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa_a$ iff there is a morhpism $(g_1,g_2,w)$ of the same type in $\spa$, with the witness $w\c R\to S$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
($\Rightarrow$): We need the axiom of choice to prove this statement. By the definition of anonymous morphisms, for every $u\in R$ there exists $v\in S$ such that $q_1(v)=p_1\comp g_1(u)$ and $q_2(v)=p_2\comp g_2(u)$. So, for each $u\in R$ there exists a set $V_u\in\powf S$ that its elements have the mentioned properties. We can form a function $h\c R\to \powf S$ that $h(u)=V_u$. The image of $h$ that we denote with $\im(h)$ is a family of non-empty sets. By the axiom of choice, there exists a function $s\c \im(h)\to S$. We define $w\c R\to S$ as $w=s\comp h$, then for every $u\in R$, we have $q_1\comp w(u)=g_1\comp p_1(u)$ and $q_2\comp w(u)=g_2\comp p_2(u)$.
|
|
|
|
($\Leftarrow$): Assuming $u\in R$, there exists $w(u)\in S$, and by the definition of onymous morphisms we have $q_1(w(u))=g_1\comp p_1(u)$ and $q_2(w(u))=g_2\comp p_2(u)$. \qed
|
|
\end{proof}
|
|
|
|
\begin{prop}\label{prop:rela-spaa}\sgnote{This becomes trivial, if we define Rel as full subcategory of Span.}
|
|
Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects existing in both categories $\rel_a$ and $\spa_a$, there is a morphism $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa_a$ iff there is a morhpism $(g_1,g_2)$ of the same type in $\rel_a$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
($\Rightarrow$): Assuming that $(g_1,g_2)$ is a morphism in $\spa_a$, then for $(x_1,x_2)\in R$ there exists $v\in S$, such that $q_1(v)=g_1(x_1)$ and $q_2(v)=g_2(x_2)$. Since $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)\in\rel_a$ then the $v$ is unique and $v=(g_1(x_1),g_2(x_2))$.
|
|
|
|
($\Leftarrow$): As $\rel_a$ is a subcategory of $\spa_a$, this is trivial.\qed
|
|
\end{proof}
|
|
|
|
\begin{prop}\label{prop:rel-spa}
|
|
Assuming that $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ and $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$, are objects existing in both categories $\rel$ and $\spa$, there is a morphism $(g_1,g_2,w)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\spa$ iff there is a morhpism $(g_1,g_2,w)$ of the same type in $\rel$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Trivial by the definitions.\qed
|
|
\end{proof}
|
|
\begin{figure}[t]
|
|
\centering
|
|
\begin{tabular}{|l|c|c|c|c|}
|
|
\hline
|
|
\quad$\Rightarrow$& $\spa$ & $\rel$ & $\spa_a$ & $\rel_a$ \\
|
|
\hline
|
|
$\spa$ & & & & \\
|
|
\hline
|
|
$\rel$ & & & & \\
|
|
\hline
|
|
$\spa_a$ & \ding{56} & & & \\
|
|
\hline
|
|
$\rel_a$ & & & & \\
|
|
\hline
|
|
\end{tabular}
|
|
\caption{Where the axiom of choice is needed to get a morphism on the top row, when a morphism in the left column exists, in $\Set$.}
|
|
\label{fig:anonymous_onymous-choice}
|
|
\end{figure}
|
|
%\begin{notation}
|
|
% In the literature it is common to see a category that has sets as objects, and binary relations as morphims. Here we denote this category $\rel'$.
|
|
%\end{notation}
|
|
|
|
\subsection{Categories of Spans and Relations as Double Categories}
|
|
As mentioned in the previous section, there is a way to define morphisms in $\spa$ and $\rel$ that can not be captured if an arbitrary category $\BC$ is replaced with $\Set$. We show that $\spa(\BC)$, $\spa_a$, $\rel(\BC)$, and $\rel_a$ are all double categories to give an abstract notion that does capture all the different notions together. Also, double categories show us a way to have $\spa_a(\BC)$ and $\rel_a(\BC)$, i.e., category of spans and category of relations over an arbitrary category $\BC$ with anonymous morphisms.
|
|
\begin{definition}[(Weak) Double Categories]\sgnote{Add citation.}
|
|
Given categories $\BC_0$ and $\BC_1$, a double category $\BC_1\rightrightarrows\BC_0$, consists of
|
|
\begin{itemize}
|
|
\item a category $\BC_0$ of objects and morphisms,
|
|
\item a category $\BC_1$ of promorphisms and cells, equipped with a source and target functor $(S\c\BC_1\to\BC_0,T\c\BC_1\to\BC_0)$, a unit functor $U\c\BC_0\to\BC_1$,
|
|
\item a category $\BC_2$ of composable promorphisms given by the pullback of $S$ and $T$, and a functor $\odot\c\BC_2\to\BC_1$, and isomorphisms imposing that $\odot$ is associative and unital up to isomorphism.
|
|
\end{itemize}
|
|
Additionally, the functors have the following properties:
|
|
\begin{gather*}
|
|
S\comp U=\Id\\
|
|
T\comp U=\Id
|
|
\end{gather*}
|
|
For $\mathcal{R}$, $\mathcal{S}$ in $\obj(\BC_1)$, if $(\mathcal{R}, \mathcal{S})$ exists in $\obj(\BC_2)$, then:
|
|
\begin{gather*}
|
|
S(\mathcal{R}\odot \mathcal{S})=S\mathcal{R}\\
|
|
T(\mathcal{R}\odot \mathcal{S})=T\mathcal{S}
|
|
\end{gather*}
|
|
\end{definition}
|
|
%
|
|
\begin{definition}
|
|
In a double category $\BC_1\rightrightarrows\BC_0$ for promorphisms $\mathcal{R}$ and $\mathcal{S}$ that $S\mathcal{R}=S\mathcal{S}$ and $T\mathcal{R}=T\mathcal{S}$, an \emph{inclusion cell} is a cell $\alpha\c\mathcal{R}\Rightarrow\mathcal{S}$ such that $S\alpha=\id_{S\mathcal{R}}$ and $T\alpha=\id_{T\mathcal{R}}$.
|
|
\end{definition}
|
|
%
|
|
\begin{example}\label{ex:spa-double-cat}
|
|
We show that assuming the category $\BC$ has pullbacks, $\spa(\BC)\rightrightarrows\BC$ is a double category. We define $S$ and $T$ as follows:
|
|
\begin{gather*}
|
|
S(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)=X_1\\
|
|
S(g_1,g_2,w)=g_1\\
|
|
T(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)=X_2\\
|
|
T(g_1,g_2,w)=g_2
|
|
\end{gather*}
|
|
Furthermore, for every object $X$ in $\BC$ we show the kernel pair of $\id_X$ with $\Delta_X$, and we define $UX=\Delta_X$ and for a morphism $f\c X\to Y$ in $\BC$, $Uf$ is the unique morphism $\Delta_f$ in the diagram below that is obtained by the universal property of pullbacks:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& {\Delta_X} \& X \\
|
|
Y \& {\Delta_Y} \& Y \\
|
|
\& Y
|
|
\arrow["f"', from=1-1, to=2-1]
|
|
\arrow[from=1-2, to=1-1]
|
|
\arrow[from=1-2, to=1-3]
|
|
\arrow["{\Delta_f}", dashed, from=1-2, to=2-2]
|
|
\arrow["f", from=1-3, to=2-3]
|
|
\arrow["{\id_Y}"', from=2-1, to=3-2]
|
|
\arrow[from=2-2, to=2-1]
|
|
\arrow[from=2-2, to=2-3]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=-45}, draw=none, from=2-2, to=3-2]
|
|
\arrow["{\id_Y}", from=2-3, to=3-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Now, we define $\odot$. For objects $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ and $(Y \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Z)$ in $\spa(\BC)$, we define
|
|
\begin{gather*}
|
|
(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\odot(Y \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Z)=(X \stackrel{p_1\comp c_R}{\leftarrow} R\odot S \stackrel{q_2\comp c_S}{\to}Z),
|
|
\end{gather*}
|
|
such that $R\odot S$, $c_R$ and $c_S$ are defined in the following pullback in $\BC$:
|
|
\begin{gather*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{R\odot S} \& S \\
|
|
R \& Y
|
|
\arrow["{c_S}", from=1-1, to=1-2]
|
|
\arrow["{c_R}"', from=1-1, to=2-1]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
|
\arrow["{q_1}", from=1-2, to=2-2]
|
|
\arrow["{p_2}"', from=2-1, to=2-2]
|
|
\end{tikzcd}
|
|
\end{gather*}
|
|
For morphisms
|
|
\begin{gather*}
|
|
(f,g,w)\c(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(Y \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Z)
|
|
\end{gather*}
|
|
and
|
|
\begin{gather*}
|
|
(g,h,v)\c(Y \stackrel{t_1}{\leftarrow} W \stackrel{t_2}{\to}Z)\to(Z \stackrel{r_1}{\leftarrow} V \stackrel{r_2}{\to}Q)
|
|
\end{gather*}
|
|
we have the following commutative diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\&\& {R\odot W} \&\& \\
|
|
X \& R \& Y \& W \& P \\
|
|
Y \& S \& Z \& V \& Q \\
|
|
\&\& {S\odot V}
|
|
\arrow["{c_R}"', from=1-3, to=2-2]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=-45}, draw=none, from=1-3, to=2-3]
|
|
\arrow["{c_W}", from=1-3, to=2-4]
|
|
\arrow["f"', from=2-1, to=3-1]
|
|
\arrow["{p_1}"', from=2-2, to=2-1]
|
|
\arrow["{p_2}", from=2-2, to=2-3]
|
|
\arrow["w", from=2-2, to=3-2]
|
|
\arrow["g", from=2-3, to=3-3]
|
|
\arrow["{t_1}"', from=2-4, to=2-3]
|
|
\arrow["{t_2}", from=2-4, to=2-5]
|
|
\arrow["v", from=2-4, to=3-4]
|
|
\arrow["h", from=2-5, to=3-5]
|
|
\arrow["{q_1}", from=3-2, to=3-1]
|
|
\arrow["{q_2}"', from=3-2, to=3-3]
|
|
\arrow["{r_1}", from=3-4, to=3-3]
|
|
\arrow["{r_2}"', from=3-4, to=3-5]
|
|
\arrow["{c_S}", from=4-3, to=3-2]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=135}, draw=none, from=4-3, to=3-3]
|
|
\arrow["{c_V}"', from=4-3, to=3-4]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Since we have
|
|
\begin{align*}
|
|
r_1\comp v\comp c_W&\\
|
|
&=g\comp t_1\comp c_w\\
|
|
&=g\comp p_2\comp c_R\\
|
|
&=q_2\comp w\comp c_R
|
|
\end{align*}
|
|
by the universal property of pullbacks, there exists a morphism $u\c R\odot W\to S\odot V$, such that the following diagram commutes:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
R \& {R\odot W} \& W \\
|
|
S \& {S\odot V} \& V
|
|
\arrow["w", from=1-1, to=2-1]
|
|
\arrow["{c_R}"', from=1-2, to=1-1]
|
|
\arrow["{c_W}", from=1-2, to=1-3]
|
|
\arrow["u"', dashed, from=1-2, to=2-2]
|
|
\arrow["v", from=1-3, to=2-3]
|
|
\arrow["{c_S}", from=2-2, to=2-1]
|
|
\arrow["{c_V}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
So, $(w,v,u)$ is a morphism of type $(R \stackrel{c_R}{\leftarrow} R\odot W \stackrel{c_W}{\to}W)\to(S \stackrel{c_S}{\leftarrow} S\odot V \stackrel{c_V}{\to}V)$. So, we define $(f,g,w)\odot(g,h,v)=(w,v,u)$. To define the natural isomorphisms is a cumbersome task, but it is a known fact in the literature. Additionally, showing that $\rel(\BC)\rightrightarrows\BC$ is also a double category is almost the same. The functors are defined the same, except that the functor $\odot$ takes the image of the pullback over its legs, instead of the pullback itself.
|
|
% The following diagrams help to see why the defined $\odot$ is a functor.
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \&\& {R\odot W} \&\& \\
|
|
% X \& R \& Y \& W \& P \\
|
|
% Y \& S \& Z \& V \& Q \\
|
|
% {Y'} \& {S'} \& {Z'} \& {V'} \& {Q'} \\
|
|
% \&\& {S'\odot V'}
|
|
% \arrow[from=1-3, to=2-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=-45}, draw=none, from=1-3, to=2-3]
|
|
% \arrow[from=1-3, to=2-4]
|
|
% \arrow["f"', from=2-1, to=3-1]
|
|
% \arrow["{p_1}"', from=2-2, to=2-1]
|
|
% \arrow["{p_2}", from=2-2, to=2-3]
|
|
% \arrow["w", from=2-2, to=3-2]
|
|
% \arrow["g", from=2-3, to=3-3]
|
|
% \arrow["{t_1}"', from=2-4, to=2-3]
|
|
% \arrow["{t_2}", from=2-4, to=2-5]
|
|
% \arrow["v", from=2-4, to=3-4]
|
|
% \arrow["h", from=2-5, to=3-5]
|
|
% \arrow["{f'}"', from=3-1, to=4-1]
|
|
% \arrow["{q_1}", from=3-2, to=3-1]
|
|
% \arrow["{q_2}"', from=3-2, to=3-3]
|
|
% \arrow["{w'}", from=3-2, to=4-2]
|
|
% \arrow["{g'}", from=3-3, to=4-3]
|
|
% \arrow["{r_1}", from=3-4, to=3-3]
|
|
% \arrow["{r_2}"', from=3-4, to=3-5]
|
|
% \arrow["{v'}", from=3-4, to=4-4]
|
|
% \arrow["{h'}", from=3-5, to=4-5]
|
|
% \arrow["{q_1'}", from=4-2, to=4-1]
|
|
% \arrow["{q_2'}"', from=4-2, to=4-3]
|
|
% \arrow["{r_1'}", from=4-4, to=4-3]
|
|
% \arrow["{r_2'}"', from=4-4, to=4-5]
|
|
% \arrow[from=5-3, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=135}, draw=none, from=5-3, to=4-3]
|
|
% \arrow[from=5-3, to=4-4]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \&\& {R\odot W} \&\&\&\&\& {S\odot V} \&\& \\
|
|
% X \& R \& Y \& W \& P \& Y \& S \& Z \& V \& Q \\
|
|
% Y \& S \& Z \& V \& Q \& {Y'} \& {S'} \& {Z'} \& {V'} \& {Q'} \\
|
|
% \&\& {S\odot V} \&\&\&\&\& {S'\odot V'}
|
|
% \arrow["{c_R}"', from=1-3, to=2-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=-45}, draw=none, from=1-3, to=2-3]
|
|
% \arrow["{c_W}", from=1-3, to=2-4]
|
|
% \arrow["{c_S}"', from=1-8, to=2-7]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=-45}, draw=none, from=1-8, to=2-8]
|
|
% \arrow["{c_V}", from=1-8, to=2-9]
|
|
% \arrow["f"', from=2-1, to=3-1]
|
|
% \arrow["{p_1}"', from=2-2, to=2-1]
|
|
% \arrow["{p_2}", from=2-2, to=2-3]
|
|
% \arrow["w", from=2-2, to=3-2]
|
|
% \arrow["g", from=2-3, to=3-3]
|
|
% \arrow["{t_1}"', from=2-4, to=2-3]
|
|
% \arrow["{t_2}", from=2-4, to=2-5]
|
|
% \arrow["v", from=2-4, to=3-4]
|
|
% \arrow["h", from=2-5, to=3-5]
|
|
% \arrow["{f'}"', from=2-6, to=3-6]
|
|
% \arrow["{q_1}"', from=2-7, to=2-6]
|
|
% \arrow["{q_2}", from=2-7, to=2-8]
|
|
% \arrow["{w'}", from=2-7, to=3-7]
|
|
% \arrow["{g'}", from=2-8, to=3-8]
|
|
% \arrow["{r_1}"', from=2-9, to=2-8]
|
|
% \arrow["{r_2}", from=2-9, to=2-10]
|
|
% \arrow["{v'}", from=2-9, to=3-9]
|
|
% \arrow["{h'}", from=2-10, to=3-10]
|
|
% \arrow["{q_1}", from=3-2, to=3-1]
|
|
% \arrow["{q_2}"', from=3-2, to=3-3]
|
|
% \arrow["{r_1}", from=3-4, to=3-3]
|
|
% \arrow["{r_2}"', from=3-4, to=3-5]
|
|
% \arrow["{q'_1}", from=3-7, to=3-6]
|
|
% \arrow["{q_2'}"', from=3-7, to=3-8]
|
|
% \arrow["{r'_1}", from=3-9, to=3-8]
|
|
% \arrow["{r'_2}"', from=3-9, to=3-10]
|
|
% \arrow["{c_S}", from=4-3, to=3-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=135}, draw=none, from=4-3, to=3-3]
|
|
% \arrow["{c_V}"', from=4-3, to=3-4]
|
|
% \arrow["{c_{S'}}", from=4-8, to=3-7]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=135}, draw=none, from=4-8, to=3-8]
|
|
% \arrow["{c_{V'}}"', from=4-8, to=3-9]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
\end{example}
|
|
\begin{example}
|
|
We show that assuming the axiom of choice, $\spa_a\rightrightarrows\Set$ is a double category. We define the functors $S$, $T$, and $U$ on objects similar to~\autoref{ex:spa-double-cat}, and on the morphisms we define them as follows:
|
|
\begin{gather*}
|
|
S(g_1,g_2)=g_1\\
|
|
T(g_1,g_2)=g_2\\
|
|
Uf=(f,f)
|
|
\end{gather*}
|
|
Similarly, the definition of the functor $\odot$ on objects is the same as~\autoref{ex:spa-double-cat}, and for morphisms
|
|
\begin{gather*}
|
|
(f,g)\c(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(Y \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Z)
|
|
\end{gather*}
|
|
and
|
|
\begin{gather*}
|
|
(g,h)\c(Y \stackrel{t_1}{\leftarrow} W \stackrel{t_2}{\to}Z)\to(Z \stackrel{r_1}{\leftarrow} V \stackrel{r_2}{\to}Q)
|
|
\end{gather*}
|
|
by the axiom of choice and~\autoref{prop:spa-spaa} there exist $c_1\c R\to S$ and $c_2\c W\to V$ such that $(f,g,c_1)$ and $(g,h,c_2)$ are morphisms of the same type in $\spa$. So, as we did show in~\autoref{ex:spa-double-cat} there exists $u\c R\odot W\to S\odot V$ such that $(c_1,c_2,u)\c (R \stackrel{c_R}{\leftarrow} R\odot W \stackrel{c_W}{\to}W)\to(S \stackrel{c_S}{\leftarrow} S\odot V \stackrel{c_V}{\to}V)$ is a morphism in $\spa$ and it is equal with $(f,g,c_1)\odot(g,h,c_2)$, so here we define $(f,g)\odot(g,h)=(c_1,c_2)$. Using $u$ it is obvious that $(c_1,c_2)$ is a morphism in $\spa_a$. Defining the natural isomorphisms are the same for~\autoref{ex:spa-double-cat}. Again, for $\rel_a$ the definitions are similar, except that to define $\odot$ we need to take the image of the pullback over its legs, similar to the case for $\rel(\BC)$.\todo{It is better to double check your last claims.}\\
|
|
Inclusion cells in these categories imply set inclusion.
|
|
\end{example}
|
|
%
|
|
\section{Coalgebraic Bisimulation}%\label{sec:}
|
|
%
|
|
%\begin{definition}[Relation Lifting]
|
|
% Assuming $F\c\BC\to\BC$ is a functor, then we call $\rel(F)\c\rel(\BC)\to\rel(\BC)$ a relation lifting of $F$, where the following diagram commutes:
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \rel(\BC) \&\& \rel(\BC) \\
|
|
% {\BC\times\BC} \&\& {\BC\times\BC}
|
|
% \arrow["{\rel(F)}", from=1-1, to=1-3]
|
|
% \arrow[from=1-1, to=2-1]
|
|
% \arrow[from=1-3, to=2-3]
|
|
% \arrow["{F\times F}"', from=2-1, to=2-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
%\end{definition}
|
|
%%
|
|
%We have notions of bisimulation that may involve relation lifting. An example of a relation lifting is obtained
|
|
%by the image factorization in regular categories. \ppnote{Initially, I wanted to give the definitions for an arbitrary relation lifting. I think it can be doable, but for simplicity I preferred to stick to this one.}
|
|
%%
|
|
%%\begin{equation*}
|
|
%% \begin{tikzcd}[ampersand replacement=\&]
|
|
% % R \& {R^\dagger} \&\& {X\times X}
|
|
% % \arrow["{e_R}"', two heads, from=1-1, to=1-2]
|
|
% % \arrow["{\brks{p_1,p_2}}", bend left=20, from=1-1, to=1-4]
|
|
% % \arrow["{\brks{p^\dagger_1,p^\dagger_2}}"', tail, from=1-2, to=1-4]
|
|
% % \end{tikzcd}
|
|
%%\end{equation*}
|
|
%We define a functor of type $(-)^\dagger\c\spa(\BC)\to\rel(\BC)$. It takes every span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to the image of its legs:
|
|
%
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% R \& {R^\dagger} \&\& {X\times Y}
|
|
% \arrow["{e_R}"', two heads, from=1-1, to=1-2]
|
|
% \arrow["{\brks{p_1,p_2}}", bend left=20, from=1-1, to=1-4]
|
|
% \arrow["{\brks{p^\dagger_1,p^\dagger_2}}"', tail, from=1-2, to=1-4]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%
|
|
%So, for every functor $F\c\BC\to\BC$ we have $(F-)^\dagger\c\rel(\BC)\to\rel(\BC)$ that takes every relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to the following relation:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \& {(FR)^\dagger} \& \\
|
|
% FX \&\& FY \\
|
|
% \& {FX\times FY}
|
|
% \arrow["{{{(Fp_1)^\dagger}}}"', from=1-2, to=2-1]
|
|
% \arrow["{{{(Fp_2)^\dagger}}}", from=1-2, to=2-3]
|
|
% \arrow["{{\brks{{(Fp_1)^\dagger},{(Fp_2)^\dagger}}}}"{description}, dashed, tail, from=1-2, to=3-2]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
|
|
%By varying from anonymous to non-anonymous morphisms and from $\rel$ to $\spa$ we
|
|
%can obtain for flavors of bisimulation (\autoref{eq:acz-mend-diag}--\autoref{def:vanila}) --
|
|
%\autoref{fig:anonymous_onymous} contains a comprehensible summary.
|
|
%
|
|
%
|
|
%
|
|
%%We take $\rel(F)\c\rel(\BC)\to\rel(\BC)$ to be the functor that for an arbitrary functor $F\c\BC\to\BC$ takes a relation $R$, where $R\in\obj(\rel)$ and $R\subseteq X_1\times X_2$, and gives the relation that is the image of the function $\brks{Fp_1,Fp_2}\c FR\to FX\times FY$.
|
|
%%\begin{definition}[Bisimulation]
|
|
%% For a functor $F\c\BC\to\BC$, a bisimulation is a $\rel(F)$-coalgebra in $\rel$.
|
|
%%\end{definition}
|
|
%
|
|
%%\begin{definition}[$F$-Relator]
|
|
%% For a set functor $F$, and for sets $X$ and $Y$, an $F$-relator $\relar$ is a map that takes every relation on $X\times Y$ to a relation on $FX\times FY$, and it is monotone with respect to inclusion.
|
|
%%\end{definition}
|
|
%%Also, for the time being, we limit the discussion to the case $\BC=\Set$. For simplicity, by $\rel$ we mean $\rel(\BC)$.
|
|
%%
|
|
%%\begin{prop}
|
|
%% Assuming that $(R,\alpha)$ is a $\rel(F)$-coalgebra, where $\alpha=\beta_1\times\beta_2$ in $\BC\times\BC$, then the following diagram commutes, and vice-versa:
|
|
%% \begin{equation*}
|
|
%% \begin{tikzcd}[ampersand replacement=\&]
|
|
%% {X_1} \& R \& {X_2} \\
|
|
%% {FX_1} \& FR \& {FX_2}
|
|
%% \arrow["{\beta_1}"', from=1-1, to=2-1]
|
|
%% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
%% \arrow["{p_2}", from=1-2, to=1-3]
|
|
%% \arrow["\beta", from=1-2, to=2-2]
|
|
%% \arrow["{\beta_2}", from=1-3, to=2-3]
|
|
%% \arrow["{Fp_1}", from=2-2, to=2-1]
|
|
%% \arrow["{Fp_2}"', from=2-2, to=2-3]
|
|
%% \end{tikzcd}
|
|
%% \end{equation*}
|
|
%%\end{prop}
|
|
%\begin{definition}[Aczel-Mendler Bisimulation]
|
|
%\todo{Explain in terms of span morphisms}.
|
|
%
|
|
%
|
|
% A relation $R\subseteq X\times Y$ is an \emph{Aczel-Mendler bisimulation} from an $F$-coalgebra $(X,\alpha)$ to an $F$-coalgebra $(Y,\beta)$ whenever there is a morphism $\gamma\c R\to FR$ called witness that commutes in the following diagram:
|
|
% \begin{equation*}\label{eq:acz-mend-diag}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X} \& R \& {Y} \\
|
|
% {FX} \& FR \& {FY}
|
|
% \arrow["{\alpha}"', from=1-1, to=2-1]
|
|
% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
% \arrow["{p_2}", from=1-2, to=1-3]
|
|
% \arrow["\gamma", from=1-2, to=2-2]
|
|
% \arrow["{\beta}", from=1-3, to=2-3]
|
|
% \arrow["{Fp_1}", from=2-2, to=2-1]
|
|
% \arrow["{Fp_2}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
%\end{definition}
|
|
%Aczel-Mendler bisimulation can be defined for an arbitrary category $\BC$
|
|
%instead of $\Set$. It is worth noting that with this definition, if $R$ is a
|
|
%relation, it does not necassirily mean that $FR$ is a relation as well.
|
|
%%
|
|
%\begin{definition}[Witnessless Bisimulation]
|
|
% A relation $R\subseteq X\times Y$ is a \emph{witnessless bisimulation} from an $F$-coalgebra $(X,\alpha)$ to an $F$-coalgebra $(Y,\beta)$ whenever for every $x\in X$ and $y\in Y$, we have $x\mathrel{R} y\Rightarrow \alpha(x)\mathrel{(FR)^\dagger}\beta(y)$.
|
|
%% \begin{equation*}
|
|
%% \begin{tikzcd}[ampersand replacement=\&]
|
|
%% {X} \& R \& {Y} \\
|
|
%% {FX} \& \rel(F)R \& {FY}
|
|
%% \arrow["{\alpha}"', from=1-1, to=2-1]
|
|
%% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
%% \arrow["{p_2}", from=1-2, to=1-3]
|
|
%% \arrow["{\beta}", from=1-3, to=2-3]
|
|
%% \arrow["{q_1}", from=2-2, to=2-1]
|
|
%% \arrow["{q_2}"', from=2-2, to=2-3]
|
|
%% \end{tikzcd}
|
|
%% \end{equation*}
|
|
%\end{definition}
|
|
%A more general version of witnessless bisimulation is given by Hughes and Jacobs, where the lifting is an arbitrary lifting not necessarily the one with image factorization that we mentioned here.
|
|
%
|
|
%\begin{definition}[Hermida-Jacobs Bisimulation]
|
|
% A relation $R\subseteq X\times Y$ is a \emph{Hermida-Jacobs bisimulation} from an $F$-coalgebra $(X,\alpha)$ to an $F$-coalgebra $(Y,\beta)$ whenever there is a morphism $\gamma\c R\to (FR)^\dagger$ called witness that commutes in the following diagram:
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X} \& R \& {Y} \\
|
|
% {FX} \& (FR)^\dagger \& {FY}
|
|
% \arrow["{\alpha}"', from=1-1, to=2-1]
|
|
% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
% \arrow["{p_2}", from=1-2, to=1-3]
|
|
% \arrow["\gamma", from=1-2, to=2-2]
|
|
% \arrow["{\beta}", from=1-3, to=2-3]
|
|
% \arrow["{(Fp_1)^\dagger}", from=2-2, to=2-1]
|
|
% \arrow["{(Fp_2)^\dagger}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
%\end{definition}
|
|
%Hermida-Jacobs bisimulation is also traditionally defined for an arbitrary category $\BC$.
|
|
%
|
|
%\begin{definition}[Vanilla Bisimulation]\label{def:vanila}
|
|
% A relation $R\subseteq X\times Y$ is a \emph{vanilla bisimulation} from an $F$-coalgebra $(X,\alpha)$ to an $F$-coalgebra $(Y,\beta)$ whenever for every $x\in X$ and $y\in Y$, we have $x\mathrel{R} y\Rightarrow \alpha(x)\mathrel{(FR)}\beta(y)$.
|
|
%% \begin{equation*}
|
|
%% \begin{tikzcd}[ampersand replacement=\&]
|
|
%% {X} \& R \& {Y} \\
|
|
%% {FX} \& FR \& {FY}
|
|
%% \arrow["{\alpha}"', from=1-1, to=2-1]
|
|
%% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
%% \arrow["{p_2}", from=1-2, to=1-3]
|
|
%% \arrow["{\beta}", from=1-3, to=2-3]
|
|
%% \arrow["{Fp_1}", from=2-2, to=2-1]
|
|
%% \arrow["{Fp_2}"', from=2-2, to=2-3]
|
|
%% \end{tikzcd}
|
|
%% \end{equation*}
|
|
%\end{definition}
|
|
%We do not know if this definition exists anywhere.
|
|
%\begin{prop}
|
|
% The following propositions hold in $\Set$:
|
|
% \todo{How about proving that 1.3,1.4,1.5 are equivalent to each other, and all together to 1.2 under axiom of choice? }
|
|
% \begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
% %\item Every Aczel-Mendler bisimulation is a vanilla bisimulation.
|
|
% \item Every vanilla bisimulation is an Aczel-Mendler bisimulation.
|
|
% \item Every Aczel-Mendler bisimulation is a Hermida-Jacobs bisimulation.
|
|
% \item Assuming the axiom of choice, every Hermida-Jacobs bisimulation is an Aczel-Mendler bisimulation.
|
|
% \item Every witnessless bisimulation is a Hermida-Jacobs bisimulation.
|
|
% %\item Every Hermida-Jacobs bisimulation is a witnessless bisimulation.
|
|
% \end{enumerate}
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% %(I): ???%Assuming $x\mathrel{R}y$, then given by~\eqref{eq:acz-mend-diag} we have $\gamma(x,y)\in FR$, $Fp_1\comp\gamma(x,y)=\alpha(x)$, and $Fp_2\comp\gamma(x,y)=\beta(y)$ that means $\alpha(x)\mathrel{(FR)}\beta(y)$.
|
|
%
|
|
% (I): Since $R$ is a vanilla bisimulation, for every $(x,y)\in R$ we have $\alpha(x)\mathrel{(FR)}\beta(y)$, so we can define $\gamma\c R\to FR$ as $\gamma(x,y)=(\alpha(x),\beta(y))$, and then $\gamma$ commutes in~\eqref{eq:acz-mend-diag}.
|
|
%
|
|
% (II):\autoref{lem:norm-simp}, and given by Staton.
|
|
%
|
|
% (III): Given by Staton, and similar to \autoref{lem:norm-simp}.
|
|
%
|
|
% (IV): Similar to (I) we can define $\gamma(x,y)=(\alpha(x),\beta(y))$ as the witness for $R$ to be a Hermida-Jacobs bisimulation.
|
|
%\end{proof}
|
|
%\begin{rem}\todo{No, these definitions are just equivalent.}
|
|
% For vanilla bisimulation to be a witnessless bisimulation (or vice-versa) for a relation $R$ we need to have $FR\subseteq (FR)^\dagger$ (or $(FR)^\dagger\subseteq FR$), which is rarely true. The condition may not even be true for other relation liftings for these functors.
|
|
%\end{rem}
|
|
%\begin{rem}
|
|
%\todo{No, this does not follow from anythying.}
|
|
% To have an Aczel-Mendler bisimulation to be a vanilla bisimulation we need to have $FR\subseteq FX\times FY$ that is a rare condition. It does not hold for powerset functor or maybe functor.
|
|
%\end{rem}
|
|
%\todo{Discuss 4 versions of bisimulation (with witness/without witness, for relations/for spans). Which are equivalent? Which do not make sense?}
|
|
%
|
|
%\todo{In next section run a similar analysis for simulation: relator-based vs. Aczel-Mendler.}\\
|
|
%--------------------------------------------------------------
|
|
\begin{definition}[Aczel-Mendler Bisimulation]
|
|
In an arbitrary category $\BC$, a span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an \emph{Aczel-Mendler bisimulation} over $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$, if there exists a morphism in $\spa(\BC)$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$.
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Relation Lifting]
|
|
Assuming $F\c\BC\to\BC$ is a functor, then we call $\rel(F)\c\rel(\BC)\to\rel(\BC)$ a relation lifting of $F$, whenever the following diagram commutes:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\rel(\BC) \&\& \rel(\BC) \\
|
|
{\BC\times\BC} \&\& {\BC\times\BC}
|
|
\arrow["{\rel(F)}", from=1-1, to=1-3]
|
|
\arrow["{U}"',from=1-1, to=2-1]
|
|
\arrow["{U}",from=1-3, to=2-3]
|
|
\arrow["{F\times F}"', from=2-1, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{definition}
|
|
\begin{definition}[Abstract Relational Bisimulation]\label{def:abs-rel-bis}
|
|
In an arbitrary category $\BC$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an \emph{abstract relational bisimulation} over $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$, if there exists a morphism in $\rel(\BC)$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{\rel(F)p_1}{\leftarrow} \rel(F)R \stackrel{\rel(F)p_2}{\to}FY)$.
|
|
\end{definition}
|
|
The given definition is highly abstract. There is a relation lifting that abstracts Barr-relators that are known to be well-behaved relators, and it also gives an interesting notion of bisimulation, called \emph{Hermida-Jacobs bisimulation}.
|
|
%
|
|
The lifting is using the image factorization in regular categories. %\ppnote{Initially, I wanted to give the definitions for an arbitrary relation lifting. I think it can be doable, but for simplicity I preferred to stick to this one.}
|
|
%
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% R \& {R^\dagger} \&\& {X\times X}
|
|
% \arrow["{e_R}"', two heads, from=1-1, to=1-2]
|
|
% \arrow["{\brks{p_1,p_2}}", bend left=20, from=1-1, to=1-4]
|
|
% \arrow["{\brks{p^\dagger_1,p^\dagger_2}}"', tail, from=1-2, to=1-4]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
For a regular category $\BC$, we define a functor of type $(-)^\clubsuit\c\spa(\BC)\to\rel(\BC)$. It takes every span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to the image of its legs:
|
|
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
R \& {R^\clubsuit} \&\& {X\times Y}
|
|
\arrow["{e_R}"', two heads, from=1-1, to=1-2]
|
|
\arrow["{\brks{p_1,p_2}}", bend left=20, from=1-1, to=1-4]
|
|
\arrow["{\brks{p^\clubsuit_1,p^\clubsuit_2}}"', tail, from=1-2, to=1-4]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Also, for every functor $F\c\BC\to\BC$ we have a trivial lifting to $\spa(\BC)$ that takes every object $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to $(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$, and every morphism $(f,g,w)$ to $(Ff,Fg,Fw)$, and we denote it with $\spa(F)$. Since $\rel(\BC)$ is a subcategory of $\spa(\BC)$, we have an inclusion functor $I\c\rel(\BC)\to\spa(\BC)$ as well.
|
|
So, given a functor $F\c\BC\to\BC$ we define its lifting $(F-)^\dagger\c\rel(\BC)\to\rel(\BC)$ as $(F-)^\dagger=(\spa(F)I-)^\clubsuit$. The functor $(F-)^\dagger$ takes every relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to the following relation:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& {(FR)^\dagger} \& \\
|
|
FX \&\& FY \\
|
|
\& {FX\times FY}
|
|
\arrow["{{{(Fp_1)^\dagger}}}"', from=1-2, to=2-1]
|
|
\arrow["{{{(Fp_2)^\dagger}}}", from=1-2, to=2-3]
|
|
\arrow["{{\brks{{(Fp_1)^\dagger},{(Fp_2)^\dagger}}}}"{description}, dashed, tail, from=1-2, to=3-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
%
|
|
\begin{definition}[Hermida-Jacobs Bisimulation]
|
|
In an arbitrary category $\BC$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a \emph{Hermida-Jacobs bisimulation} over $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$, if there exists a morphism in $\rel(\BC)$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{(Fp_1)^\dagger}{\leftarrow} (FR)^\dagger \stackrel{(Fp_2)^\dagger}{\to}FY)$.
|
|
\end{definition}
|
|
%
|
|
\begin{lemma}\label{lem:morph-spa-rel}
|
|
In a regular category $\BC$ with the axiom of choice, assuming that $(Y_1 \stackrel{q^\dagger_1}{\leftarrow} S^\dagger \stackrel{q^\dagger_2}{\to}Y_2)$ is an object in $\spa(\BC)$, if for an object $A$ in $\BC$ we have a morphism $w\c A\to S^\dagger$, then there exist a morphism $v\c A\to S$ such that for $i\in\{1,2\}$, we have $q_i\comp v=q_i^\dagger\comp w$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Having the axiom of choice in a regular category $\BC$ means that for $e_S\c S\to S^\dagger$ there exist a section $s$. We define $v=s\comp w$, then for $i\in\{1,2\}$ we have:
|
|
\begin{align*}
|
|
q_i\comp v&\\
|
|
=&q^\dagger_i\comp e_s\comp v\\
|
|
=&q^\dagger_i\comp e_s\comp s\comp w\\
|
|
=&q^\dagger_i\comp w
|
|
\end{align*}
|
|
\qed
|
|
\end{proof}
|
|
%
|
|
%\begin{prop}
|
|
% For a regular category $\BC$ with the axiom of choice, assuming that in $\rel(\BC)$, there is a morphism $(g_1,g_2,w)$ from an object $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ to $(Y_1 \stackrel{q^\dagger_1}{\leftarrow} S^\dagger \stackrel{q^\dagger_2}{\to}Y_2)$, then there exist a morphism $(g_1,g_2,v)$ from $(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)$ to $(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% Having the axiom of choice in a regular category $\BC$ means that for $e_S\c S\to S^\dagger$ there exist a section $s$. We define $v=s\comp w$, then for $i\in\{1,2\}$ we have:
|
|
% \begin{align*}
|
|
% q_i\comp v&\\
|
|
% =&q^\dagger_i\comp e_s\comp v\\
|
|
% =&q^\dagger_i\comp e_s\comp s\comp w\\
|
|
% =&q^\dagger_i\comp w\\
|
|
% =&g_i\comp p_i
|
|
% \end{align*}
|
|
% \qed
|
|
%\end{proof}
|
|
%
|
|
\begin{prop}\label{prop:HeJ-AM}
|
|
\begin{enumerate}
|
|
\item Assuming $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an AM-bisimulation on coalgebras $(X,\alpha)$ and $(Y,\beta)$ then it is a HJ-bisimulation.
|
|
\item In a regular category $\BC$ with the axiom of choice, assuming $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a HJ-bisimulation on coalgebras $(X,\alpha)$ and $(Y,\beta)$ then it is an AM-bisimulation.
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
$(1)$: Trivial.\\
|
|
$(2)$: It is obvious using~\autoref{lem:morph-spa-rel}.\qed
|
|
\end{proof}
|
|
%
|
|
\subsection{Coalgebraic Bisimulation in Set}
|
|
We have two more notions for coalgebraic bisimulation in $\Set$, that is to define them in $\spa_a$ and $\rel_a$, respectively called \emph{span-based bisimulation} and \emph{Hughes-Jacobs bisimulation}.
|
|
%
|
|
\begin{definition}[Hughes-Jacobs Bisimulation]
|
|
A relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ in $\rel_a$ is a \emph{Hughes-Jacobs bisimulation} over $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$, whenever there exists a morhpism in $\rel_a$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{(Fp_1)^\dagger}{\leftarrow} (FR)^\dagger \stackrel{(Fp_2)^\dagger}{\to}FY)$.
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Span-Based Bisimulation]
|
|
A span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ in $\spa_a$ is a \emph{span-based bisimulation} over $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$, whenever there exists a morhpism in $\spa_a$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$.
|
|
\end{definition}
|
|
%
|
|
\begin{prop}
|
|
In $\Set$, for $F$-coalgebras $(X,\alpha)$ and $(Y,\beta)$,
|
|
\begin{enumerate}
|
|
\item Assuming the axiom of choice span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an Aczel-Mendler bisimulation iff it is a span-based bisimulation.
|
|
\item A relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a Hermida-Jacobs bisimulation iff it is a Hughes-Jacobs bisimulation.
|
|
\item A relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a span-based bisimulation iff it is a Hughes-Jacobs bisimulation.
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
(1): Follows from~\autoref{prop:spa-spaa}.
|
|
|
|
(2): Follows from~\autoref{prop:rel-rela}.
|
|
|
|
(3): ($\Rightarrow$): Assuming there is a morphism in $\spa_a$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$ we need to prove that exists a morphism of type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{(Fp_1)^\dagger}{\leftarrow} (FR)^\dagger \stackrel{(Fp_2)^\dagger}{\to}FY)$ in $\rel_a$. Assuming $(x,y)\in R$ there exists $v$ such that $Fp_1(v)=\alpha(x)$ and $Fp_2(v)=\beta(y)$, and it exactly means that $(\alpha(x),\beta(y))\in(FR)^\dagger$, so $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a Hughes-Jacobs bisimulation.\\
|
|
($\Leftarrow$): Assuming there is a morphism in $\rel_a$ of the type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{(Fp_1)^\dagger}{\leftarrow} (FR)^\dagger \stackrel{(Fp_2)^\dagger}{\to}FY)$ we need to prove that exists a morphism of type $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)\to(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$ in $\spa_a$. Assuming $(x,y)\in R$ there then $(\alpha(x),\beta(y))\in(FR)^\dagger$ that means that exists $u\in FR$ such that $Fp_1(u)=\alpha(x)$ and $Fp_2(u)=\beta(u)$, and it exactly means that $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a span-based bisimulation.
|
|
\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Recalling~\autoref{prop:HeJ-AM}, all the four introduced definitions for bisimulation (\autoref{fig:anonymous_onymous}) are equivalent under the axiom of choice.
|
|
\end{cor}
|
|
\begin{figure}[t]
|
|
\centering
|
|
\begin{tabular}{|l|c|c|}
|
|
\hline
|
|
& \textbf{Anonymous} & \textbf{Onymous} \\
|
|
\hline
|
|
\textbf{Relations} & Hughes-Jacobs & Hermida-Jacobs \\
|
|
\hline
|
|
\textbf{Spans} & Span-based & Aczel-Mendler \\
|
|
\hline
|
|
\end{tabular}
|
|
\caption{Comparison of anonymous and onymous settings for relations and spans in $\Set$.}
|
|
\label{fig:anonymous_onymous}
|
|
\end{figure}
|
|
%
|
|
\begin{figure}[t]
|
|
\centering
|
|
\begin{tabular}{|l|c|c|c|c|}
|
|
\hline
|
|
\qquad Hughes-Jacobs$\Rightarrow$& Aczel-Mendler & Hermida-Jacobs & Span-based & Hughes-Jacobs \\
|
|
\hline
|
|
Aczel-Mendler & & & & \\
|
|
\hline
|
|
Hermida-Jacobs & \ding{56} & & & \\
|
|
\hline
|
|
Span-based & \ding{56} & & & \\
|
|
\hline
|
|
Hughes-Jacobs & \ding{56} & & & \\
|
|
\hline
|
|
\end{tabular}
|
|
\caption{Where the axiom of choice is needed to say one bisimulation based on one notion is also a bisimulation with respect to another notion, in $\Set$.}
|
|
\label{fig:bisim-choice}
|
|
\end{figure}
|
|
\section{Coalgebraic Simulation}
|
|
%\todo{Give an introduction of the definitions for $\spa(\BC)$ and $\rel(\BC)$ that are AM-simulation and HJ-simulation, then open up the discussion about relators.}
|
|
\begin{definition}[Aczel-Mendler Simulation]
|
|
Assuming that $\appr$ is a natural order structure on a functor $F\c\BC\to\BC$, an object $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ of $\spa(\BC)$ is an \emph{Aczel-Mendler simulation} from a coalgebra $(X,\alpha)$ to $(Y,\beta)$, whenever the following diagram commutes laxly:
|
|
\begin{equation}\label{eq:diag-am-sim}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y \\
|
|
FX \& FR \& FY
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["{{Fp_1}}", from=2-2, to=2-1]
|
|
\arrow["{{Fp_2}}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation}
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Hermida-Jacobs Simulation]
|
|
Assuming that $\appr$ is a natural order structure on a functor $F\c\BC\to\BC$, an $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ of $\rel(\BC)$ is a \emph{Hermida-Jacobs simulation} from a coalgebra $(X,\alpha)$ to $(Y,\beta)$, whenever the following diagram commutes laxly:
|
|
\begin{equation*}\label{eq:diag-hj-sim}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y \\
|
|
FX \& (FR)^\dagger \& FY
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["{{(Fp_1)^\dagger}}", from=2-2, to=2-1]
|
|
\arrow["{{(Fp_2)^\dagger}}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{definition}
|
|
%
|
|
\begin{prop}
|
|
\begin{enumerate}
|
|
\item Assuming $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$\sgnote{Is it a relation or a general span?} is an AM-simulation from a coalgebra $(X,\alpha)$ to $(Y,\beta)$ then it is a HJ-simulation.
|
|
\item In a regular category $\BC$ with the axiom of choice, assuming $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a HJ-simulation from a coalgebra $(X,\alpha)$ to $(Y,\beta)$ then it is an AM-simulation.
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
$(1)$: Trivial.\sgnote{Does not look so trivial.}\\
|
|
$(2)$: It is obvious using~\autoref{lem:morph-spa-rel}.\sgnote{Add more details.}\qed
|
|
\end{proof}
|
|
%
|
|
\subsection{Simulations in Set}
|
|
|
|
%\begin{figure}[ht]
|
|
% \centering
|
|
% \begin{tabular}{|c|c|c|c|c|}
|
|
% \hline
|
|
% & \textbf{Hughes-Jacobs} & \textbf{Hermida-Jacobs} & \textbf{Span-based} & \textbf{Aczel-Mendler} \\
|
|
% \hline
|
|
% $\appr\cdot-$ & ? & ? & ? & ? \\
|
|
% \hline
|
|
% $-\cdot\appr$ & ? & ? & ? & ? \\
|
|
% \hline
|
|
% $\appr\cdot-\cdot\appr$ & ? & ? & ? & ? \\
|
|
% \hline
|
|
% \end{tabular}
|
|
% \caption{Comparison of anonymous and onymous settings for relations and spans in $\Set$.}
|
|
% \label{fig:anonymous_onymous_transposed}
|
|
%\end{figure}
|
|
|
|
|
|
\begin{tikzpicture}[
|
|
scale=0.5, every node/.style={transform shape}
|
|
]
|
|
\matrix (m) [matrix of nodes,
|
|
nodes={draw, ellipse, minimum width=2.6cm, minimum height=1cm,
|
|
align=center, font=\scriptsize, inner sep=2pt},
|
|
row sep=6mm, column sep=6mm]
|
|
{
|
|
$\appr\cdot-$-Hughes-Jacobs & $\appr\cdot-$-Hermida-Jacobs & $\appr\cdot-$-Span-based & $\appr\cdot-$-Aczel-Mendler \\
|
|
$-\cdot\appr$-Hughes-Jacobs & $-\cdot\appr$-Hermida-Jacobs & $-\cdot\appr$-Span-based & $-\cdot\appr$-Aczel-Mendler \\
|
|
$\appr\cdot-\cdot\appr$-Hughes-Jacobs & $\appr\cdot-\cdot\appr$-Hermida-Jacobs & $\appr\cdot-\cdot\appr$-Span-based & $\appr\cdot-\cdot\appr$-Aczel-Mendler \\
|
|
};
|
|
|
|
% Example arrows (uncomment / edit as needed):
|
|
\draw[-{Latex[length=2mm]}] (m-3-1) to[bend right=30] node[midway, above] {AC} (m-3-2);
|
|
\draw[-{Latex[length=2mm]}] (m-3-2) to[bend right=30] (m-3-1);
|
|
|
|
\end{tikzpicture}
|
|
|
|
Traditionally, simulations in $\Set$ are defined using relators. In this section we make a comparison on this concept with the other notions of simulation that we mentioned.
|
|
%\begin{notation}
|
|
% We show the category of sets and binary relations with $\rels$, and we show a morphism in this category as $R\c X\rto Y$ that is a relation $R$.
|
|
%\end{notation}
|
|
%\begin{lemma}\label{lem:set-rel-span-equiv}
|
|
% $R\c X\rto Y$ is a morphism in $\rels$ iff there is an object $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ in $\rel$.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% ($\Rightarrow$): $R\c X\rto Y$ being a morphism in $\rels$ means that in $\Set$ there exist an object $R$ with a unique mono of type $R\to X\times Y$ that is a pairing that we show with $\brks{p_1,p_2}$. So, $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an object in $\rel$.
|
|
%
|
|
% ($\Leftarrow$): If $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an object in $\rel$, then $R$ is a binary relation from $X$ to $Y$ so, it is a morphism of type $X\rto Y$ in $\rels$.\qed
|
|
%\end{proof}
|
|
%The above translation seems to be true in a more general case, where $\spa$ and $\rel$ are defined on an arbitrary category (the latter is called an allegory then).
|
|
\begin{definition}[Relator]\label{def:relator}
|
|
Assuming $F$ is a functor on $\Set$, an $F$-relator or simply a relator $\relar$ is a monotone map that sends an object $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ of $\Rel$ to $(FX \stackrel{q_1}{\leftarrow} \relar R \stackrel{q_2}{\to}FY)$.
|
|
\end{definition}
|
|
%
|
|
%\begin{definition}[Hermida-Jacobs Simulation]\label{def:hej-sim-rela}
|
|
% For a relator $\relar$ on a functor $F$ a HJ-simulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ is a relation $r$ for which there exists a morphism $\sigma\c r\to\relar r$ called \emph{witness} such that the following diagram commutes ($;$ is the relation composition):
|
|
% \begin{equation}\label{eq:hej-sim}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% X \& r \& Y \\
|
|
% {FX} \& {\relar r} \& {FY}
|
|
% \arrow["\alpha"', from=1-1, to=2-1]
|
|
% \arrow["{p_1}"', from=1-2, to=1-1]
|
|
% \arrow["{p_2}", from=1-2, to=1-3]
|
|
% \arrow["\sigma", from=1-2, to=2-2]
|
|
% \arrow["\beta", from=1-3, to=2-3]
|
|
% \arrow["{{(Fp_1)}^\relar}", from=2-2, to=2-1]
|
|
% \arrow["{{(Fp_2)}^\relar}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
% \end{equation}
|
|
%\end{definition}
|
|
|
|
\begin{definition}[Relator-based Simulation]
|
|
Given a relator $\relar$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a $\relar$-simulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ if there is a morphism in $\rel_a$ from $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to $(FX \stackrel{q_1}{\leftarrow} \relar R \stackrel{q_2}{\to}FY)$, i.e, if $(x,y)\in R$ entails $(\alpha(x),\beta(y))\in\relar R$, for all $x\in X$ and $y\in Y$.
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Symmetric Relator]
|
|
A relator $\relar$ is symmetric if and only if for every relation $R$ we have $\relar(R^\op)=(\relar R)^\op$.
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Relator-based Bisimulation]
|
|
Given a relator $\relar$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an $\relar$-bisimulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ whenever $R$ is an $\relar$-simulation, and $\relar$ is a symmetric relator.
|
|
\end{definition}
|
|
%
|
|
\begin{example}
|
|
Given $F\c\Set\to\Set$ the best example of a relator is the Barr relator. We have already introduced Barr relator, but we did not note it as a relator. It sends every relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to $(FX \stackrel{(Fp_1)^\dagger}{\leftarrow} (FR)^\dagger \stackrel{(Fp_2)^\dagger}{\to}Y)$. It is a symmetric relator. So, every simulation with this relator is actually a bisimulation.
|
|
\end{example}
|
|
%
|
|
\begin{example}
|
|
Given $F\c\Set\to\Set$ with an order structure $\appr$ on it the relator that sends $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to $(FX \stackrel{q_1}{\leftarrow} \appr\comp(FR)^\dagger\comp\appr \stackrel{q_2}{\to}FY)$ is a relator. We call it \emph{bi-lax Barr Relator}. There are other variations of this: left-lax ($\appr\comp(FR)^\dagger$) and right-lax ($(FR)^\dagger\comp\appr$). We denote this relator with $\tilde{F}$.
|
|
\end{example}
|
|
%
|
|
\begin{prop}\label{prop:HeJ-HuJ}
|
|
Given $F\c\Set\to\Set$ and an object $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ in $\rel$,
|
|
\begin{enumerate}
|
|
\item $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a $\tilde{F}$-simulation from a coalgebra $(X,\alpha)$ to a coalgebra $(Y,\beta)$ if it is a Hermida-Jacobs simulation.
|
|
\item $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is a Hermida-Jacobs simulation from a coalgebra $(X,\alpha)$ to a coalgebra $(Y,\beta)$ if it is a $\tilde{F}$-simulation, assuming the axiom of choice.
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
(1): $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ being a Hermida-Jacobs simulation means that for every $(x,y)\in R$, we have $\alpha\comp p_1(x,y)\appr(Fp_1)^\dagger\comp\sigma(x,y)$ and $(Fp_2)^\dagger\comp\sigma(x,y)\appr\beta\comp p_2(x,y)$. So, $x\mathrel{R}y$ gives that $\alpha(x)\mathrel{(\appr\comp(FR)^\dagger\comp\appr)}\beta(y)$ that means that $R$ is a $\tilde{F}$-simulation.\\
|
|
(2): $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ being a $\tilde{F}$-simulation means that $x\mathrel{R}y$ gives $\alpha(x)\mathrel{(\appr\comp(FR)^\dagger\comp\appr)}\beta(y)$ that means for every $(x,y)\in R$, there exist $(u,v)\in(FR)^\dagger$ such that $\alpha(x)\appr u$ and $v\appr\beta(y)$. We form a function $f\c R\to\powf(FR)^\dagger$ such that takes every $(x,y)$ to the set of the mentioned existing pairs $(u,v)$ in $(FR)^\dagger$. By the axiom of choice there exist a function $s\c\im_f\to(FR)^\dagger$. So, assuming that $f$ has the epi-mono factorization $(e,m)$, then we define $\sigma\c R\to(FR)^\dagger$ as $\sigma=s\comp e$. Now, the diagram~\eqref{eq:diag-hj-sim} commutes laxly for the defined $\sigma$.\qed
|
|
\end{proof}
|
|
\begin{rem}
|
|
The proposition entails that Hermida-Jacobs simulation subsumes simulation relations defined with bi-lax Barr relators.
|
|
\end{rem}
|
|
Having lax versions of a symmetric relator, allows us to have simulation relations that are related to bisimulation. Also, we did show in the previous proposition that making the commuting diagram lax, with the relator that is not laxed, we get an equivalent definition under the axiom of choice. But if the relator is not symmetric, it is already giving us a notion of simulation, even though we have not laxed it!
|
|
%\begin{figure}[t]
|
|
% \centering
|
|
% \begin{tabular}{|l|c|c|}
|
|
% \hline
|
|
% & \textbf{Anonymous} & \textbf{Onymous} \\
|
|
% \hline
|
|
% \textbf{Relations} & Hughes-Jacobs & Hermida-Jacobs \\
|
|
% \hline
|
|
% \textbf{Spans} & Span-based & Aczel-Mendler \\
|
|
% \hline
|
|
% \end{tabular}
|
|
% \caption{Comparison of anonymous and onymous settings for relations and spans in $\Set$.}
|
|
% \label{fig:anonymous_onymous-sim}
|
|
%\end{figure}
|
|
%\begin{figure}[t]
|
|
% \centering
|
|
% \begin{tabular}{|l|c|c|c|c|}
|
|
% \hline
|
|
% \qquad Hughes-Jacobs$\Rightarrow$& Aczel-Mendler & Hermida-Jacobs & Span-based & Hughes-Jacobs \\
|
|
% \hline
|
|
% Aczel-Mendler & & & & \\
|
|
% \hline
|
|
% Hermida-Jacobs & \ding{56} & & & \\
|
|
% \hline
|
|
% Span-based & \ding{56} & & & \\
|
|
% \hline
|
|
% Hughes-Jacobs & \ding{56} & & & \\
|
|
% \hline
|
|
% \end{tabular}
|
|
% \caption{Where the axiom of choice is needed to say one bisimulation based on one notion is also a bisimulation with respect to another notion, in $\Set$.}
|
|
% \label{fig:sim-choice}
|
|
%\end{figure}
|
|
%
|
|
%We show the category of partially ordered sets with monotone functions between them with $\poset$.
|
|
%\begin{definition}[A Partial Order Over a Functor]
|
|
% Assuming $F\c\Set\to\Set$ is a functor, we call $\appr\c\Set\to\preord$ an order over the functor $F$ iff the following diagram commutes:
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \& \preord \\
|
|
% \Set \& \Set
|
|
% \arrow["U", from=1-2, to=2-2]
|
|
% \arrow["\appr", from=2-1, to=1-2]
|
|
% \arrow["F"', from=2-1, to=2-2]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
%\end{definition}
|
|
%
|
|
%
|
|
%We gave an introduction to Hughes and Jacobs paper. They also have a way to represent simulation relations. In the following, we try to find a suitable formalization for simulation relations, inspired by Hughes and Jacobs.
|
|
%\subsection{Relations as \cancel{Pullbacks} Spans(?)}
|
|
%We can not show every relation by pullbacks, but we can just show relations of the form
|
|
%\begin{gather*}
|
|
% \{(a,b)\mid f(a)=g(b)\}
|
|
%\end{gather*}
|
|
%for some functions $f$ and $g$, when we are in $\Set$, so we can not show every object in $\rel$ using this approach, including $\rel_\appr(F)(R)\appr_{X_2};\rel(F)(R);\appr_{X_1}$ that is the target of simulation. Although we can show $\rel_\appr(F)(R)\appr_{X_2};\rel(F)(R);\appr_{X_1}$ as a span.
|
|
%
|
|
%Assuming, we have a category $\BC$, an object of the category of spans over $\BC$ is $(R,X_1,X_2,p_1,p_2)$ in the form of the following diagram:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% \& R \\
|
|
% {X_1} \&\& {X_2}
|
|
% \arrow["{p_1}"', from=1-2, to=2-1]
|
|
% \arrow["{p_2}", from=1-2, to=2-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%A morhpism from a span $(R,X_1,X_2,p_1,p_2)$ to a span $(S,Y_1,Y_2,q_1,q_2)$ is a morphism $f\c R\to S$ in $\BC$, for which exist $f_1\c X_1\to Y_i$ and $f_2\c X_2\to Y_j$, where $i,j\in\{1,2\}$ and $i\neq j$, and they are in $\BC$, that take part in the following commuting diagram:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_2} \& R \& {X_1} \\
|
|
% {Y_j} \& S \& {Y_i}
|
|
% \arrow["{f_2}"', from=1-1, to=2-1]
|
|
% \arrow["{p_2}"', from=1-2, to=1-1]
|
|
% \arrow["{p_1}", from=1-2, to=1-3]
|
|
% \arrow["f", from=1-2, to=2-2]
|
|
% \arrow["{f_1}", from=1-3, to=2-3]
|
|
% \arrow["{q_j}", from=2-2, to=2-1]
|
|
% \arrow["{q_i}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%We define a $F$-simulation as the coalgebra of the object $\appr_{X_1};\rel(F)(R);\appr_{X_2}$ that has the following structure in $\BC$:
|
|
%\begin{tikzcd}[ampersand replacement=\&]
|
|
% {\appr_{X_1};\rel(F)(R);\appr_{X_2}} \& {\appr_{X_1};\rel(F)(R)} \& {\appr_{X_1}} \& {FX_1} \\
|
|
% \& FR \& {FX_1} \\
|
|
% {\appr_{X_2}} \& {FX_2} \\
|
|
% {FX_2}
|
|
% \arrow["{{{{{{{\pi_1}}}}}}}", from=1-1, to=1-2]
|
|
% \arrow["{{{{{{{\pi_2}}}}}}}"', from=1-1, to=3-1]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-2]
|
|
% \arrow["{{{{{{{\varphi_1}}}}}}}", from=1-2, to=1-3]
|
|
% \arrow["{{{{{{{\varphi_2}}}}}}}", from=1-2, to=2-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=2-3]
|
|
% \arrow["{{{{{{{i^1_1}}}}}}}", from=1-3, to=1-4]
|
|
% \arrow["{{{{{{{i^1_2}}}}}}}", from=1-3, to=2-3]
|
|
% \arrow["{{{{{{{Fp_1}}}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{{{Fp_2}}}}}}}"', from=2-2, to=3-2]
|
|
% \arrow["{{{{{{{i^2_1}}}}}}}"', from=3-1, to=3-2]
|
|
% \arrow["{{{{{{{i^2_2}}}}}}}"', from=3-1, to=4-1]
|
|
%\end{tikzcd}
|
|
%
|
|
%For $F=B(\mS,-)$ this object has the following structure in $\Set$:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\appr_{X_2};\rel(B(\mS,-))(E);\appr_{X_1}} \& {\appr_{X_2};\rel(B(\mS,-))(E)} \& {\appr_{X_2}} \& {B(\mS,X_2)} \\
|
|
% \& {B(\mS,E)} \& {B(\mS,FX_2)} \\
|
|
% {\appr_{X_1}} \& {B(\mS,X_1)} \\
|
|
% {B(\mS,X_1)}
|
|
% \arrow["{\pi_1}", from=1-1, to=1-2]
|
|
% \arrow["{\pi_2}"', from=1-1, to=3-1]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=3-2]
|
|
% \arrow["{\varphi_1}", from=1-2, to=1-3]
|
|
% \arrow["{\varphi_2}", from=1-2, to=2-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=45}, draw=none, from=1-2, to=2-3]
|
|
% \arrow["{i^2_1}", from=1-3, to=1-4]
|
|
% \arrow["{i^2_2}", from=1-3, to=2-3]
|
|
% \arrow["{B(\mS,p_2)}", from=2-2, to=2-3]
|
|
% \arrow["{B(\mS,p_1)}"', from=2-2, to=3-2]
|
|
% \arrow["{i^1_1}"', from=3-1, to=3-2]
|
|
% \arrow["{i^1_2}"', from=3-1, to=4-1]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%
|
|
%So, we call $\sigma\c E\to \appr_{X_2};\rel(B(\mS,-))(E);\appr_{X_1}$ a $B(\mS,-)$-simulation iff the following diagram commutes:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_2} \& E \& {X_1} \\
|
|
% {B(\mS,X_2)} \& {\appr_{X_2};\rel(B(\mS,-))(E);\appr_{X_1}} \& {B(\mS,X_1)} \\
|
|
% {\appr_{X_1}} \& {\appr_{X_2};\rel(B(\mS,-))(E)} \& {\appr_{X_2}}
|
|
% \arrow["\beta"', from=1-1, to=2-1]
|
|
% \arrow["{p_2}"', from=1-2, to=1-1]
|
|
% \arrow["{p_1}", from=1-2, to=1-3]
|
|
% \arrow["\sigma", from=1-2, to=2-2]
|
|
% \arrow["\alpha", from=1-3, to=2-3]
|
|
% \arrow["{\pi_2}", from=2-2, to=3-1]
|
|
% \arrow["{\pi_1}"', from=2-2, to=3-2]
|
|
% \arrow["{i_2^1}", from=3-1, to=2-1]
|
|
% \arrow["{\varphi_1}"', from=3-2, to=3-3]
|
|
% \arrow["{i_1^2}"', from=3-3, to=2-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%
|
|
%We show that if we consider a relation $R$ and its opposite are both simulation relations, then $R$ is a bisimulation. To reach to that goal, we give a formal definition of what we mean by the opposite of $R$ in our categorical setting that we show with $R^\op$. $(R^\op,p'_1,p'_2)$ is a span, that is isomorphic to $R$ via morphism $s\c R\to R^\op$ in $\rel$ that we call swap, and it commutes in the following commutative diagram:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_2} \& R \& {X_1} \\
|
|
% {X_2} \& {R^\op} \& {X_1}
|
|
% \arrow["\id"', from=1-1, to=2-1]
|
|
% \arrow["{p_2}"', from=1-2, to=1-1]
|
|
% \arrow["{p_1}", from=1-2, to=1-3]
|
|
% \arrow["s"', from=1-2, to=2-2]
|
|
% \arrow["\id", from=1-3, to=2-3]
|
|
% \arrow["{p'_1}", from=2-2, to=2-1]
|
|
% \arrow["{p'_2}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%\begin{lemma}
|
|
% The relation $(\appr_{X_1};\rel(F)(R);\appr_{X_2})^\op$ is isomorphic to $\appr_{X^\op_2};\rel(F)(R^\op);\appr_{X^\op_1}$.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% We set $s_1\c\appr_{X_1}\to\appr^\op_{X_1}$ and $s_2\c\appr_{X_2}\to\appr^\op_{X_2}$ to be the swaps of $\appr_{X_1}$ and $\appr_{X_2}$, respectively. Since we have
|
|
% \begin{align*}
|
|
% i'^{1}_{1}\comp s_1\comp\varphi_1&\\
|
|
% &=i^{1}_{2}\comp\varphi_2\\
|
|
% &=Fp_1\comp\varphi_2\\
|
|
% &=Fp'_2\comp Fs\comp \varphi_2,
|
|
% \end{align*}
|
|
% there exists the morphism $s''\c\appr_{X_1};\rel(F)(R)\to\rel(F)(R^\op);\appr_{X_1}^\op$ depicted in the following commutative diagram:
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\appr_{X_1};\rel(F)(R)} \& {\appr_{X_1}} \\
|
|
% FR \& {\rel(F)(R^\op);\appr_{X_1}^\op} \& {\appr_{X_1}^\op} \\
|
|
% \& {FR^\op} \& {FX_1}
|
|
% \arrow["{\varphi_1}", from=1-1, to=1-2]
|
|
% \arrow["{\varphi_2}"', from=1-1, to=2-1]
|
|
% \arrow["{s''}", dashed, from=1-1, to=2-2]
|
|
% \arrow["{s_1}"', from=1-2, to=2-3]
|
|
% \arrow["Fs"', from=2-1, to=3-2]
|
|
% \arrow["{{{{{\varphi'_2}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{\varphi'_1}}}}}"', from=2-2, to=3-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=3-3]
|
|
% \arrow["{{{{{{{{i'^{1}_1}}}}}}}}"', from=2-3, to=3-3]
|
|
% \arrow["{{{{{{{{Fp'_2}}}}}}}}", from=3-2, to=3-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
% Similarily, we get $s''^\mone\c\appr_{X_1};\rel(F)(R)\to\rel(F)(R^\op);\appr_{X_1}^\op$ since
|
|
% \begin{align*}
|
|
% i^{1}_{2}\comp s_1^\mone\comp\varphi'_2&\\
|
|
% &=i'^{1}_{1}\comp\varphi_2'\\
|
|
% &=Fp'_2\comp\varphi'_1\\
|
|
% &=Fp_1\comp Fs_1^\mone\comp \varphi'_1,
|
|
% \end{align*}
|
|
% and it is depicted in the following diagram:
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\rel(F)(R^\op);\appr_{X_1}^\op} \& {\appr_{X_1}} \\
|
|
% {FR^\op} \& {\appr_{X_1};\rel(F)(R)} \& {\appr_{X_1}} \\
|
|
% \& FR \& {FX_1}
|
|
% \arrow["{{\varphi'_2}}", from=1-1, to=1-2]
|
|
% \arrow["{{\varphi'_1}}"', from=1-1, to=2-1]
|
|
% \arrow["{{s''^\mone}}"', dashed, from=1-1, to=2-2]
|
|
% \arrow["{s_1^\mone}", from=1-2, to=2-3]
|
|
% \arrow["{Fs^\mone}"', from=2-1, to=3-2]
|
|
% \arrow["{{\varphi_1}}", from=2-2, to=2-3]
|
|
% \arrow["{{\varphi_2}}"', from=2-2, to=3-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=3-3]
|
|
% \arrow["{i_2^1}", from=2-3, to=3-3]
|
|
% \arrow["{Fp_1}"', from=3-2, to=3-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
% Obviously, $s''$ and $s''^\mone$ are each other's inverse, thus $\appr_{X_1};\rel(F)(R)$ and $\rel(F)(R^\op);\appr_{X_1}^\op$ are isomorphic.
|
|
% \begin{align*}
|
|
% Fp'_1\comp\varphi'_1\comp s''\comp\pi_1&\\
|
|
% &=Fp'_1\comp Fs\comp\varphi_2\comp\pi_1\\
|
|
% &=Fp_2\comp\varphi_2\comp\pi_1\\
|
|
% &=i^2_1\comp\pi_2\\
|
|
% &=i'^{2}_{2}\comp s_2\comp \pi_2
|
|
% \end{align*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\appr_{X_1};\rel(F)(R);\appr_{X_2}} \& {\appr_{X_1};\rel(F)(R)} \\
|
|
% \& {\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op} \& {\rel(F)(R^\op);\appr_{X_1}^\op} \\
|
|
% {\appr_{X_2}} \&\& {FR^\op} \\
|
|
% \& {\appr_{X_2}^\op} \& {FX_2}
|
|
% \arrow["{{{{{{{\pi_1}}}}}}}", from=1-1, to=1-2]
|
|
% \arrow["{s'}", dashed, from=1-1, to=2-2]
|
|
% \arrow["{{{{{{{\pi_2}}}}}}}"', from=1-1, to=3-1]
|
|
% \arrow["{s''}", from=1-2, to=2-3]
|
|
% \arrow["{{{{{{\pi'_2}}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{{\pi'_1}}}}}}"', from=2-2, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-3]
|
|
% \arrow["{\varphi'_1}", from=2-3, to=3-3]
|
|
% \arrow["{{{s_2}}}", from=3-1, to=4-2]
|
|
% \arrow["{Fp'_1}", from=3-3, to=4-3]
|
|
% \arrow["{i'^{2}_{2}}", from=4-2, to=4-3]
|
|
% \end{tikzcd}
|
|
%
|
|
% \begin{align*}
|
|
% Fp_2\comp\varphi_2\comp s''^\mone\comp\pi'_2&\\
|
|
% &=Fp_2\comp Fs^\mone\comp\varphi'_1\comp\pi'_2\\
|
|
% &=Fp'_1\comp\varphi'_1\comp\pi'_2\\
|
|
% &=i'^{2}_{2}\comp\pi'_1\\
|
|
% &=i^{2}_{1}\comp s_2^\mone\comp\pi'_1
|
|
% \end{align*}
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op} \&\& {\rel(F)(R^\op);\appr_{X_1}^\op} \\
|
|
% \& {\appr_{X_1};\rel(F)(R);\appr_{X_2}} \& {\appr_{X_1};\rel(F)(R)} \\
|
|
% \&\& FR \\
|
|
% {\appr_{X_2}^\op} \& {\appr_{X_2}} \& {FX_2}
|
|
% \arrow["{{{{{{\pi'_2}}}}}}", from=1-1, to=1-3]
|
|
% \arrow["{s'^\mone}"', dashed, from=1-1, to=2-2]
|
|
% \arrow["{{{{{{\pi'_1}}}}}}"', from=1-1, to=4-1]
|
|
% \arrow["{s''^\mone}", from=1-3, to=2-3]
|
|
% \arrow["{{{{{{{\pi_1}}}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{{{\pi_2}}}}}}}"', from=2-2, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-3]
|
|
% \arrow["{{{{{{{\varphi_2}}}}}}}", from=2-3, to=3-3]
|
|
% \arrow["{{{{{{{Fp_2}}}}}}}", from=3-3, to=4-3]
|
|
% \arrow["{{{s_2}^\mone}}"', from=4-1, to=4-2]
|
|
% \arrow["{{{{{{{i^2_1}}}}}}}"', from=4-2, to=4-3]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
% So, we could prove that $\appr_{X_1};\rel(F)(R);\appr_{X_2}$ and $\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op$ are isomorphic. $(\appr_{X_1};\rel(F)(R);\appr_{X_2})^\op$ is isomorphic to $\appr_{X_1};\rel(F)(R);\appr_{X_2}$ by definition, so it is also isomorphic with $\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op$.\qed
|
|
% %{\tiny
|
|
% % \begin{equation*}
|
|
% % \begin{tikzcd}[ampersand replacement=\&]
|
|
% % {\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op} \& {\rel(F)(R^\op);\appr_{X_1}^\op} \& {\appr_{X_1}^\op} \& {FX_1} \\
|
|
% % \& {FR^\op} \& {FX_1} \\
|
|
% % {\appr_{X_2}^\op} \& {FX_2} \\
|
|
% % {FX_2}
|
|
% % \arrow["{{{{{{\pi'_2}}}}}}"', from=1-1, to=1-2]
|
|
% % \arrow["{{{{{{\pi'_1}}}}}}", from=1-1, to=3-1]
|
|
% % \arrow["{{{{\varphi'_2}}}}", from=1-2, to=1-3]
|
|
% % \arrow[""{name=0, anchor=center, inner sep=0}, "{{{{\varphi'_1}}}}"', from=1-2, to=2-2]
|
|
% % \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-2, to=2-3]
|
|
% % \arrow["{{{{{{{i'^{1}_2}}}}}}}", from=1-3, to=1-4]
|
|
% % \arrow["{{{{{{{i'^{1}_1}}}}}}}"', from=1-3, to=2-3]
|
|
% % \arrow["{{{{{{{Fp'_2}}}}}}}", from=2-2, to=2-3]
|
|
% % \arrow["{{{{{{{Fp'_1}}}}}}}", from=2-2, to=3-2]
|
|
% % \arrow[""{name=1, anchor=center, inner sep=0}, from=3-1, to=3-2]
|
|
% % \arrow["{{{{{{{i'^{2}_1}}}}}}}", from=3-1, to=4-1]
|
|
% % \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=45}, draw=none, from=1-1, to=0]
|
|
% % \arrow["{{{{{{{i'^{2}_2}}}}}}}", from=1, to=3-2]
|
|
% % \end{tikzcd}
|
|
% % \end{equation*}
|
|
% %}
|
|
%\end{proof}
|
|
%\begin{prop}
|
|
% Having $\sigma\c R\to\appr_{X_2};\rel(F)(R);\appr_{X_1}$ and $\sigma^\op\c R^{\op}\to\appr_{X_1};\rel(F)(R^{\op});\appr_{X_2}$ gives rise to a morphism $\gamma\c R\to\rel(F)(R)$, and vice-versa.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% {\tiny
|
|
% \begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% R \&\&\&\& {X_1} \\
|
|
% \& {\appr_{X_1};\rel(F)(R);\appr_{X_2}} \& {\appr_{X_1};\rel(F)(R)} \& {\appr_{X_1}} \& {FX_1} \& {X_1} \\
|
|
% \&\& FR \& {FX_1} \& {\appr_{X_1}^\op} \\
|
|
% \& {\appr_{X_2}} \& {FX_2} \& {FR^\op} \& {\rel(F)(R^\op);\appr_{X_1}^\op} \\
|
|
% {X_2} \& {FX_2} \& {\appr_{X_2}^\op} \&\& {\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op} \\
|
|
% \& {X_2} \&\&\&\& {R^\op}
|
|
% \arrow["{{{{{p_1}}}}}", from=1-1, to=1-5]
|
|
% \arrow["\sigma"', from=1-1, to=2-2]
|
|
% \arrow["{{{{{p_2}}}}}"', from=1-1, to=5-1]
|
|
% \arrow["\alpha", from=1-5, to=2-5]
|
|
% \arrow["{{{{{{\pi_1}}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{{\pi_2}}}}}}"', from=2-2, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-3]
|
|
% \arrow["{{{{{{\varphi_1}}}}}}", from=2-3, to=2-4]
|
|
% \arrow["{{{{{{\varphi_2}}}}}}", from=2-3, to=3-3]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-3, to=3-4]
|
|
% \arrow["{{{{{{i^1_1}}}}}}", from=2-4, to=2-5]
|
|
% \arrow["{{{{{{i^1_2}}}}}}", from=2-4, to=3-4]
|
|
% \arrow["{{s_1}}", from=2-4, to=3-5]
|
|
% \arrow["\alpha", from=2-6, to=2-5]
|
|
% \arrow["{{{{{{Fp_1}}}}}}", from=3-3, to=3-4]
|
|
% \arrow["{{{{{{Fp_2}}}}}}"', from=3-3, to=4-3]
|
|
% \arrow["Fs"', from=3-3, to=4-4]
|
|
% \arrow["{{{{{{i'^{1}_2}}}}}}", from=3-5, to=2-5]
|
|
% \arrow["{{{{{{i'^{1}_1}}}}}}"', from=3-5, to=3-4]
|
|
% \arrow["{{{{{{i^2_1}}}}}}"', from=4-2, to=4-3]
|
|
% \arrow["{{{{{{i^2_2}}}}}}"', from=4-2, to=5-2]
|
|
% \arrow["{{s_2}}"', from=4-2, to=5-3]
|
|
% \arrow["{{{{{{Fp'_2}}}}}}", from=4-4, to=3-4]
|
|
% \arrow["{{{{{{Fp'_1}}}}}}", from=4-4, to=4-3]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=225}, draw=none, from=4-5, to=3-4]
|
|
% \arrow["{{{\varphi'_2}}}", from=4-5, to=3-5]
|
|
% \arrow[""{name=0, anchor=center, inner sep=0}, "{{{\varphi'_1}}}"', from=4-5, to=4-4]
|
|
% \arrow["\beta"', from=5-1, to=5-2]
|
|
% \arrow[""{name=1, anchor=center, inner sep=0}, from=5-3, to=4-3]
|
|
% \arrow["{{{{{{i'^{2}_1}}}}}}", from=5-3, to=5-2]
|
|
% \arrow["{{{{{\pi'_2}}}}}"', from=5-5, to=4-5]
|
|
% \arrow["{{{{{\pi'_1}}}}}", from=5-5, to=5-3]
|
|
% \arrow["\beta"', from=6-2, to=5-2]
|
|
% \arrow["{{{{{p'_2}}}}}", from=6-6, to=2-6]
|
|
% \arrow["{{{{{\sigma^\op}}}}}"', from=6-6, to=5-5]
|
|
% \arrow["{{{{{p'_1}}}}}"', from=6-6, to=6-2]
|
|
% \arrow["{{{{{{i'^{2}_2}}}}}}", from=1, to=4-3]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125, rotate=180}, draw=none, from=5-5, to=0]
|
|
% \end{tikzcd}
|
|
% \end{equation*}
|
|
% }
|
|
% ($\Leftarrow$): We assume that we have the morphism $\gamma\c R\to FR$ such that the following diagram commutes:
|
|
% \begin{equation}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_2} \& R \& {X_1} \\
|
|
% {FX_2} \& FR \& {FX_1}
|
|
% \arrow["\beta"', from=1-1, to=2-1]
|
|
% \arrow["{p_2}"', from=1-2, to=1-1]
|
|
% \arrow["{p_1}", from=1-2, to=1-3]
|
|
% \arrow["\gamma", from=1-2, to=2-2]
|
|
% \arrow["\alpha", from=1-3, to=2-3]
|
|
% \arrow["{Fp_2}", from=2-2, to=2-1]
|
|
% \arrow["{Fp_1}"', from=2-2, to=2-3]
|
|
% \end{tikzcd}
|
|
% \end{equation}
|
|
% Since $\appr_{X_1}$ and $\appr_{X_1}$ preorders, they each have a morphism $\refl$ that pre-composed with their projections gives identity. As it is depicted in the following diagram the pullback property of ${\appr_{X_1};\rel(F)(R)}$ gives us $\sigma'\c R\to{\appr_{X_1};\rel(F)(R)}$ in the following commutative diagram:
|
|
% \begin{equation}\label{eq:diag-thm-sig'}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% R \& {X_1} \& {FX_1} \\
|
|
% \& {\appr_{X_1};\rel(F)(R)} \& {\appr_{X_1}} \\
|
|
% \& FR \& {FX_1}
|
|
% \arrow["{p_1}", from=1-1, to=1-2]
|
|
% \arrow["{\sigma'}", dashed, from=1-1, to=2-2]
|
|
% \arrow["\gamma"', bend right=30, from=1-1, to=3-2]
|
|
% \arrow["\alpha", from=1-2, to=1-3]
|
|
% \arrow["\refl", from=1-3, to=2-3]
|
|
% \arrow["{\varphi_1}", from=2-2, to=2-3]
|
|
% \arrow["{\varphi_2}", from=2-2, to=3-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=3-3]
|
|
% \arrow["{i_2^1}", from=2-3, to=3-3]
|
|
% \arrow["{Fp_1}"', from=3-2, to=3-3]
|
|
% \end{tikzcd}
|
|
% \end{equation}
|
|
% Then the pullback property of $\appr_{X_1};\rel(F)(R);\appr_{X_2}$ gives us the existence of $\sigma\c R\to \appr_{X_1};\rel(F)(R);\appr_{X_2}$ in the following commutative diagram:
|
|
% \begin{equation}\label{eq:diag-thm-sig}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {X_2} \&\&\& R \\
|
|
% \& {\appr_{X_1};\rel(F)(R);\appr_{X_2}} \& {\appr_{X_1};\rel(F)(R)} \\
|
|
% \&\& FR \\
|
|
% {FX_2} \& {\appr_{X_2}} \& {FX_2}
|
|
% \arrow["\beta"', from=1-1, to=4-1]
|
|
% \arrow["{p_2}"', from=1-4, to=1-1]
|
|
% \arrow["\sigma"', dashed, from=1-4, to=2-2]
|
|
% \arrow["{\sigma'}", from=1-4, to=2-3]
|
|
% \arrow["\gamma", bend left=40, from=1-4, to=3-3]
|
|
% \arrow["{\pi_1}", from=2-2, to=2-3]
|
|
% \arrow["{\pi_2}"', from=2-2, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=4-3]
|
|
% \arrow["{\varphi_2}", from=2-3, to=3-3]
|
|
% \arrow["{Fp_2}", from=3-3, to=4-3]
|
|
% \arrow["\refl"', from=4-1, to=4-2]
|
|
% \arrow["{i_1^2}"', from=4-2, to=4-3]
|
|
% \end{tikzcd}
|
|
% \end{equation}
|
|
% Now, we show that $\sigma$ is a simulation:
|
|
% \begin{align*}
|
|
% i^1_1\comp\varphi_1\comp\pi_1\comp\sigma&&\\
|
|
% &=i^1_1\comp\varphi_1\comp\sigma'&\by{\eqref{eq:diag-thm-sig}}\\
|
|
% &=i^1_1\comp\refl\comp\alpha\comp p_1&\by{\eqref{eq:diag-thm-sig'}}\\
|
|
% &=\alpha\comp p_1&
|
|
% \end{align*}
|
|
% \begin{align*}
|
|
% i^2_2\comp\pi_2\comp\sigma&&\\
|
|
% &=i^2_2\comp\refl\comp\beta\comp p_2&\by{\eqref{eq:diag-thm-sig}}\\
|
|
% &=\beta\comp p_2
|
|
% \end{align*}
|
|
% Considering that $s\c R\to R^\op$ and $s'\c\appr_{X_1};\rel(F)(R);\appr_{X_2}\to\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op$ are swapping isomorphisms, We set $\sigma^\op\c R^\op\to\appr_{X_2}^\op;\rel(F)(R^\op);\appr_{X_1}^\op$ to be $\sigma^\op=s'\comp\sigma\comp s^\mone$. Now, we show that $\sigma'$ is a simulation:
|
|
% \begin{align*}
|
|
% i'^{1}_{2}\comp\varphi'_2\comp\pi'_2\comp \sigma^\op&\\
|
|
% &=i'^{1}_{2}\comp\varphi'_2\comp\pi'_2\comp s'\comp\sigma\comp s^\mone\\
|
|
% &=i^{1}_{1}\comp\varphi_1\comp\pi_1\comp\sigma\comp s^\mone\\
|
|
% &=\alpha\comp p_1\comp s^\mone\\
|
|
% &=\alpha\comp p'_2
|
|
% \end{align*}
|
|
% \begin{align*}
|
|
% i'^{2}_{1}\comp\pi'_1\comp\sigma^\op=&\\
|
|
% &=i'^{2}_{1}\comp\pi'_1\comp s'\comp\sigma\comp s^\mone\\
|
|
% &=i_2^2\comp\pi_2\comp\sigma\comp s^\mone\\
|
|
% &=\beta\comp p_2\comp s^\mone\\
|
|
% &=\beta\comp p'_1
|
|
% \end{align*}
|
|
%\end{proof}
|
|
%%
|
|
%\subsection{Simulation with one relation composition}
|
|
%We recall everything we had in the previous section. Although we want to work with the functor that takes $R\subseteq X_1\times X_2$ and gives $\rel(F)(R);\appr_{X_2}$.
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {\rel(F)(R);\appr_{X_2}} \& FR \& {FX_1} \\
|
|
% {\appr_{X_2}} \& {FX_2} \\
|
|
% {FX_2}
|
|
% \arrow["{{{{{{{{\pi_1}}}}}}}}", from=1-1, to=1-2]
|
|
% \arrow["{{{{{{{{\pi_2}}}}}}}}"', from=1-1, to=2-1]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
|
% \arrow["{Fp_1}", from=1-2, to=1-3]
|
|
% \arrow["{{{{{{{{Fp_2}}}}}}}}", from=1-2, to=2-2]
|
|
% \arrow["{{{{{{{{i_1}}}}}}}}"', from=2-1, to=2-2]
|
|
% \arrow["{{{{{{{{i_2}}}}}}}}"', from=2-1, to=3-1]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%%
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% R \&\&\& {X_1} \\
|
|
% \& {\rel(F)(R);\appr_{X_2}} \& FR \& {FX_1} \& {X_1} \\
|
|
% \& {\appr_{X_2}} \& {FX_2} \& FR \\
|
|
% {X_2} \& {FX_2} \& {\appr_{X_2}^\op} \& {\rel(F)(R);\appr_{X_2}^\op} \\
|
|
% \& {X_2} \&\&\& R
|
|
% \arrow["{{p_1}}", from=1-1, to=1-4]
|
|
% \arrow["\sigma", from=1-1, to=2-2]
|
|
% \arrow["{{p_2}}"', from=1-1, to=4-1]
|
|
% \arrow["\alpha", from=1-4, to=2-4]
|
|
% \arrow["{{{{{{{{{\pi_1}}}}}}}}}", from=2-2, to=2-3]
|
|
% \arrow["{{{{{{{{{\pi_2}}}}}}}}}"', from=2-2, to=3-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=2-2, to=3-3]
|
|
% \arrow["{{Fp_1}}", from=2-3, to=2-4]
|
|
% \arrow["{{Fp_2}}", from=2-3, to=3-3]
|
|
% \arrow["\alpha"', from=2-5, to=2-4]
|
|
% \arrow["{{{{{{{{{i_1}}}}}}}}}"', from=3-2, to=3-3]
|
|
% \arrow["{{{{{{{{{i_2}}}}}}}}}"', from=3-2, to=4-2]
|
|
% \arrow["{Fp_1}"', from=3-4, to=2-4]
|
|
% \arrow["{Fp_2}", from=3-4, to=3-3]
|
|
% \arrow["\beta"', from=4-1, to=4-2]
|
|
% \arrow["{i'_1}", from=4-3, to=3-3]
|
|
% \arrow["{i'_2}", from=4-3, to=4-2]
|
|
% \arrow["{\pi'_1}"', from=4-4, to=3-4]
|
|
% \arrow["{\pi'_2}", from=4-4, to=4-3]
|
|
% \arrow["\beta", from=5-2, to=4-2]
|
|
% \arrow[from=5-5, to=2-5]
|
|
% \arrow["{\sigma^\op}"', from=5-5, to=4-4]
|
|
% \arrow["{p_2}", from=5-5, to=5-2]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%%
|
|
%\begin{prop}
|
|
% Assuming $R\subseteq X\times X$, then if we have $\sigma\c R\to\rel(F)(R);\appr_{X}$ as a simulation for $R$, and $R$ is reflexive, then we have $\gamma\c R\to\rel(F)(R)$ as a bisimulation for $R$, and vice-versa.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% $(\Rightarrow):$
|
|
% \begin{align*}
|
|
% Fp_2\comp\pi_1\comp\sigma&&\\
|
|
% &=i_1\comp\pi_2\comp\sigma\\
|
|
% &=i'_2\comp s\comp\pi_2\comp\sigma\\
|
|
% &=
|
|
% \end{align*}
|
|
%\end{proof}
|
|
%\subsection{Using Lax Pullbacks (Comma Objects) to Model Simulation}
|
|
%A big concern with this approach is that Comma Objects are defined in a 2-category, so we can not define them in $\Set$, while our main inspirational example is coming from $\Set$.
|
|
%
|
|
%\subsection{Working in $\Set$ First, Like Hughes and Jacobs}
|
|
%
|
|
%\subsection{Choosing a suitable order for our setting}
|
|
%Maybe we can first choose a suitable order on $T(\Sigma_\val\mS\times D(\mS,\mS))$ and then prove that if a relation and its inverse is a simulation then it is a bisimulation as well. Maybe $T$ being $\omega$-continuous can give the ordering. It can be something easier that relates to termination as well! That if a term has a big-step evaluation, then it is bigger than or equal to any other term, and if it does not, then it is less than or equal to any other term.
|
|
\section{Simulations and Bisimulations in Double Categories}
|
|
%
|
|
%
|
|
%
|
|
\begin{definition}[Double Relator]\label{def:doub-rela}
|
|
Given a double category $\BC_1\rightrightarrows\BC_0$, for an endofunctor $F$, an $F$-double relator $\relar$ is an endomap on objects of category $\BC_1$, for which the following equations hold:
|
|
\begin{enumerate}
|
|
\item $S\comp\relar=F\comp S$
|
|
\item $T\comp\relar=F\comp T$
|
|
\end{enumerate}
|
|
Additionally, it has the following condition called monotonicity: If there is an inclusion cell from a promorphisms $\mathcal{R}$ to $\mathcal{S}$ then there must be an inclusion promorphism from $\relar\mathcal{R}$ to $\relar\mathcal{S}$ as well.
|
|
\end{definition}
|
|
\begin{remark}
|
|
In the above definition we are describing an endomap on objects of a category, and we form equations with compositions of this map. The composition that we mean, is the one over functions and not the functors. Indeed, it does not make a functor, but it makes another function. Additionally, we will the endomap in the future with another endomap on morphisms as well, but we use the tradition in denoting functors for it that is to use one letter to refer to both maps. %Basically, one can think of a relator $\relar$ as a functor that does not necessarily preserve identities and compositions.
|
|
\end{remark}
|
|
%
|
|
\begin{definition}[Double Coalgebra]
|
|
Given a double category $\BC_1\rightrightarrows\BC_0$, and a $F$-double relator $\relar$, a pair $(\mathcal{R},\delta)$ that $\delta\c\mathcal{R}\Rightarrow\relar\mathcal{R}$ is an $\relar$-double coalgebra on $\BC_1\rightrightarrows\BC_0$.
|
|
\end{definition}
|
|
%
|
|
\begin{example}
|
|
For every relator $\relar$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an $\relar$-simulation from a coalgebra $(X,f)$ to $(Y,g)$ if and only if $((X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y),(f,g))$ is an $\relar$-double coalgebra in $\rel_a\rightrightarrows\Set$.
|
|
\end{example}
|
|
%
|
|
%\subsection{Properties of Double Relators}
|
|
\begin{definition}[Normal Double Relator]
|
|
An $F$-double relator $\relar$ is normal, whenever the following equation holds:
|
|
\begin{gather*}
|
|
\relar\comp U=U\comp F
|
|
\end{gather*}
|
|
\end{definition}
|
|
\begin{definition}
|
|
An $F$-relator $\relar$ is normal, whenever for every set $X$, we have $\relar\id_X=\id_{FX}$.
|
|
\end{definition}
|
|
%
|
|
It is obvious to see that normal relators and double relators in $\rel_a\rightrightarrows\Set$ coincide.\\
|
|
%
|
|
\begin{definition}[Natural Double Relator]
|
|
An $F$-double relator $\relar$ is natural, whenever it takes every cell of type $\mathcal{R}\Rightarrow\mathcal{S}$ to a cell of type $\relar\mathcal{R}\Rightarrow\relar\mathcal{S}$.
|
|
\end{definition}
|
|
Given an $F$-relator $\relar$ defined in~\autoref{def:relator} we have a canonical way to extend its definition over morphisms of $\rel_a$ as well:
|
|
\begin{definition}[Natural Relator]
|
|
An $F$-relator $\relar$ is called \emph{natural}, whenever for every relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$, and arbitrary functions $f\c A\to X$, and $g\c B\to Y$, we have $\relar (g^\op\comp R\comp f)=(Fg)^\op\comp\relar R\comp Ff$.
|
|
\end{definition}
|
|
%
|
|
\begin{definition}[Extendable Relator]
|
|
Assuming $\relar$ is an $F$-relator, it is an \emph{extendable} relator whenever for every morphims $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\rel_a$ there exists the morphism $(Fg_1,Fg_2)\c(FX_1 \stackrel{Fp_1}{\leftarrow} \relar R \stackrel{Fp_2}{\to}FX_2)\to(FY_1 \stackrel{Fq_1}{\leftarrow} \relar S \stackrel{Fq_2}{\to}FY_2)$ in $\rel_a$.
|
|
\end{definition}
|
|
%
|
|
It is obvious that every extendable relator is a double relator in $\rel_a\rightrightarrows\Set$ and vice-versa.
|
|
%
|
|
\begin{prop}
|
|
Every natural $F$-relator $\relar$ is extendable.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming that there exists $(g_1,g_2)\c(X_1 \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}X_2)\to(Y_1 \stackrel{q_1}{\leftarrow} S \stackrel{q_2}{\to}Y_2)$ in $\rel_a$ we need to show that exists the morphism $(Fg_1,Fg_2)\c(FX_1 \stackrel{Fp_1}{\leftarrow} \relar R \stackrel{Fp_2}{\to}FX_2)\to(FY_1 \stackrel{Fq_1}{\leftarrow} \relar S \stackrel{Fq_2}{\to}FY_2)$ in $\rel_a$. It is equivalent with showing that $\relar R\subseteq (Fg_2)^\op\comp\relar S\comp Fg_1$. Since $\relar$ is natural we have $(Fg_2)^\op\comp\relar S\comp Fg_1=\relar(g_2^\op\comp S\comp g_1)$. Existence of $(g_1,g_2)$ is equivalent with $R\subseteq g_2^\op\comp S\comp g_1$, so by the monotonicity of $\relar$ as a relator we have $\relar R\subseteq \relar(g_2^\op\comp S\comp g_1)$. So, we have $\relar R\subseteq Fg_2^\op\comp \relar S\comp Fg_1$.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Every natural relator is a natural double relator in $\rel_a\rightrightarrows\Set$.
|
|
\end{cor}
|
|
%
|
|
\begin{definition}[Two-sided Double Category]
|
|
Assuming that in the double category $\BC_1\rightrightarrows\BC_0$ for every promorphism $\mathcal{R}$ there exists an up to isomorphism unique promorphism $\mathcal{R}^\op$ such that $S\mathcal{R}=T\mathcal{R}^\op$, $T\mathcal{R}=S\mathcal{R}^\op$, and there is an inclusion cell between the two promorphisms that is an isomorphism in $\BC_1$, then $\BC_1\rightrightarrows\BC_0$ is a \emph{two-sided} double category.
|
|
\end{definition}
|
|
\begin{example}
|
|
For an arbitrary category $\BC$ all the categories $\spa(\BC)$, $\rel(\BC)$, $\spa_a$, and $\rel_a$, are two-sided double categories.
|
|
\end{example}
|
|
%
|
|
\begin{definition}[Symmetric Double Relator]
|
|
An $F$-double relator $\relar$ over a two-sided double category $\BC_1\rightrightarrows\BC_0$ is a \emph{symmetric} double relator, whenever for every promorphism $\mathcal{R}$, the promorphism $\relar(\mathcal{R}^\op)$ is isomorphic to $(\relar\mathcal{R})^\op$.
|
|
\end{definition}
|
|
%
|
|
\begin{example}
|
|
Every relator $\relar$ is symmetric if and only if it is a symmetric double relator on $\rel_a\rightrightarrows\Set$.
|
|
Additionally, for every symmetric relator $\relar$, a relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an $\relar$-bisimulation in coalgebras $(X,f)$ and $(Y,g)$ if and only if $((X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y),(f,g))$ is an $\relar$-double coalgebra on $\rel_a\rightrightarrows\Set$.
|
|
\end{example}
|
|
%
|
|
\begin{example}
|
|
For a category $\BC_0$ with pullbacks, every Aczel-Mendler bisimulation is an $\relar$-double coalgebra that $\relar$ is the map that takes every span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to a span $(FX \stackrel{Fp_1}{\leftarrow} FR \stackrel{Fp_2}{\to}FY)$.
|
|
\end{example}
|
|
%
|
|
\begin{definition}[Horizontal Lifting]
|
|
In a double category $\BC_1\rightrightarrows\BC_0$, assuming that $F$ is an endofunctor on $\BC_0$, a lifting of $F$ that we denote with $\bar{F}$ is an endofunctor on $\BC_1$ such that the following diagrams commute:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{\BC_1} \& {\BC_1} \& {\BC_1} \& {\BC_1} \\
|
|
{\BC_0} \& {\BC_0} \& {\BC_0} \& {\BC_0} \\
|
|
\& {\BC_0} \& {\BC_0} \\
|
|
\& {\BC_1} \& {\BC_1} \\
|
|
{\BC_2} \&\&\& {\BC_1} \\
|
|
\&\&\& {\BC_1}
|
|
\arrow["{\bar{F}}", from=1-1, to=1-2]
|
|
\arrow["S"', from=1-1, to=2-1]
|
|
\arrow["S", from=1-2, to=2-2]
|
|
\arrow["{\bar{F}}", from=1-3, to=1-4]
|
|
\arrow["T"', from=1-3, to=2-3]
|
|
\arrow["T", from=1-4, to=2-4]
|
|
\arrow["F"', from=2-1, to=2-2]
|
|
\arrow["F"', from=2-3, to=2-4]
|
|
\arrow["F", from=3-2, to=3-3]
|
|
\arrow["U"', from=3-2, to=4-2]
|
|
\arrow["U", from=3-3, to=4-3]
|
|
\arrow["{\bar{F}}"', from=4-2, to=4-3]
|
|
\arrow["{-\odot-}", from=5-1, to=5-4]
|
|
\arrow["{\bar{F}-\odot \bar{F}-}"', from=5-1, to=6-4]
|
|
\arrow["{\bar{F}}", from=5-4, to=6-4]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{definition}
|
|
\begin{remark}
|
|
Every horizontal lifting of a vertical endofunctor $F$ is an $\bar{F}$-double relator. Actually, these relators are highly well-behaved, and satisfy some of the mentioned properties of double relators. For example they are natural, normal (both are obvious) and symmetric.
|
|
\end{remark}
|
|
\begin{prop}
|
|
Every horizontal lifting is a symmetric double relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
%\begin{definition}[Barr Double Relator]
|
|
% \todo{Finish}
|
|
%\end{definition}
|
|
\begin{example}
|
|
For a category $\BC_0$ with pullbacks and an endofunctor $F$ on it, every abstract relational bisimulation (\autoref{def:abs-rel-bis}) is an $\rel(F)$-double coalgebra that $\rel(F)$ is the map that takes every relation $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to a relation $(FX \stackrel{\rel(F)p_1}{\leftarrow} \rel(F)R \stackrel{\rel(F)p_2}{\to}FY)$. It entails that every Hermida-Jacobs bisimulation is also an $(F-)^\dagger$-double coalgebra if we conceive $(F-)^\dagger$ as the proper double relator.
|
|
\end{example}
|
|
%
|
|
Using the basic fact in $\Set$ that a relation $R\subseteq X\times X$ is a poset iff $R\comp R=R$, we define poset enrichment abstractly as follows:
|
|
\begin{definition}[Poset-enrichment Functor]
|
|
On a double category $\BC_1\rightrightarrows\BC_0$, for a functor $F\c\BC_0\to\BC_0$ a functor $P_F\c\BC_0\to\BC_1$ is a \emph{poset-enrichment} functor over $F$ such that for every objects $X$ and mrphisms $f$, the following equations hold ($\Delta$ is the diagonal functor):
|
|
\begin{gather*}
|
|
SP_F=F\\
|
|
TP_F=F\\
|
|
P_F=\odot\comp\Delta\comp P_F
|
|
\end{gather*}
|
|
\end{definition}
|
|
%
|
|
\begin{example}
|
|
We abstract Hughes-Jacobs simulation using poset-enrichment functor. On $\BC_1\rightrightarrows\BC_0$ with an endofunctor $F\c\BC_0\to\BC_0$ we have an $F$-double relator that takes every promorphism $\mathcal{R}$ to $P_FS\mathcal{R}\odot\mathcal{R}\odot P_FT\mathcal{R}$. More concretely, if $\BC_0$ is just an arbitrary category with pullbacks $\BC$, and $\BC_1=\rel(\BC)$ (it can also be $\spa(\BC)$ to give a different notion), then we have a double relator that takes $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ to $(FX \stackrel{c_{PF_X}}{\leftarrow} P_FX\odot (FR)^\dagger\odot P_FY \stackrel{c_{PF_X}}{\to}Y)$.\todo{Check this!}
|
|
\end{example}
|
|
%
|
|
\begin{prop}
|
|
Assuming that $\BC$ is a regular category with choice, then the concrete notion above is equivalent with Hermida-Jacobs simulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Fisith!}
|
|
\end{proof}
|
|
%
|
|
|
|
\todo{Talk more about the poset enrichment here. You can say that in $\rel(\BC)\rightrightarrows\BC$ and $\spa(\BC)\rightrightarrows\BC$ an order enrichment on $F$ can be represented as a functor of type $\BC\to\spa(\BC)$ or $\BC\to\rel(\BC)$!}
|
|
\begin{definition}[Bi-laxed Barr Double Relator]
|
|
\todo{Finish}
|
|
\end{definition}
|
|
\begin{prop}
|
|
A span $(X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y)$ is an Aczel-Mendler simulation from a coalgebra $(X,g)$ to $(Y,h)$ with the morhpism $(g,h,w)$ if and only if it is a $((X \stackrel{p_1}{\leftarrow} R \stackrel{p_2}{\to}Y),(g,h,w))$ is an $F^\leftrightarrow$-double coalgebra, and $F^\leftrightarrow$ is a bi-laxed Barr double relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Finish}
|
|
\end{proof}
|
|
|
|
\todo{Investigate if double coalgebras form a (possibly double) category. If it is true find out what is the final object their!}
|
|
\subsection{Bisimulation in Double Categories}
|
|
\todo{Start writing this after the section for double categories in the previous section. You should first introduce your definition, and then give all the definitions as examples of your "double coalgebra".}
|
|
\subsection{Simulations in Double Categories}
|
|
\todo{Try to define simulation in a given double category. Perhaps you need to order enrichment over the endofunctor on $\BC_0$. Then inspired by~\autoref{prop:HeJ-HuJ} you may be able to prove a general theorem for an arbitrary lifting!}
|
|
\section{Symmetric Simulation is a Bisimulation}
|
|
\todo{Obviously, this chapter should be changed. All the definitions should be moved to somewhere else. You should start the chapter by giving your counter examples, and then presenting your proofs.}
|
|
\todo{I think that these "double coalgebra"s for a category. Perhaps a double category! I wonder what is the final object there.
|
|
Additionally, you can think of defining "double algebras" and see the relevance with congruence relations!}
|
|
\begin{definition}[Graph]
|
|
In a category $\BC$ a graph is a tuple $(R,X)$ of the following form:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& R \& \\
|
|
X \&\& X
|
|
\arrow["{p_1}"', from=1-2, to=2-1]
|
|
\arrow["{p_2}", from=1-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Graphs over $\BC$ form a category that we denote by $\gra(\BC)$.
|
|
\end{definition}
|
|
\begin{definition}[Symmetric Graph]
|
|
A graph $(R,X)$ is symmetric iff there exists an endomorphism $s\c R\to R$, such that the following diagram commutes
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
X \& R \& X
|
|
\arrow["\id"', from=1-1, to=2-1]
|
|
\arrow["{p_2}"', from=1-2, to=1-1]
|
|
\arrow["{p_1}", from=1-2, to=1-3]
|
|
\arrow["s", from=1-2, to=2-2]
|
|
\arrow["\id", from=1-3, to=2-3]
|
|
\arrow["{p_1}", from=2-2, to=2-1]
|
|
\arrow["{p_2}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
and $s\comp s=\id$. We call $s$ a \emph{swap} for $R$.
|
|
\end{definition}
|
|
\begin{lemma}\label{lem:gra-sym}
|
|
Symmetry of a graphs over preserved a functor.
|
|
\end{lemma}
|
|
\begin{definition}[Relation]
|
|
A relation in a category $\BC$ is a graph $(R,X)$ where $\brks{p_1,p_2}\c R \to X\times Y$ is monic. Relations over $\BC$ form a category that we denote by $\rel(\BC)$.
|
|
\end{definition}
|
|
\begin{definition}[Jointly Monic]
|
|
A pair of morphisms $p_1,p_2\c R\to X$ is jointly monic iff for every pair of morphisms $f,g\c A \to R$ assuming that $p_1\comp f=p_1\comp g$ and $p_2\comp f=p_2\comp g$ then $f=g$.
|
|
\end{definition}
|
|
\begin{prop}\label{prop:rel-joi-mon}
|
|
A graph $(R,X)$ is a relation iff $p_1$ and $p_2$ are jointly monic.
|
|
\end{prop}
|
|
\begin{proof}
|
|
($\Rightarrow$): We assume that for morphisms $f,g\c A\to R$ we have $p_1\comp f=p_1\comp g$ and $p_2\comp f=p_2\comp g$, and we want to prove that $f=g$. Assuming that $\pi_1,\pi_2\c X\times X\to X$ are projections of $X\times X$, then we have:
|
|
\begin{align*}
|
|
\brks{p_1,p_2}\comp f&\\
|
|
=&\brks{p_1\comp f,p_2\comp f}\\
|
|
=&\brks{p_1\comp g,p_2\comp g}\\
|
|
=&\brks{p_1,p_2}\comp g
|
|
\end{align*}
|
|
Since $\brks{p_1,p_2}$ is moinc, from $\brks{p_1,p_2}\comp f=\brks{p_1,p_2}\comp g$ we get $f=g$.
|
|
|
|
($\Leftarrow$): Assuming for some morphisms $f,g\c A\to R$ we have $\brks{p_1,p_2}\comp f=\brks{p_1,p_2}\comp g$ we need to prove $f=g$. From $\brks{p_1,p_2}\comp f=\brks{p_1,p_2}\comp g$ we get $\brks{p_1\comp f,p_2\comp f}=\brks{p_1\comp g,p_2\comp g}$. Assuming that $\pi_1,\pi_2\c X\times X\to X$ are projections of $X\times X$, then we have $\pi_1\comp\brks{p_1\comp f,p_2\comp f}=\pi_1\comp\brks{p_1\comp g,p_2\comp g}$, and then $p_1\comp f=p_1\comp g$. Similarly we also get $p_2\comp f=p_2\comp g$. So, since $p_1$ and $p_2$ are jointly monic, then we have $f=g$.\qed
|
|
\end{proof}
|
|
We need to work with endofunctors over $\BC$ that are lifted over $\rel(\BC)$, for which we need to first define endofunctors lifted over $\gra(\BC)$.
|
|
Lifting from $\BC$ to $\gra(\BC)$ is easy. For $F\c\BC\to\BC$ we define $F_\gra\c\gra(\BC)\to\gra(\BC)$ as a functor that takes a graph $(R,X)$, and gives $(FR,FX)$, were $F$ is also applied on legs of the graph, i.e., $p_1,p_2\c R\to X$, so, we get the following graph:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& FR \& \\
|
|
FX \&\& FX
|
|
\arrow["{Fp_1}"', from=1-2, to=2-1]
|
|
\arrow["{Fp_2}", from=1-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
This lifting does not work for $\rel$. As an example, if we set $F$ to be the powerset functor $\powf$, then $(\powf R,\powf X)$ is not necessarily a relation anymore. For example, if we take $R=\{(1,0),(0,1),(0,0),(1,1)\}$, then taking $\{\{(1,0),(0,1),(0,0),(1,1)\}\}$ and $\{\{(1,0),(0,1),(0,0)\}\}$ as elements of $\powf R$, the morphism $\brks{\powf p_1,\powf p_2}$ maps them both to $(\{0,1\},\{0,1\})$ so it is not monic.
|
|
|
|
To cope with this, we assume the following epi-mono decomposition for $(R,X)\in\rel(\BC)$:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
R \& {R^\dagger} \&\& {X\times X}
|
|
\arrow["{e_R}"', two heads, from=1-1, to=1-2]
|
|
\arrow["{\brks{p_1,p_2}}", bend left=20, from=1-1, to=1-4]
|
|
\arrow["{\brks{p^\dagger_1,p^\dagger_2}}"', tail, from=1-2, to=1-4]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
We can define $(-)^\dagger$ as a functor from $\gra(\BC)\to\rel(\BC)$, then we define $F_\rel\c\rel(\BC)\to\rel(\BC)$ to take $(R,X)$ to the following relation:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& {(FR)^\dagger} \& \\
|
|
FX \&\& FX \\
|
|
\& {FX\times FX}
|
|
\arrow["{{{(Fp_1)^\dagger}}}"', from=1-2, to=2-1]
|
|
\arrow["{{{(Fp_2)^\dagger}}}", from=1-2, to=2-3]
|
|
\arrow["{{\brks{{(Fp_1)^\dagger},{(Fp_2)^\dagger}}}}"{description}, dashed, tail, from=1-2, to=3-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\begin{lemma}\label{lem:norm-simp}
|
|
Assuming that we have the following commutative diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
FX \& FR \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{{p_1}}"', from=1-2, to=1-1]
|
|
\arrow["{{p_2}}", from=1-2, to=1-3]
|
|
\arrow["{{\sigma}}", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["{Fp_1}", from=2-2, to=2-1]
|
|
\arrow["{Fp_2}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Then there exists $\sigma^\dagger\c R\to(FR)^\dagger$ in the following diagram that is also commutative:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
FX \& {(FR)^\dagger} \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{{p_1}}"', from=1-2, to=1-1]
|
|
\arrow["{{p_2}}", from=1-2, to=1-3]
|
|
\arrow["{{\sigma^\dagger}}", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["{Fp_1^\dagger}", from=2-2, to=2-1]
|
|
\arrow["{Fp_2^\dagger}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
The proof is trivial considering that $\sigma^\dagger=e_{FR}\comp\sigma$, where $e_{FR}$ is the epimorphism in the epi-mono factorization of $\brks{Fp_1,Fp_2}$, as depicted in the following diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
FX \& FR \& FX \\
|
|
FX \& {(FR)^\dagger} \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{{p_1}}"', from=1-2, to=1-1]
|
|
\arrow["{{p_2}}", from=1-2, to=1-3]
|
|
\arrow["{{\sigma}}", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["\id"', from=2-1, to=3-1]
|
|
\arrow["{Fp_1}", from=2-2, to=2-1]
|
|
\arrow["{Fp_2}"', from=2-2, to=2-3]
|
|
\arrow["{e_{FR}}", from=2-2, to=3-2]
|
|
\arrow["\id", from=2-3, to=3-3]
|
|
\arrow["{Fp_1^\dagger}", from=3-2, to=3-1]
|
|
\arrow["{Fp_2^\dagger}"', from=3-2, to=3-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\qed
|
|
\end{proof}
|
|
We denote this relation with $F_\rel(R,X)$.
|
|
\begin{definition}[Simulation]\label{def:sim}
|
|
A coalgebra $\sigma\c R\to (FR)^\dagger$ is a simulation over the $F$-coalgebra $\alpha\c X\to FX$ iff the following diagram is lax-commutative:
|
|
\begin{equation}\label{eq:diag-lax-sim}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
FX \& (FR)^\dagger \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["{{(Fp_1)^\dagger}}", from=2-2, to=2-1]
|
|
\arrow["{{(Fp_2)^\dagger}}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation}
|
|
\end{definition}
|
|
\todo{Prove that simulation in this sense of this definition is a standard simulation for powerset.}
|
|
\begin{definition}[Bisimulation]\label{def:bisim}
|
|
The morphism $\sigma$ in~\autoref{def:sim} is a bisimulation iff the mentioned diagram is fully commutative.
|
|
\end{definition}
|
|
\begin{remark}
|
|
The mentioned definition of bisimulation is actually, the classical one in the literature that is to have the following commutative diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \\
|
|
FX \& (FR)^\dagger \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["{{(Fp_1)^\dagger}}", from=2-2, to=2-1]
|
|
\arrow["{{(Fp_2)^\dagger}}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
It may look different because we have $FX$ and not $(FX)^\dagger$, but they are the same.
|
|
An object $X\in\obj(\BC)$ is $(X,X)\in\rel(\BC)$ having $\id$ as its legs. Meaning that the $(FX)^\dagger=FX$.
|
|
\end{remark}
|
|
%
|
|
\begin{prop}\label{prop:iff-sim-bsim}
|
|
Assuming that we have a bisimulation $\sigma$ for $R$, we have the following equation:
|
|
\begin{gather*}
|
|
\sigma\comp s=(Fs)^\dagger\comp\sigma
|
|
\end{gather*}
|
|
\end{prop}
|
|
\begin{proof}
|
|
We recall that by~\autoref{lem:gra-sym}, $F_\rel(R,X)$ is symmetric with the swap $(Fs)^\dagger$. Assuming that $\sigma$ is a bisimulation, we have the following commutative diagram:
|
|
\begin{equation}\label{eq:diag-sym-rel}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& R \& \\
|
|
X \& R \& X \\
|
|
FX \& (FR)^\dagger \& FX \\
|
|
\& (FR)^\dagger
|
|
\arrow["{p_2}"', bend right=20, from=1-2, to=2-1]
|
|
\arrow["s", from=1-2, to=2-2]
|
|
\arrow["{p_1}", bend left=20, from=1-2, to=2-3]
|
|
\arrow["\alpha"', from=2-1, to=3-1]
|
|
\arrow["{p_1}"', from=2-2, to=2-1]
|
|
\arrow["{p_2}", from=2-2, to=2-3]
|
|
\arrow["\sigma", from=2-2, to=3-2]
|
|
\arrow["\alpha", from=2-3, to=3-3]
|
|
\arrow["{(Fp_2)^\dagger}", bend left=20, from=4-2, to=3-1]
|
|
\arrow["{(Fp_1)^\dagger}", from=3-2, to=3-1]
|
|
\arrow["{(Fp_2)^\dagger}"', from=3-2, to=3-3]
|
|
\arrow["(Fs)^\dagger", from=3-2, to=4-2]
|
|
\arrow["{(Fp_1)^\dagger}"', bend right=20, from=4-2, to=3-3]
|
|
\end{tikzcd}
|
|
\end{equation}
|
|
And it entails that the following diagrams are also commutative:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& X \& X \& R \& X \\
|
|
FX \& (FR)^\dagger \& FX \& FX \& (FR)^\dagger \& FX
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_2}"', from=1-2, to=1-1]
|
|
\arrow["{p_1}", from=1-2, to=1-3]
|
|
\arrow["{\sigma\comp s}", from=1-2, to=2-2]
|
|
\arrow["\alpha", from=1-3, to=2-3]
|
|
\arrow["\alpha"', from=1-4, to=2-4]
|
|
\arrow["{p_2}"', from=1-5, to=1-4]
|
|
\arrow["{p_1}", from=1-5, to=1-6]
|
|
\arrow["{(Fs)^\dagger\comp \sigma}", from=1-5, to=2-5]
|
|
\arrow["\alpha", from=1-6, to=2-6]
|
|
\arrow["{(Fp_1)^\dagger}", from=2-2, to=2-1]
|
|
\arrow["{(Fp_2)^\dagger}"', from=2-2, to=2-3]
|
|
\arrow["{(Fp_1)^\dagger}", from=2-5, to=2-4]
|
|
\arrow["{(Fp_2)^\dagger}"', from=2-5, to=2-6]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
So, since $(Fp_1)^\dagger$ and $(Fp_2)^\dagger$ are jointly monic (because $F_\rel(R,X)$ is a relation and~\autoref{prop:rel-joi-mon}) we have $\sigma\comp s=(Fs)^\dagger\comp\sigma$.\qed
|
|
\end{proof}
|
|
%
|
|
\begin{cor}
|
|
Assuming $\sigma_1$ and $\sigma_2$ are simulations of type $R\to (FR)^\dagger$, and $R$ is symmetric and both $\sigma_1$ and $\sigma_2$ satisfy the following property:
|
|
\begin{gather*}
|
|
(Fs)^\dagger\comp\sigma=\sigma\comp s
|
|
\end{gather*}
|
|
Then $\sigma_1=\sigma_2$.
|
|
\end{cor}
|
|
\begin{proof}
|
|
As the mentioned property is equivalent with $\sigma$ being a bisimulation, and bisimulation is unique, then $\sigma_1=\sigma_2$.
|
|
\end{proof}
|
|
%
|
|
\begin{prop}
|
|
Assuming that the ordering on the functor $F$ is natural, and gives posets, given witness $\sigma$ for a symmetric object $R$ to be an HJ-simulation, if $(Fs)^\dagger\comp\sigma=\sigma\comp s$, then $\sigma$ is a witness for $R$ to be an $HJ$-bisimulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Since $\sigma$ is a witness that $R$ is an $HJ$-simulation we have
|
|
\begin{gather*}
|
|
\alpha\comp p_1\appr (Fp_1)^\dagger\comp\sigma,\\
|
|
(Fp_2)^\dagger\comp\sigma\appr\alpha\comp p_2.
|
|
\end{gather*}
|
|
So, we have:
|
|
\begin{align*}
|
|
\alpha\comp p_1\appr (Fp_1)^\dagger\comp\sigma&\\
|
|
\Rightarrow&\alpha\comp p_1\comp s\appr (Fp_1)^\dagger\comp\sigma\comp s&\by{naturality of $\appr$}\\
|
|
\Rightarrow&\alpha\comp p_1\comp s\appr (Fp_1)^\dagger\comp (Fs)^\dagger\comp\sigma&\by{\eqref{eq:diag-sym-rel}}\\
|
|
\Rightarrow&\alpha\comp p_2\appr (Fp_1)^\dagger\comp (Fs)^\dagger\comp\sigma&\by{assumption}\\
|
|
\Rightarrow&\alpha\comp p_2\appr (Fp_2)^\dagger\comp\sigma&\by{\eqref{eq:diag-sym-rel}}\\
|
|
\Rightarrow&\alpha\comp p_2= (Fp_2)^\dagger\comp\sigma&\by{anti symmetry of $\appr$}
|
|
\end{align*}
|
|
Similarly we have:
|
|
\begin{align*}
|
|
(Fp_2)^\dagger\comp\sigma\appr\alpha\comp p_2&\\
|
|
\Rightarrow&(Fp_2)^\dagger\comp\sigma\comp s\appr\alpha\comp p_2\comp s&\by{naturality of $\appr$}\\
|
|
\Rightarrow&(Fp_2)^\dagger\comp(Fs)^\dagger\comp\sigma\appr\alpha\comp p_2\comp s&\by{\eqref{eq:diag-sym-rel}}\\
|
|
\Rightarrow&(Fp_2)^\dagger\comp(Fs)^\dagger\comp\sigma\appr\alpha\comp p_1&\by{assumption}\\
|
|
\Rightarrow&(Fp_1)^\dagger\comp\sigma\appr\alpha\comp p_1&\by{\eqref{eq:diag-sym-rel}}\\
|
|
\Rightarrow&(Fp_1)^\dagger\comp\sigma=\alpha\comp p_1&\by{anti symmetry of $\appr$}
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
%
|
|
Now, we give a counter example of a symmetric relation on $\Set$ that is a simulation according to~\autoref{def:sim}, i.e, exists the morphism $\sigma$ that commutes laxly in~\eqref{eq:diag-lax-sim}, but $\sigma$ is not a coalgebraic bisimulation, although the relation that we give is clearly a bisimulation in the classic sense.
|
|
We set $R=\{(A,B),(B,A),(C_1,C_2),(C_2,C_1),(C'_2,C_2),(C_2,C'_2),(C_2,C_2)\}$, $F=\mathbf{Id}$, $\appr=\Delta\cup\{(C_1,C_2),(C_2,C'_2)\}$, and the coalgebra $\alpha$ is defined with the following set of reductions:
|
|
\begin{gather*}
|
|
A\to C_1\qquad B\to C_2\qquad C_1\to C_1\qquad C_2\to C_2\qquad C'_2\to C_2
|
|
\end{gather*}
|
|
And finally, we define $\sigma$ as follows:
|
|
\begin{gather*}
|
|
\sigma(w)=
|
|
\begin{cases}
|
|
(\alpha\comp p_1(w),\alpha\comp p_2(w)) & w\neq (B,A) \\
|
|
(C'_2,C_2) & w=(B,A)
|
|
\end{cases}
|
|
\end{gather*}
|
|
|
|
It is easy to check that the conditions $\alpha\comp p_1\appr (Fp_1)^\dagger \comp\sigma$ and $(Fp_2)^\dagger \comp\sigma \appr \alpha\comp p_2$ are satisfied. For every $w\in R$ if $w\neq (B,A)$ then for $i\in\{1,2\}$, we have $\alpha\comp p_i= (Fp_i)^\dagger\comp\sigma$, and for $w=(B,A)$ we have $\alpha\comp p_1(B,A)=C_2\appr C'_2=(Fp_1)^\dagger \comp\sigma(B,A)$, and $\alpha\comp p_2(B,A)=C_1\appr C_2=(Fp_2)^\dagger \comp\sigma(B,A)$.
|
|
And $\sigma$ is not a coalgebraic bisimulation as $\alpha\comp p_1 (B,A)=C_2\neq C'_2=(Fp_1)^\dagger \comp\sigma(B,A)$.
|
|
|
|
An interesting question would be to find out what conditions $\sigma$ should have (maybe we have the answer to this!~\autoref{prop:iff-sim-bsim}), or how it should be constructed (perhaps based on a given poset) so that it will also be a coalgebraic bisimulation if $R$ is symmetric.
|
|
Another avenue would be to give another definition for simulation that does not have this issue.
|
|
|
|
Well! This counter example does not work! Because the described order $\appr$ does not satisfy the condition mentioned in Jacobs's paper. The condition is that the order on $FX$ should satisfy the property that for a morphism $f\c X\to Y$ the morphism $Ff\c FX\to FY$ preserves $\appr$. Probably, the only poset that has this property for $\mathbf{Id}$ is $\Delta$. If there is a counter-example, it is true for another functor.
|
|
|
|
(But still!)We have a counter-example for a symmetric relation $R$ that has a witness to be a simulation, but that morphism does not serve as a witness for $R$ to be a bisimulation. In the category of sets we assume that $F=\mathcal{P}$, and take $R=\{(1,2),(2,1),(1,3),(3,1)\}$, and $X=\{1,2,3\}$. $\alpha(x)=X$ for every $x\in X$, and $\sigma$ is defined as below:
|
|
\begin{gather*}
|
|
\sigma(w)=
|
|
\begin{cases}
|
|
(X,X) & w\neq (1,3) \\
|
|
(X,X\setminus\{2\}) & w=(1,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
In this scenario, $\sigma$ is a witness for $R$ to be a simulation, but it is not a witness for $R$ to be a bisimulation. $\sigma$ is a witness for $R$ to be a simulation since for every $w\in R$ we have $\alpha(p_1(w))\subseteq((\mathcal{P}p_1)^\ddagger(\sigma(w)))=X$. Also, for every $w\in R$, $((\mathcal{P}p_2)^\ddagger(\sigma(w)))\subseteq\alpha(p_2(w))=X$. But it is not a bisimulation, since $\alpha(p_2(1,3))=\alpha(3)=X\neq(\mathcal{P}p_2)^\ddagger(\sigma(1,3))=X\setminus\{2\}$.
|
|
%
|
|
\begin{example}
|
|
And another counter-example!!! Assume that $F=\powf $, and take $R=X\times X\setminus\{(1,3),(3,1)\}$, and $X=\{1,2,3\}$. $\alpha$ is defined as below:
|
|
\begin{gather*}
|
|
\alpha(x)=
|
|
\begin{cases}
|
|
\{1,2\} & x=1 \\
|
|
\{2,3\} & x=2\\
|
|
\{3\} & x=3
|
|
\end{cases}
|
|
\end{gather*}
|
|
And $\sigma_1$ is defined as below:
|
|
\begin{gather*}
|
|
\sigma_1(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2)\} & w=(1,2) \\
|
|
\{(2,1),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w\in\{(2,2),(3,2)\} \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_1(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w\in\{(2,2),(3,2)\} \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_1(w)=
|
|
\begin{cases}
|
|
\{2\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w\in\{(2,2),(3,2)\} \\
|
|
\{3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
\sigma'_1(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,2),(3,3),(3,2)\} & w=(3,2) \\
|
|
\{(2,3),(2,2),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma'_1(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma'_1(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
$\sigma'_1$ is not a simulation!
|
|
\begin{gather*}
|
|
\sigma''_1(w)=
|
|
\begin{cases}
|
|
\{(1,2)\} & w=(1,2) \\
|
|
\{(2,1)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
|
|
\begin{gather*}
|
|
\sigma_1\appr\sigma'_1\\
|
|
\sigma_3\appr\sigma'_1\\
|
|
\beta=\sigma'_1
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf s)^\dagger\comp\sigma_1\comp s(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(3,2),(3,3)\} & w=(3,2) \\
|
|
\{(2,2),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma_1\comp s(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma_1\comp s(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
In this scenario, $\sigma_1$ is a simulation, but it is not a bisimulation. $\sigma'_1$, $\sigma''_1$ and $(\powf s)^\dagger\comp\sigma_1\comp s$ are neither. We can not make $\sigma_1$ bigger here to make it a bisimulation as $\alpha\comp p_1(3,2)=\{3\}\subsetneq\{2,3\}=(\powf p_1)^\dagger\comp\sigma_1(3,2)$.
|
|
|
|
The following is also a simulation and not a bisimulation:
|
|
\begin{gather*}
|
|
\sigma_2(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2)\} & w=(1,2) \\
|
|
\{(2,1),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w\in\{(3,2),(3,3)\}
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_2(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w\in\{(3,2),(3,3)\}
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_2(w)=
|
|
\begin{cases}
|
|
\{2\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{3\} & w\in\{(3,2),(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
\sigma'_2(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(3,3),(3,2)\} & w=(3,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma'_2(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma'_2(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf s)^\dagger\comp\sigma_2\comp s(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(3,3),(3,2)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma_2\comp s(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{2\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma_2\comp s(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
$\sigma_2$ is a simulation, $\sigma'_2$ is a bisimulation, and $(\powf s)^\dagger\comp\sigma_2\comp s$ is neither.
|
|
The following is both a simulation and a bisimulation:
|
|
\begin{gather*}
|
|
\sigma_3(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,2),(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_3(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_3(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
The following is also a simulation and not a bisimulation:
|
|
\begin{gather*}
|
|
\sigma_4(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2),(3,2),(2,3),(3,3)\} & w=(1,2) \\
|
|
\{(1,1),(2,1),(1,2),(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(1,2),(2,2),(3,2),(2,3),(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_4(w)=
|
|
\begin{cases}
|
|
\{1,2,3\} & w=(1,2) \\
|
|
\{1,2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{1,2,3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_4(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
\sigma''_4(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,2),(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma''_4(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma''_4(w)=
|
|
\begin{cases}
|
|
\{2,3\} & w=(1,2) \\
|
|
\{1,2\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf s)^\dagger\comp\sigma_4\comp s(w)=
|
|
\begin{cases}
|
|
\{(1,1),(1,2),(2,1),(2,2),(2,3)\} & w=(1,2) \\
|
|
\{(2,1),(2,2),(2,3),(3,2),(3,3)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,3)\} & w=(2,2) \\
|
|
\{(2,1),(2,2),(2,3),(3,2),(3,3)\} & w=(2,3) \\
|
|
\{(3,2),(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma_4\comp s(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma_4\comp s(w)=
|
|
\begin{cases}
|
|
\{1,2,3\} & w=(1,2) \\
|
|
\{1,2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{1,2,3\} & w=(2,3) \\
|
|
\{2,3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
$\sigma_4$ is a simulation, $\sigma''_4$ is a bisimulation, and $(\powf s)^\dagger\comp\sigma_4\comp s$ is neither.
|
|
|
|
The following is also a simulation and not a bisimulation:
|
|
\begin{gather*}
|
|
\sigma_5(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2)\} & w=(1,2) \\
|
|
\{(2,2),(3,2)\} & w=(2,1) \\
|
|
\{(1,1),(2,1)\} & w=(1,1) \\
|
|
\{(2,2),(3,2)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,3)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_5(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_5(w)=
|
|
\begin{cases}
|
|
\{2\} & w=(1,2) \\
|
|
\{2\} & w=(2,1) \\
|
|
\{1\} & w=(1,1) \\
|
|
\{2\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
The following is also a simulation and not a bisimulation:
|
|
\begin{gather*}
|
|
\sigma_6(w)=
|
|
\begin{cases}
|
|
\{(1,2),(2,2)\} & w=(1,2) \\
|
|
\{(2,1),(3,1)\} & w=(2,1) \\
|
|
\{(1,2),(2,1)\} & w=(1,1) \\
|
|
\{(2,3),(3,3)\} & w=(2,2) \\
|
|
\{(2,3),(3,3)\} & w=(2,3) \\
|
|
\{(3,2)\} & w=(3,2) \\
|
|
\{(3,3)\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
(\powf p_1)^\dagger\comp\sigma_6(w)=
|
|
\begin{cases}
|
|
\{1,2\} & w=(1,2) \\
|
|
\{2,3\} & w=(2,1) \\
|
|
\{1,2\} & w=(1,1) \\
|
|
\{2,3\} & w=(2,2) \\
|
|
\{2,3\} & w=(2,3) \\
|
|
\{3\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
(\powf p_2)^\dagger\comp\sigma_6(w)=
|
|
\begin{cases}
|
|
\{2\} & w=(1,2) \\
|
|
\{1\} & w=(2,1) \\
|
|
\{2\} & w=(1,1) \\
|
|
\{3\} & w=(2,2) \\
|
|
\{3\} & w=(2,3) \\
|
|
\{2\} & w=(3,2) \\
|
|
\{3\} & w=(3,3)
|
|
\end{cases}
|
|
\end{gather*}
|
|
\begin{gather*}
|
|
\sigma'_2=\sigma'_5=\sigma'_6=\sigma_3=\sigma''_4
|
|
\end{gather*}
|
|
\end{example}
|
|
%
|
|
If we define $\appr$ on simulations as
|
|
\begin{gather*}
|
|
\sigma_1\appr\sigma_2\iff \forall x_1,x_2\in X, (\powf p_i)^\dagger\comp\sigma_1(x_1,x_2)\subseteq(\powf p_i)^\dagger\comp\sigma_2(x_1,x_2)
|
|
\end{gather*}
|
|
\begin{lemma}
|
|
$(Hom(R,(\powf R)^\dagger),\appr)$ is a poset.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Reflexivity and transitivity are obvious. We need to prove anti-symmetry.\\
|
|
\todo{Finish!}
|
|
\end{proof}
|
|
Then we have
|
|
\begin{equation*}
|
|
\begin{tikzcd}
|
|
{\sigma_6} && {\sigma_1} \\
|
|
{\sigma_5} & {\sigma_2} & {\sigma_3} & {\sigma_4}
|
|
\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=1-3, to=2-4]
|
|
\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=2-1, to=2-2]
|
|
\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["\sqsubseteq"{description}, draw=none, from=2-2, to=2-3]
|
|
\arrow["\sqsubseteq"{description}, draw=none, from=2-3, to=2-4]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
We recall that in the above diagram $\sigma_3$ is a bisimulation, and the rest are simulations.
|
|
\todo{So, $\sigma_3$ is a unique witness of bisimulation. How can we characterize it among all witnesses of simulation.}
|
|
%\begin{definition}\label{def:join-meet}
|
|
%We define $\join$ and $\meet$ on morphisms as follows:
|
|
%\begin{gather*}
|
|
% \forall x_1,x_2\in X,\\
|
|
% \sigma_1 \join \sigma_2 (x_1,x_2)= \sigma_1(x_1,x_2) \cup \sigma_2(x_1,x_2),\\
|
|
% \sigma_1 \meet \sigma_2 (x_1,x_2)= (\powf p_1)^\dagger\comp\sigma_1(x_1,x_2) \cap (\powf p_1)^\dagger\comp\sigma_2(x_1,x_2)\times(\powf p_2)^\dagger\comp\sigma_1(x_1,x_2) \cap (\powf p_2)^\dagger\comp\sigma_2(x_1,x_2).
|
|
%\end{gather*}
|
|
%\end{definition}
|
|
%\begin{lemma}\label{lem:proj-dist-set}
|
|
% For relations $R_1$ and $R_2$ the following equation holds:
|
|
% \begin{gather*}
|
|
% %(\powf p_i)^\dagger(R_1\cup R_2)=(\powf p_i)^\dagger(R_1)\cup(\powf p_i)^\dagger(R_2)
|
|
% (\powf p_i)(R_1\cup R_2)=(\powf p_i)(R_1)\cup(\powf p_i)(R_2)
|
|
% \end{gather*}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% We prove the lemma for the case that $i=1$. The proof is the same for $i=2$.
|
|
% Assuming $x_1\in(\powf p_1)^\dagger(R_1\cup R_2)$ then exists $x_2$ that $(x_1,x_2)\in R_1\cup R_2$, thus either $(x_1,x_2)\in R_1$ or $(x_1,x_2)\in R_2$, so we have $x_1\in(\powf p_1)^\dagger(R_1)$ or $x_1\in(\powf p_1)^\dagger(R_2)$, respectively. So, we have $x_1\in (\powf p_1)^\dagger(R_1)\cup(\powf p_1)^\dagger(R_2)$.
|
|
%
|
|
% Now, assuming that $x_1\in(\powf p_1)^\dagger(R_1)\cup(\powf p_1)^\dagger(R_2)$ either $x_1\in(\powf p_1)^\dagger(R_1)$ or $x_1\in(\powf p_1)^\dagger(R_2)$. Without loss of generality, we can assume $x_1\in(\powf p_1)^\dagger(R_j)$, where $j\in\{1,2\}$.
|
|
% Then there exists $x_2$ that $(x_1,x_2)\in R_j$, then we have $(x_1,x_2)\in R_1\cup R_2$ that gives $x_1\in(\powf p_1)^\dagger(R_1\cup R_2)$.
|
|
% % We prove the lemma for the case that $i=1$. The proof is the same for $i=2$.
|
|
% %
|
|
% % First, we prove $(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
% % Assuming $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$ then exists $y_2$ that we have either $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $(y_1,y_2)\in\sigma(x_1,x_2)$. So, we have either $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ that means that we have $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
% %
|
|
% % Now, we prove $(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$. Assuming $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ then we have:
|
|
% % \begin{itemize}
|
|
% % \item $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% % \item $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in\sigma(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% % \end{itemize} \qed
|
|
% % \todo{Rewrite the proof according to the statement!}
|
|
%\end{proof}
|
|
%\begin{lemma}
|
|
% Assuming that $\sigma_1$ and $\sigma_2$ are simulation structures of type $R\to(\powf R)^\dagger$, then $\sigma_1 \join \sigma_2$ and $\sigma_1 \meet \sigma_2$ are also simulation structures of the same type.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% Since $\sigma_1$ and $\sigma_2$ are simulation structures, for every $(x_1,x_2)\in R$, for $i\in\{1,2\}$ we have:
|
|
% \begin{gather}
|
|
% \alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma_i(x_1,x_2),\\
|
|
% (\powf p_2)^\dagger\comp\sigma_i(x_1,x_2)\subseteq\alpha(x_2).
|
|
% \end{gather}
|
|
% First, we prove the case for $\join$. Since $\alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma_i(x_1,x_2)$ we have the following:
|
|
% \begin{gather*}
|
|
% \alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma_1(x_1,x_2)\cup (\powf p_1)^\dagger\comp\sigma_2(x_1,x_2)
|
|
% \end{gather*}
|
|
% So, by~\autoref{lem:proj-dist-set} we have $\alpha(x_1)\subseteq(\powf p_1)^\dagger(\sigma_1(x_1,x_2)\cup \sigma_2(x_1,x_2))$.
|
|
% Similarly, we have $(\powf p_2)^\dagger\comp\sigma_i(x_1,x_2)\subseteq\alpha(x_2)$ that gives the following:
|
|
% \begin{gather*}
|
|
% (\powf p_2)^\dagger\comp\sigma_1(x_1,x_2)\cup (\powf p_2)^\dagger\comp\sigma_2(x_1,x_2)\subseteq\alpha(x_2)
|
|
% \end{gather*}
|
|
% So, by~\autoref{lem:proj-dist-set} we have $(\powf p_2)^\dagger(\sigma_1(x_1,x_2)\cup \sigma_2(x_1,x_2))\subseteq\alpha(x_2)$.
|
|
%
|
|
% Now, we prove the case for $\meet$. For $\meet$ unlike $\join$ we need to prove that $\sigma_1\meet\sigma_2(x_1,x_2)\in(\powf R)^\dagger$. To achieve this, we need to show that assuming $\pi_1$, $\pi_2$ are projections of $\sigma_1\meet\sigma_2(x_1,x_2)$, then for $j\in\{1,2\}$ we have $\pi_j\comp(\sigma_1\meet\sigma_2)(x_1,x_2)\subseteq\powf p_j(R)$. Since $(\powf p_j)^\dagger\comp\sigma_i(x_1,x_2)\subseteq\powf p_j(R)$, we have $\pi_j\comp(\sigma_1 \meet \sigma_2) (x_1,x_2)\subseteq\powf p_j(R)$, so we have $\sigma_1\meet\sigma_2(x_1,x_2)\in(\powf R)^\dagger$, meaning that $\pi_j\comp(\sigma_1\meet\sigma_2)(x_1,x_2)=(\powf p_j)^\dagger\comp(\sigma_1\meet\sigma_2)(x_1,x_2)$.\ppnote{The last part of the proof is necessary because the type of the codomain of the definition of $\meet$ is not $(\powf R)^\dagger$, but it is $\powf X\times\powf X$. Perhaps the epi-mono factorization must be used to cope with this in the abstract case.}
|
|
%
|
|
% For $j\in\{1,2\}$ we have
|
|
% \begin{gather}\label{eq:proj-meet}
|
|
% (\powf p_j)^\dagger\comp(\sigma_1 \meet \sigma_2 (x_1,x_2))=(\powf p_j)^\dagger\comp\sigma_1(x_1,x_2) \cap (\powf p_j)^\dagger\comp\sigma_2(x_1,x_2).
|
|
% \end{gather}
|
|
% Since $\alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma_i(x_1,x_2)$, we have
|
|
% \begin{gather*}
|
|
% \alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma_1(x_1,x_2)\cap (\powf p_1)^\dagger\comp\sigma_2(x_1,x_2),
|
|
% \end{gather*}
|
|
% so by~\eqref{eq:proj-meet} we have $\alpha(x_1)\subseteq(\powf p_1)^\dagger\comp(\sigma_1 \meet \sigma_2 (x_1,x_2))$. Similarly, since $(\powf p_2)^\dagger\comp\sigma_i(x_1,x_2)\subseteq\alpha(x_2)$, we have
|
|
% \begin{gather*}
|
|
% (\powf p_2)^\dagger\comp\sigma_1(x_1,x_2)\cap (\powf p_2)^\dagger\comp\sigma_2(x_1,x_2)\subseteq\alpha(x_2),
|
|
% \end{gather*}
|
|
% so by~\eqref{eq:proj-meet} we have $(\powf p_2)^\dagger\comp(\sigma_1 \meet \sigma_2 (x_1,x_2))\subseteq\alpha(x_2)$.\qed
|
|
%\end{proof}
|
|
%\begin{lemma}\label{lem:sim-opsim-inc}
|
|
% Assuming that $\sigma\c R\to(\powf R)^\dagger$ is a simulation structure, and $R$ is symmetric, then for all $(x_1,x_2)\in R$ we have:
|
|
% \begin{enumerate}
|
|
% \item $(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)\subseteq (\powf p_1)^\dagger\comp\sigma(x_1,x_2)$
|
|
% \item $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq (\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$
|
|
% \end{enumerate}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% We prove the second clause.
|
|
% By~\eqref{eq:diag-lax-sim} for every $(x_1,x_2)\in R$ we have
|
|
% \begin{gather*}
|
|
% \alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma(x_1,x_2),\\
|
|
% (\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq\alpha(x_2).
|
|
% \end{gather*}
|
|
% From $\alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ since $R$ is symmetric we get $\alpha(x_2)\subseteq(\powf p_1)^\dagger\comp\sigma(x_2,x_1)$, where
|
|
% \begin{gather*}
|
|
% (\powf p_1)^\dagger\comp\sigma(x_2,x_1)=(\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2).
|
|
% \end{gather*}
|
|
% So, from $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq\alpha(x_2)$ we have $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq (\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$. Similarly, we can get the other inequation.\qed
|
|
%\end{proof}
|
|
%\begin{lemma}\label{lem:sim-bisim-inc}
|
|
% Assuming that $\sigma\c R\to(\powf R)^\dagger$ is a simulation structure, and $\beta\c R\to(\powf R)^\dagger$ is a bisimulation structure,
|
|
% \begin{enumerate}
|
|
% \item if $\sigma\appr\beta$ then we have:
|
|
% \begin{gather*}
|
|
% \alpha(x_1)=(\powf p_1)^\dagger\comp\sigma(x_1,x_2),
|
|
% \end{gather*}
|
|
% and if $R$ is symmetric we have
|
|
% \begin{gather*}
|
|
% (\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)=\alpha(x_2).
|
|
% \end{gather*}
|
|
% \item if $\beta\appr\sigma$ then we have:
|
|
% \begin{gather*}
|
|
% (\powf p_2)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_2)
|
|
% \end{gather*}
|
|
% and if $R$ is symmetric we have
|
|
% \begin{gather*}
|
|
% \alpha(x_1)=(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2).
|
|
% \end{gather*}
|
|
% \end{enumerate}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% \begin{enumerate}
|
|
% \item Since $\sigma$ is a simulation structure for an arbitrary $(x_1,x_2)\in R$ we have $\alpha(x_1)\subseteq(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$. Since $\sigma\appr\beta$ we have $(\powf p_1)\comp\sigma(x_1,x_2)\subseteq(\powf p_1)\comp\beta(x_1,x_2)$, while $(\powf p_1)\comp\beta(x_1,x_2)=\alpha(x_1)$ by definition of bisimulation. So we have $\alpha(x_1)=(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$. Then because of the symmetry of $R$ the second clause is easily achievable by using the equations in~\eqref{eq:diag-sym-rel}.
|
|
% \item This clause can be proven similar to (1).
|
|
% \end{enumerate}\qed
|
|
%\end{proof}
|
|
%\begin{prop}
|
|
% Assuming that $\sigma\c R\to(\powf R)^\dagger$ is a simulation structure, and $\beta\c R\to(\powf R)^\dagger$ is a bisimulation structure,
|
|
% \begin{enumerate}
|
|
% \item if $\sigma\appr\beta$ then we have:
|
|
% \begin{gather*}
|
|
% \beta=\sigma\join ((\powf s)^\dagger\comp\sigma\comp s)
|
|
% \end{gather*}
|
|
% \item if $\beta\appr\sigma$ then we have:
|
|
% \begin{gather*}
|
|
% \beta=\sigma\meet((\powf s)^\dagger\comp\sigma\comp s)
|
|
% \end{gather*}
|
|
% \end{enumerate}
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% 1. We need to prove that $\sigma\join((\powf s)^\dagger\comp\sigma\comp s)$ is the bisimulation structure.
|
|
% By~\autoref{lem:sim-opsim-inc}.(1), for every $(x_1,x_2)\in R$, we have $(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$, and by~\autoref{lem:sim-bisim-inc}.(1), we have $(\powf p_1)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_1)$. So, we have $(\powf p_1)^\dagger\comp\sigma\join(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)=\alpha(x_1)$, then by~\autoref{lem:proj-dist-set} we have $(\powf p_1)^\dagger\comp(\sigma\join((\powf s)^\dagger\comp\sigma\comp s))(x_1,x_2)=\alpha(x_1)$.
|
|
%
|
|
% Also, by~\autoref{lem:sim-bisim-inc}.(1) we have $(\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_2)$. So, since we already have $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq\alpha(x_2)$ then by~\autoref{lem:proj-dist-set} we have $(\powf p_2)^\dagger\comp(\sigma\join((\powf s)^\dagger\comp\sigma\comp s))(x_1,x_2)=\alpha(x_2)$.
|
|
%
|
|
% 2. We need to prove that $\sigma\meet((\powf s)^\dagger\comp\sigma\comp s)$ is the bisimulation structure.
|
|
% For $i\in\{1,2\}$, for every $(x_1,x_2)\in R$, we have:
|
|
% {\small
|
|
% \begin{align*}
|
|
% &(\powf p_i)^\dagger\comp(\sigma\meet((\powf s)^\dagger\comp\sigma\comp s))(x_1,x_2)\\
|
|
% &=(\powf p_i)^\dagger\comp(((\powf p_1)^\dagger\comp\sigma(x_1,x_2) \cap (\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2))\times((\powf p_2)^\dagger\comp\sigma(x_1,x_2) \cap (\powf p_2)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2)))\\
|
|
% &=(\powf p_i)^\dagger\comp\sigma(x_1,x_2) \cap (\powf p_i)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2)
|
|
% \end{align*}
|
|
% }
|
|
%
|
|
% By~\autoref{lem:sim-opsim-inc}.(1), $(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2)\subseteq (\powf p_1)^\dagger\comp\sigma(x_1,x_2)$, and by~\autoref{lem:sim-bisim-inc}.(2) we have $(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2)=\alpha(x_1)$, so we have $(\powf p_1)^\dagger\comp(\sigma\meet((\powf s)^\dagger\comp\sigma\comp s))(x_1,x_2)=\alpha(x_1)$.
|
|
%
|
|
% Also, by~\autoref{lem:sim-bisim-inc}.(2) we have $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_2)$, so, since by~\autoref{lem:sim-opsim-inc}.(2), we have $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq (\powf p_2)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$, so we have $(\powf p_2)^\dagger\comp(\sigma\meet((\powf s)^\dagger\comp\sigma\comp s)(x_1,x_2))=\alpha(x_2)$.\qed
|
|
%\end{proof}
|
|
%\begin{cor}
|
|
% Assuming that $R$ is a symmetric relation, and $S\neq\emptyset$ is the set of all simulation structures of the type $R\to (\powf R)^\dagger$, then if the bisimulation morphism exists, it is equal with the following morphism:
|
|
% \begin{gather*}
|
|
% (\bigjoin_{\sigma\in S}\sigma)\meet(\powf s)^\dagger\comp(\bigjoin_{\sigma\in S}\sigma)\comp s
|
|
% \end{gather*}
|
|
%\end{cor}
|
|
%
|
|
%\begin{lemma}
|
|
% For every $S\in \powf R$,
|
|
% \begin{gather*}
|
|
% ((\powf p_1)(S),(\powf p_2)(S))\in(\powf R)^\dagger\Leftrightarrow(\powf p_1)(S)\subseteq(\powf p_1)(R),(\powf p_2)(S)\subseteq(\powf p_2)(R)
|
|
% \end{gather*}
|
|
%\end{lemma}
|
|
%\begin{lemma}\label{prop:alph-prod}
|
|
% Assuming that $R$ is a symmetric relation, and $S\neq\emptyset$ is the set of all simulation structures of the type $R\to (\powf R)^\dagger$, then there there exists a simulation structure $\sigma\in S$ that for every $(x_1,x_2)$, $(\powf p_1)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_1)$.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% Since $S\neq\emptyset$ there exists $\delta\in S$. We define $\sigma$ for every $(x_1,x_2)$ as the following:
|
|
% \begin{gather*}
|
|
% \sigma(x_1,x_2)=(\alpha(x_1), (\powf p_2)^\dagger\comp\delta(x_1,x_2))
|
|
% \end{gather*}
|
|
% We have $\sigma(x_1,x_2)\in(\powf R)^\dagger$, as $\alpha(x_1)\subseteq\powf p_1(R)$ and $(\powf p_2)^\dagger\comp\delta(x_1,x_2)\subseteq\powf p_2(R)$ are inherited from $\delta$ being a simulation structure.
|
|
% Also, it obviously is a simulation as $(\powf p_1)^\dagger\comp\sigma(x_1,x_2)=\alpha(x_1)$ and $(\powf p_2)^\dagger\comp\sigma(x_1,x_2)\subseteq\alpha(x_2)$ as $(\powf p_2)^\dagger\comp\delta(x_1,x_2)\subseteq\alpha(x_2)$.
|
|
%\end{proof}
|
|
%\begin{prop}\label{prop:sym-rel-bisim}
|
|
% Assuming that $R$ is a symmetric relation, and $S\neq\emptyset$ is the set of all simulation structures of the type $R\to (\powf R)^\dagger$, then the following morphism is the bisimulation structure:
|
|
% \begin{gather*}
|
|
% (\bigmeet_{\sigma\in S}\sigma)\join(\powf s)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)\comp s
|
|
% \end{gather*}
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% For every $(x_1,x_2)\in R$ we have
|
|
% \begin{gather*}
|
|
% (\powf p_1)^\dagger\comp((\bigmeet_{\sigma\in S}\sigma)\join(\powf s)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)\comp s)(x_1,x_2)=
|
|
% (\powf p_1)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)(x_1,x_2),
|
|
% \end{gather*}
|
|
% and
|
|
% \begin{gather*}
|
|
% (\powf p_2)^\dagger\comp((\bigmeet_{\sigma\in S}\sigma)\join(\powf s)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)\comp s)(x_1,x_2)=
|
|
% (\powf p_2)^\dagger\comp((\powf s)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)\comp s)(x_1,x_2).
|
|
% \end{gather*}
|
|
% By~\autoref{prop:alph-prod} there exists a simulation $\delta\in S$ for which we have $(\powf p_1)^\dagger\comp\delta(x_1,x_2)=\alpha(x_1)$. So, $(\powf p_1)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)(x_1,x_2)=\alpha(x_1)$. Then by the equations in~\eqref{eq:diag-sym-rel} we also get $(\powf p_2)^\dagger\comp((\powf s)^\dagger\comp(\bigmeet_{\sigma\in S}\sigma)\comp s)(x_1,x_2)=\alpha(x_2)$.\qed
|
|
%\end{proof}
|
|
%\section{Symmetric Simulation in Quantaloids}
|
|
%We generalize~\autoref{prop:sym-rel-bisim} in regular quantaloids. A quantaloid is a category enriched with suplattices.
|
|
%Abstractly, first we define an operation that we need on morphisms that takes two simulation witnesses of type $R\to(FR)^\dagger$ to a morphism of type $R\to FX\times FX$:
|
|
%\begin{gather*}
|
|
% \sigma_1\simeet\sigma_2=(Fp_1)^\dagger\comp\sigma_1\meet(Fp_1)^\dagger\comp\sigma_2\times(Fp_2)^\dagger\comp\sigma_1\meet(Fp_2)^\dagger\comp\sigma_2
|
|
%\end{gather*}
|
|
%\begin{lemma}\label{prop:alph-prod-abs}
|
|
% Assuming that $R$ is a symmetric relation, and $S\neq\emptyset$ is the set of all simulation witnesses of the type $R\to (FR)^\dagger$, then there exists a simulation witness $\sigma\in S$ that, $(Fp_1)^\dagger\comp\sigma=\alpha\comp p_1$.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% Since $S\neq\emptyset$ there exists $\delta\in S$. We define $\sigma$ as the following:
|
|
% \begin{gather*}
|
|
% \sigma=\brks{(\alpha\comp p_1), (Fp_2)^\dagger\comp\delta}
|
|
% \end{gather*}
|
|
% We have $(Fp_1)^\dagger\comp\sigma$.
|
|
%\end{proof}
|
|
\begin{example}\label{ex:sym-sim-bisim-count}
|
|
Assuming that the functor is the powerset endofunctor over the category of sets and injective maps. Let us call this functor $\powfi$. For every $X$, we define the order on $\powfi X$ as $A\appr B$ whenever $|A|\leq |B|$, where $|A|$ and $|B|$ are just cardinalities of $|A|$ and $|B|$ respectively. This is a preorder. Then we define the order over every $\Hom(X,\powfi Y)$ pointwise. We have chosen the category of sets with injective maps because we could not define a functor from $\Set$ to $\preord$ with the mentioned ordering.\\
|
|
We take $R=\{(1,1),(1,2),(2,1),(2,2)\}$, and $X=\{1,2,3\}$. $\alpha$ is defined as below:
|
|
\begin{gather*}
|
|
\alpha(x)=
|
|
\begin{cases}
|
|
\{1,2\} & x=1 \\
|
|
\{2,3\} & x=2\\
|
|
\{3\} & x=3
|
|
\end{cases}\qquad
|
|
\begin{tikzpicture}[scale=0.1]
|
|
\tikzstyle{every node}+=[inner sep=0pt]
|
|
\draw [black] (25.6,-20.9) circle (3);
|
|
\draw (25.6,-20.9) node {$1$};
|
|
\draw [black] (44.9,-20.9) circle (3);
|
|
\draw (44.9,-20.9) node {$2$};
|
|
\draw [black] (34.5,-32) circle (3);
|
|
\draw (34.5,-32) node {$3$};
|
|
\draw [black] (24.277,-18.22) arc (234:-54:2.25);
|
|
\fill [black] (26.92,-18.22) -- (27.8,-17.87) -- (26.99,-17.28);
|
|
\draw [black] (43.577,-18.22) arc (234:-54:2.25);
|
|
\fill [black] (46.22,-18.22) -- (47.1,-17.87) -- (46.29,-17.28);
|
|
\draw [black] (33.177,-29.32) arc (234:-54:2.25);
|
|
\fill [black] (35.82,-29.32) -- (36.7,-28.97) -- (35.89,-28.38);
|
|
\draw [black] (42.85,-23.09) -- (36.55,-29.81);
|
|
\fill [black] (36.55,-29.81) -- (37.46,-29.57) -- (36.73,-28.89);
|
|
\draw [black] (28.6,-20.9) -- (41.9,-20.9);
|
|
\fill [black] (41.9,-20.9) -- (41.1,-20.4) -- (41.1,-21.4);
|
|
\end{tikzpicture}
|
|
\end{gather*}
|
|
$\sigma$ is defined as below:
|
|
\begin{gather*}
|
|
\forall w\in R,\quad\sigma(w)=\{(1,2),(2,1)\}
|
|
\end{gather*}
|
|
$R$ is symmetric, and $\sigma$ is a witness for $R$ to be an AM-simulation, but $R$ is not a bisimulation in the traditional sense because $(2,1)\in R$, and $2\to 3$, but $(3,1)$ or $(3,2)$ are not in $R$. It is easy to see that it is not an AM-bisimulation because we can not define a function that can serve as an evidence for it as $3$ does not appear in any pair in $R$, while it exists in $\alpha(2)$.
|
|
|
|
This counter-example also works as a counter-example for Hughes-Jacobs definition of simulation. Actually, $\appr\comp(FR)^\dagger\comp\appr=\powf X\times \powf X$, so $R\subseteq( \alpha\comp\appr\comp(FR)^\dagger\comp\appr\comp\alpha)$ that means that $R$ is a simulation. Worth noting that they claim that their setting works for an arbitrary category. Worth noting that Hughes and Jacobs give their definition in $\Set$, and they claim that it is easy to generalize to other categories.
|
|
|
|
The mentioned ordering is not liftable. Assuming $h\in\Hom(\nats,\powfi \nats)$, $g\c \nats\to \nats$, and $k\in\Hom(\nats,\powfi \nats)$, and they are defined for every $n$ in $\nats$ as $h(n)=\{2\times n\}$, $g(n)=3\times n$, and $k(n)=\{n\}$, then $|h(n)|=|\powfi g(k(n))|=1$ that means $h\appr\powfi g\comp k$ is satisfied, but there is no $k'$ that $h=\powfi g\comp k$ because we can never have $h(1)=\powfi g\comp k'(1)$, as assuming $k'(1)=\{n\}$, and $n$ must be a natural number, then we should have $2=3\times n$ that is impossible.
|
|
|
|
It is the same for HJ-simulation and HJ-bisimulation. We define $\sigma^\dagger\c R\to (\powf R)^\dagger$ as:
|
|
\begin{gather*}
|
|
\forall w\in R,\quad\sigma^\dagger(w)=(\{1,2\},\{1,2\})
|
|
\end{gather*}
|
|
Indeed, $\sigma^\dagger$ is a witness for $R$ to be an HJ-simulation, but it is not a witness for $R$ to be an HJ-bisimulation. Similar to the case for AM-bisimulation, we can not have a witness for $R$ to be an HJ-bisimulation.
|
|
\end{example}
|
|
|
|
\begin{example}
|
|
In~\autoref{ex:sym-sim-bisim-count} we gave a counter-example with an ordering that gives preorders. Now, we slightly change the ordering so that it gives posets. We define $\appr$ as follows:
|
|
\begin{gather*}
|
|
A\appr B \qquad\iff\qquad A=B \quad or\quad |A|<|B|
|
|
\end{gather*}
|
|
For the defined $\alpha$ and $R$ in~\autoref{ex:sym-sim-bisim-count}, we define $\sigma\c R\to \powfi R$ as follows as a witness for $R$ to be an AM-simulation:
|
|
\begin{gather*}
|
|
\forall w\in R,\quad\sigma(w)=\{(1,1),(2,1),(3,1)\}
|
|
\end{gather*}
|
|
Additionally, we define $\sigma^\dagger$ as follows as a witness for $R$ to be an HJ-simulation:
|
|
\begin{gather*}
|
|
\forall w\in R,\quad\sigma^\dagger(w)=(\{1,2,3\},\{1\})
|
|
\end{gather*}
|
|
However, $R$ is not an AM-bisimulation nor an HJ-bisimulation regardless of the choice for the order structure.
|
|
\end{example}
|
|
|
|
\subsection{From Symmetric Simulation To Bisimulation (Aczel-Mendler)}
|
|
%\begin{lemma}\label{lem:sim-opsim-inc1}\ppnote{Actually, this lemma holds for every functor in an arbitrary category.}
|
|
% Assuming that $\sigma\c R\to\powf R$ is witness for a symmetric relation $R$ to be an AM simulation on $\powf$-coalgebra $(X,\alpha)$, then for all $(x_1,x_2)\in R$ we have:
|
|
% \begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
% \item $\powf p_1\comp\powf s\comp\sigma\comp s(x_1,x_2)\subseteq \powf p_1\comp\sigma(x_1,x_2)$\label{item:sim-opsim-inc:I1}
|
|
% \item $\powf p_2\comp\sigma(x_1,x_2)\subseteq \powf p_2\comp\powf s\comp\sigma\comp s(x_1,x_2)$\label{item:sim-opsim-inc:II1}
|
|
% \end{enumerate}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% By~\eqref{eq:diag-lax-sim} for every $(x_1,x_2)\in R$ we have
|
|
% \begin{align}
|
|
% \alpha(x_1)\subseteq&\;\powf p_1\comp\sigma(x_1,x_2),\label{eq:alpha_x_11}\\
|
|
% \powf p_2\comp\sigma(x_1,x_2)\subseteq&\;\alpha(x_2).\label{eq:alpha_x_21}
|
|
% \end{align}
|
|
% (I): Since $R$ is symmetric $(x_2,x_1)\in R$, so from \eqref{eq:alpha_x_21} we get $\powf p_2\comp\sigma(x_2,x_1)\subseteq\alpha(x_1)$.
|
|
% Therefore:
|
|
% \begin{align*}
|
|
% \powf p_1\comp\powf s\comp\sigma\comp s(x_1,x_2)=&\; \powf p_2\comp\sigma\comp s(x_1,x_2)\\
|
|
% =&\; \powf p_2\comp\sigma (x_2,x_1)\\
|
|
% \subseteq&\;\alpha(x_1)\\
|
|
% \subseteq&\; \powf p_1\comp\sigma(x_1,x_2) & \by{\eqref{eq:alpha_x_11}}
|
|
% \end{align*}
|
|
% %
|
|
% (II): Analogously, from~\eqref{eq:alpha_x_11} by the symmetry of $R$ we have $(x_2,x_1)\in R$, so we get $\alpha(x_2)\subseteq\powf p_1\comp\sigma(x_2,x_1)$.
|
|
% Therefore:
|
|
% \begin{align*}
|
|
% \powf p_2\comp\sigma(x_1,x_2)\subseteq&\; \alpha(x_2) &\by{\eqref{eq:alpha_x_21}}\\
|
|
% \subseteq&\; \powf p_1\comp\sigma(x_2,x_1) \\
|
|
% =&\; \powf p_1\comp\sigma\comp s(x_1,x_2) \\
|
|
% =&\;\powf p_2\comp\powf s\comp\sigma\comp s(x_1,x_2).&
|
|
% \end{align*}
|
|
% \qed
|
|
%\end{proof}
|
|
\begin{lemma}\label{lem:sim-opsim-inc}
|
|
In a category $\BC$, assuming that $F$ has a natural order structure $\appr$, and $\sigma\c R\to F R$ is witness for a symmetric relation $R$ to be an AM-simulation on $F$-coalgebra $(X,\alpha)$, then we have:
|
|
\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
\item $F p_1\comp F s\comp\sigma\comp s\appr F p_1\comp\sigma$\label{item:sim-opsim-inc:I}
|
|
\item $F p_2\comp\sigma\appr F p_2\comp F s\comp\sigma\comp s$\label{item:sim-opsim-inc:II}
|
|
\end{enumerate}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
By~\eqref{eq:diag-lax-sim} for every $(x_1,x_2)\in R$ we have
|
|
\begin{align}
|
|
\alpha\comp p_1\appr&\;F p_1\comp\sigma,\label{eq:alpha_x_1}\\
|
|
F p_2\comp\sigma\appr&\;\alpha\comp p_2.\label{eq:alpha_x_2}
|
|
\end{align}
|
|
(I): Since $R$ is symmetric and $\appr$ is a natural order structure, from \eqref{eq:alpha_x_2} we get $F p_2\comp\sigma\comp s\appr\alpha\comp p_2\comp s$.
|
|
Therefore:
|
|
\begin{align*}
|
|
F p_1\comp F s\comp\sigma\comp s=&\; F p_2\comp\sigma\comp s\\
|
|
\appr&\;\alpha\comp p_2\comp s\\
|
|
=&\;\alpha\comp p_1\\
|
|
\appr&\; F p_1\comp\sigma & \by{\eqref{eq:alpha_x_1}}
|
|
\end{align*}
|
|
%
|
|
(II): Analogously, from~\eqref{eq:alpha_x_1} since $R$ is symmetric and $\appr$ is a natural order structure, we get $\alpha\comp p_1\comp s\appr F p_1\comp\sigma\comp s$.
|
|
Therefore:
|
|
\begin{align*}
|
|
F p_2\comp\sigma\appr&\; \alpha\comp p_2 &\by{\eqref{eq:alpha_x_2}}\\
|
|
=&\; \alpha\comp p_1\comp s\\
|
|
\appr&\; F p_1\comp\sigma\comp s \\
|
|
=&\;F p_2\comp F s\comp\sigma\comp s.&
|
|
\end{align*}
|
|
\qed
|
|
\end{proof}
|
|
We define $\join$ on each $\Hom(X,\powf Y)$ for every sets $X$ and $Y$:
|
|
\begin{gather*}
|
|
\forall x_1,x_2\in X,\\
|
|
\sigma_1 \join \sigma_2 (x_1,x_2)= \sigma_1(x_1,x_2) \cup \sigma_2(x_1,x_2).
|
|
\end{gather*}
|
|
%\begin{lemma}
|
|
% Assuming that $\sigma_1$ and $\sigma_2$ are witnesses that $R$ is an Aczel-Mendler simulation from a coalgebra $(X,\alpha)$ to another coalgebra $(Y,\beta)$, then $\sigma_1\join\sigma_2$ is also a witness that $R$ is an Aczel-Mendler simulation.
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% \todo{Finish.}
|
|
%\end{proof}
|
|
\subsection{Powerset Functor}
|
|
%\begin{lemma}\label{lem:proj-dist-set}
|
|
% For relations $R_1$ and $R_2$ the following equation holds:
|
|
% \begin{gather*}
|
|
% %(\powf p_i)^\dagger(R_1\cup R_2)=(\powf p_i)^\dagger(R_1)\cup(\powf p_i)^\dagger(R_2)
|
|
% \powf p_i(R_1\cup R_2)=\powf p_i(R_1)\cup(\powf p_i)(R_2)
|
|
% \end{gather*}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% We prove the lemma for the case that $i=1$. The proof is the same for $i=2$.
|
|
% Assuming $x_1\in\powf p_1(R_1\cup R_2)$ then exists $x_2$ that $(x_1,x_2)\in R_1\cup R_2$, thus either $(x_1,x_2)\in R_1$ or $(x_1,x_2)\in R_2$, so we have $x_1\in\powf p_1R_1$ or $x_1\in\powf p_1R_2$, respectively. So, we have $x_1\in \powf p_1R_1\cup\powf p_1R_2$.
|
|
%
|
|
% Now, assuming that $x_1\in\powf p_1R_1\cup\powf p_1R_2$ either $x_1\in\powf p_1R_1$ or $x_1\in\powf p_1R_2$. Without loss of generality, we can assume $x_1\in\powf p_1R_j$, where $j\in\{1,2\}$.
|
|
% Then there exists $x_2$ that $(x_1,x_2)\in R_j$, then we have $(x_1,x_2)\in R_1\cup R_2$ that gives $x_1\in\powf p_1(R_1\cup R_2)$.\qed
|
|
% % We prove the lemma for the case that $i=1$. The proof is the same for $i=2$.
|
|
% %
|
|
% % First, we prove $(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
% % Assuming $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$ then exists $y_2$ that we have either $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $(y_1,y_2)\in\sigma(x_1,x_2)$. So, we have either $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ that means that we have $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
% %
|
|
% % Now, we prove $(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$. Assuming $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ then we have:
|
|
% % \begin{itemize}
|
|
% % \item $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% % \item $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in\sigma(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% % \end{itemize} \qed
|
|
% % \todo{Rewrite the proof according to the statement!}
|
|
%\end{proof}
|
|
%%\begin{rem}
|
|
%% Assuming that $\sigma_1$ and $\sigma_2$ are witnesses that $R$ is an Aczel-Mendler simulation from a coalgebra $(X,\alpha)$ to another coalgebra $(Y,\beta)$, then $\sigma_1\meet\sigma_2$ is not necessarily a witness that $R$ is an Aczel-Mendler simulation.
|
|
%%\end{rem}
|
|
%Since $\subseteq$ is a liftable order (\autoref{def:liftable-ord}), we have the following lemma. The liftability is not used in the proof, but if $\subseteq$ was not liftable, perhaps we could not prove this.
|
|
%\begin{prop}\label{prop:alph-prod}
|
|
% Assuming that $R$ is a relation, and $\sigma\c R\to\powf R$ is a witness for $R$ to be an AM simulation, then there exists $\sigma'\c R\to\powf R$ that is another witness for $R$ to be an AM simulation, such that $\powf p_1\comp\sigma'=\alpha\comp p_1$.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% We define $\sigma'(x_1,x_2)=\{(x'_1,x'_2)\mid x'_1\in\alpha\comp p_1(x_1,x_2)\;,\;(x'_1,x'_2)\in\sigma(x_1,x_2)\}$. We have $\sigma'\subseteq\sigma$ that gives $\powf p_2\comp\sigma'\subseteq\powf p_2\comp\sigma$. Additionally, we have $\powf p_1\comp\sigma'\subseteq\alpha\comp p_1$.
|
|
% Furthermore, if $x'_1\in\alpha\comp p_1(x_1,x_2)$, since $\alpha\comp p_1\subseteq \powf p_1\comp\sigma$ then $x'_1\in\powf p_1\comp\sigma(x_1,x_2)$. Let $(x'_1,x'_2)\in\sigma(x_1,x_2)$. By definition of $\sigma'$, we have $(x'_1,x'_2)\in\sigma'(x_1,x_2)$, so $x'_1\in\powf p_1\comp\sigma'(x_1,x_2)$ that means $\alpha\comp p_1\subseteq \powf p_1\comp\sigma'$ as well. So, $\sigma'$ is another witness for $R$ to be an AM simulation, and we have $\alpha\comp p_1=\powf p_1\comp\sigma'$.
|
|
% \qed
|
|
%\end{proof}
|
|
%A more abstract version of the following proposition is given by Dubut:
|
|
%\begin{prop}\label{prop:alph-prod-dubut}
|
|
% Assuming that $R$ is a relation, and $\sigma\c R\to FR$ is a witness for $R$ to be an AM-simulation, then there exists $\sigma'\c R\to FR$ that is another witness for $R$ to be an AM-simulation, such that $Fp_1\comp\sigma'=\alpha\comp p_1$.
|
|
%\end{prop}\qed
|
|
%Now, we prove our main statement.
|
|
%\begin{prop}\label{prop:sym-rel-bisim}
|
|
% Assuming that $R$ is a symmetric relation, and $\sigma\c R\to \powf R$ is a witness for $R$ to be a simulation, for which $\powf p_1\comp\sigma=\alpha\comp p_1$, then the following morphism is a witness for $R$ to be a bisimulation:
|
|
% \begin{gather*}
|
|
% \sigma\join(\powf s\comp\sigma\comp s)
|
|
% \end{gather*}
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% For every $(x_1,x_2)\in R$ by~\autoref{lem:proj-dist-set} and~\autoref{lem:sim-opsim-inc}.\ref{item:sim-opsim-inc:I} we have
|
|
% \begin{gather*}
|
|
% \powf p_1\comp(\sigma\join(\powf s\comp\sigma\comp s))(x_1,x_2)=
|
|
% \powf p_1\comp\sigma(x_1,x_2).
|
|
% \end{gather*}
|
|
%% and by~\autoref{prop:alph-prod},
|
|
% Recall that $\powf p_1\comp\sigma(x_1,x_2)=\alpha(x_1)$. By~\autoref{lem:proj-dist-set} and~\autoref{lem:sim-opsim-inc}.\ref{item:sim-opsim-inc:II} we have
|
|
% \begin{gather*}
|
|
% \powf p_2\comp(\sigma\join(\powf s\comp\sigma\comp s))(x_1,x_2)=
|
|
% \powf p_2\comp(\powf s\comp\sigma\comp s)(x_1,x_2).
|
|
% \end{gather*}
|
|
% Since $\powf p_1\comp\sigma=\alpha\comp p_1$ by precomposing $s$ to the both sides of the equation we get $\powf p_2\comp(\powf s\comp\sigma\comp s)=\alpha\comp p_2$. So, $\sigma\join(\powf s\comp\sigma\comp s)$ is a witness for $R$ to be an AM bisimulation.\qed
|
|
%\end{proof}
|
|
%\begin{cor}
|
|
% Considering~\autoref{prop:alph-prod}, assuming that $R$ is a symmetric relation and it is an AM simulation, then $R$ is an AM bisimulation as well.
|
|
%\end{cor}
|
|
%Now, we make the proof more abstract. We prove the statement for set-functors of the form $\powf F$, where $F$ is an arbitrary set-functor, and $\powf$ is the powerset functor.
|
|
We prove the stronger statement for $\powf$, where $F$ is an arbitrary endofunctor on $\Set$. The ordering that we consider on this functor is the set inclusion. We recall the following lemma:
|
|
\begin{lemma}\label{lem:func-dist-set}
|
|
For sets $X$ and $Y$ in $\powf A$, and a function $f\c A\to B$ the following equation holds:
|
|
\begin{gather*}
|
|
%(\powf p_i)^\dagger(R_1\cup R_2)=(\powf p_i)^\dagger(R_1)\cup(\powf p_i)^\dagger(R_2)
|
|
\powf f(X_1\cup X_2)=\powf f(X_1)\cup \powf f(X_2)
|
|
\end{gather*}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
Assuming $b\in\powf f(X_1\cup X_2)$ then exists $z$ that $z\in X_1\cup X_2$ and $f(z)=b$, thus either $z\in X_1$ or $z\in X_2$, so we have $b\in\powf f(X_1)$ or $b\in\powf f(X_2)$, respectively. So, we have $b\in \powf f(X_1)\cup\powf f(X_2)$.
|
|
|
|
Now, assuming that $b\in\powf f(X_1)\cup\powf f(X_2)$ then we either have $b\in\powf f(X_1)$ or $b\in\powf f(X_2)$. Without loss of generality, we assume $b\in\powf f(X_j)$, where $j\in\{1,2\}$.
|
|
Then there exists $z$ that $z\in X_j$ and $f(z)=b$, so we have $z\in X_1\cup X_2$ that gives $b\in\powf f(X_1\cup X_2)$.\qed
|
|
% We prove the lemma for the case that $i=1$. The proof is the same for $i=2$.
|
|
%
|
|
% First, we prove $(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
% Assuming $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$ then exists $y_2$ that we have either $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $(y_1,y_2)\in\sigma(x_1,x_2)$. So, we have either $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$ or $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ that means that we have $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$.
|
|
%
|
|
% Now, we prove $(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)\subseteq(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$. Assuming $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s\join(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$ then we have:
|
|
% \begin{itemize}
|
|
% \item $y_1\in(\powf p_1)^\dagger\comp(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% \item $y_1\in(\powf p_1)^\dagger\comp\sigma(x_1,x_2)$: Then there exists $y_2$ such that $(y_1,y_2)\in\sigma(x_1,x_2)$. So, $(y_1,y_2)\in(\powf s)^\dagger\comp\sigma\comp s\join\sigma(x_1,x_2)$, thus $y_1\in(\powf p_1)^\dagger\comp((\powf s)^\dagger\comp\sigma\comp s\join\sigma)(x_1,x_2)$.
|
|
% \end{itemize} \qed
|
|
% \todo{Rewrite the proof according to the statement!}
|
|
\end{proof}
|
|
The following statement is proven by Dubut:
|
|
\begin{prop}\label{prop:alph-prod-dubut-pf}
|
|
Assuming that $R$ is a relation, and $\sigma\c R\to FR$ is a witness for $R$ to be an AM-simulation, then there exists $\sigma'\c R\to FR$ that is another witness for $R$ to be an AM-simulation, such that $Fp_1\comp\sigma'=\alpha\comp p_1$.
|
|
\end{prop}\qed
|
|
For arbitrary sets $X$ and $Y$, and functions $f,g\in\Hom(X,\powf FY)$, we define $f\join g$ as follows:
|
|
\begin{gather*}
|
|
(f\join g)(x)=f(x)\cup g(x)
|
|
\end{gather*}
|
|
\begin{prop}\label{prop:sym-rel-bisim-pf}
|
|
Assuming that $R$ is a symmetric relation, and $\sigma\c R\to \powf FR$ is a witness for $R$ to be an AM-simulation, then the following morphism is a witness for $R$ to be an AM-bisimulation:
|
|
\begin{gather*}
|
|
\sigma\join(\powf Fs\comp\sigma\comp s)
|
|
\end{gather*}
|
|
\end{prop}
|
|
\begin{proof}
|
|
For every $(x_1,x_2)\in R$ by~\autoref{lem:func-dist-set} and~\autoref{lem:sim-opsim-inc}.\ref{item:sim-opsim-inc:I} we have
|
|
\begin{gather*}
|
|
\powf Fp_1\comp(\sigma\join(\powf Fs\comp\sigma\comp s))(x_1,x_2)=
|
|
\powf Fp_1\comp\sigma(x_1,x_2).
|
|
\end{gather*}
|
|
% and by~\autoref{prop:alph-prod},
|
|
Recall that by~\autoref{prop:lift-gen-func}, set inclusion is a liftable ordering, so by~\autoref{prop:alph-prod-dubut-pf}, we have $\powf Fp_1\comp\sigma(x_1,x_2)=\alpha(x_1)$. By~\autoref{lem:func-dist-set} and~\autoref{lem:sim-opsim-inc}.\ref{item:sim-opsim-inc:II} we have
|
|
\begin{gather*}
|
|
\powf Fp_2\comp(\sigma\join(\powf Fs\comp\sigma\comp s))(x_1,x_2)=
|
|
\powf Fp_2\comp(\powf Fs\comp\sigma\comp s)(x_1,x_2).
|
|
\end{gather*}
|
|
Since $\powf Fp_1\comp\sigma=\alpha\comp p_1$ by precomposing $s$ to the both sides of the equation we get $\powf Fp_2\comp(\powf Fs\comp\sigma\comp s)=\alpha\comp p_2$. So, $\sigma\join(\powf Fs\comp\sigma\comp s)$ is a witness for $R$ to be an AM bisimulation.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Assuming that $R$ is a symmetric relation and it is an AM-simulation on a $\powf F$ coalgebra, then $R$ is an AM-bisimulation as well.
|
|
\end{cor}
|
|
|
|
\subsection{Maybe Functor}
|
|
We prove that symmetric simulation is a bisimulation for the case that $FX=X+1$. First, we prove it for $\Set$.\\
|
|
Proven by Dubut, for every AM-simulation relation over a coalgebra $(X,\alpha)$ of a functor with a liftable order, we have a witness $\sigma\c R\to R+1$ such that $\alpha\comp p_1=Fp_1\comp\sigma$. We have shown that the ordering on the maybe functor is liftable and coliftalbe in~\autoref{prop:maybe-lif} and~\autoref{prop:maybe-colif}
|
|
%\begin{lemma}\label{lem:maybe-func-set}
|
|
% Assuming that $R$ is a symmetric AM-simulation over an $F$-coalgebra $(X,\alpha)$ that $FX=X+1$, then for every $(x_1,x_2)\in R$
|
|
% \begin{gather*}
|
|
% \alpha(x_1)\in X\Rightarrow \alpha(x_2)\in X\\
|
|
% \alpha(x_1)=\bot \Rightarrow \alpha(x_2)=\bot
|
|
% \end{gather*}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% By~\autoref{prop:alph-prod-dubut} and~\autoref{prop:maybe-lif} there exists $\sigma\c R\to R+1$ that is a witness for $R$ to be an AM-simulation, and $Fp_1\comp\sigma=\alpha\comp p_1$. %Since $R$ is symmetric, for every $(x_1,x_2)\in R$ we have the following\sgnote{Consider noting which facts come from the pair $(x_1,x_2)$ (namely \eqref{eq:maybe-func-set-1} and \eqref{eq:maybe-func-set-4}) and which from $(x_2,x_1)$ (namely \eqref{eq:maybe-func-set-2} and \eqref{eq:maybe-func-set-3}); it saves the reader from reconstructing it.}:
|
|
%% \begin{enumerate}
|
|
%% \item $\alpha(x_1)=Fp_1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1}
|
|
%% \item $\alpha(x_2)=Fp_1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2}
|
|
%% \item $\alpha(x_1)\sappr Fp_2\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3}
|
|
%% \item $\alpha(x_2)\sappr Fp_2\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4}
|
|
%% \end{enumerate}
|
|
% For every $(x_1,x_2)$ we have
|
|
% \begin{enumerate}
|
|
% \item $\alpha(x_1)=Fp_1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
|
|
% \item $\alpha(x_2)\sappr Fp_2\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
|
|
%% \end{enumerate}
|
|
% and since $R$ is symmetric, we have
|
|
%% \begin{enumerate}
|
|
% \item $\alpha(x_2)=Fp_1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
|
|
% \item $\alpha(x_1)\sappr Fp_2\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
|
|
% \end{enumerate}
|
|
% Now, we have two cases:
|
|
% \begin{itemize}
|
|
% \item Assuming $\alpha(x_1)\in X$ then by~\eqref{eq:maybe-func-set-1} we have $Fp_1\comp\sigma(x_1,x_2)\in X$, thus $\sigma(x_1,x_2)\in R$. So, we have $Fp_2\comp\sigma(x_1,x_2)\in X$ that by~\eqref{eq:maybe-func-set-4} means $\alpha(x_2)\in X$.
|
|
% \item Assuming $\alpha(x_1)=\bot$ then by~\eqref{eq:maybe-func-set-3} we have $Fp_2\comp\sigma(x_2,x_1)=\bot$, thus $\sigma(x_2,x_1)=\bot$. So, we have $Fp_1\comp\sigma(x_2,x_1)=\bot$ that by~\eqref{eq:maybe-func-set-2} means $\alpha(x_2)=\bot$ as well.\qed
|
|
% \end{itemize}
|
|
%\end{proof}
|
|
%%\begin{proof}
|
|
%% We assume that $\sigma\c R\to R+1$ is the witness that we have for $R$ to be an AM-simulation. We prove the statement by contradiction. If the statement is false, then we either have
|
|
%% \begin{gather*}
|
|
%% \alpha(x_1)\in X\quad\&\quad \alpha(x_2)=\bot,
|
|
%% \end{gather*}
|
|
%% or
|
|
%% \begin{gather*}
|
|
%% \alpha(x_1)=\bot\quad\&\quad \alpha(x_2)\in X.
|
|
%% \end{gather*}
|
|
%% Assuming $\alpha(x_1)\in X\;\&\; \alpha(x_2)=\bot$, then since $R$ is an AM-simulation, by~\eqref{eq:diag-lax-sim}$(p_2+1)\comp \sigma(x_1,x_2)\appr\alpha(x_2)$, we have $(p_2+1)\comp \sigma(x_1,x_2)=\bot$ that entails $\sigma(x_1,x_2)=\bot$. So, we have $(p_1+1)\comp \sigma(x_1,x_2)=\bot$, while $\alpha(x_1)\not\sqsubseteq\bot$ that means that $\sigma$ is not a witness for $R$ to be an AM-simulation.\textreferencemark
|
|
%%
|
|
%% Assuming $\alpha(x_1)=\bot\;\&\; \alpha(x_2)\in X$, since $R$ is symmetric, then we have $(x_2,x_1)\in R$ as well. So, by~\eqref{eq:diag-lax-sim} we have $(p_2+1)\comp \sigma(x_2,x_1)\appr\alpha(x_1)$ that entails $\sigma(x_2,x_1)=\bot$. So, we have $(p_1+1)\comp\sigma(x_2,x_1)=\bot$, while $\alpha(x_2)\not\sqsubseteq\bot$ that means that $\sigma$ is not a witness for $R$ to be an AM-simulation.\textreferencemark\qed
|
|
%%\end{proof}
|
|
%
|
|
%\begin{prop}\label{prop:sym-sim-bis-maybe}
|
|
% For a $F$-coalgebra $(X,\alpha)$ in $\Set$, such that $FX=X+1$, a symmetric AM-simulation relation $R$ on $X$ is an AM-bisimulation.
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% By~\autoref{lem:maybe-func-set} for every $(x_1,x_2)\in R$, we either have
|
|
% \begin{gather*}
|
|
% \alpha(x_1)=\bot\quad\&\quad \alpha(x_2)=\bot,
|
|
% \end{gather*}
|
|
% or
|
|
% \begin{gather*}
|
|
% \alpha(x_1)\in X\quad\&\quad \alpha(x_2)\in X.
|
|
% \end{gather*}
|
|
% We define $\beta$ as a witness for $R$ to be an AM-bisimulation as follows:
|
|
% \begin{gather*}
|
|
% \beta(x_1,x_2)=
|
|
% \begin{cases}
|
|
% \bot & \alpha(x_1)=\bot\;\&\; \alpha(x_2)=\bot \\
|
|
% (\alpha(x_1),\alpha(x_2)) & \alpha(x_1)\in X\;\&\;\alpha(x_2)\in X
|
|
% \end{cases}
|
|
% \end{gather*}
|
|
% Assuming $\alpha(x_1)=\bot\;\&\; \alpha(x_2)=\bot$ we have $Fp_i\comp\beta(x_1,x_2)=\bot=\alpha(x_i)$, and assuming $\alpha(x_1)\in X\;\&\;\alpha(x_2)\in X$ we have $Fp_i\comp\beta(x_1,x_2)=\alpha(x_i)\in X$.
|
|
% Now, we are left to prove that if $\alpha(x_1)\in X$ and $\alpha(x_2)\in X$ then $\beta(x_1,x_2)\in R$ that means that the codomain of $\beta$ is indeed $R+1$. We assume $\alpha(x_i)\in X$. Since $Fp_i\comp\beta(x_1,x_2)=\alpha(x_i)\in X$ we have $\beta(x_1,x_2)\in R$.
|
|
% \qed
|
|
%\end{proof}
|
|
%%%%%%Second proof:
|
|
\begin{prop}\label{prop:sym-sim-bis-may}
|
|
For a $F$-coalgebra $(X,\alpha)$ in $\Set$, such that $FX=X+1$, a symmetric AM-simulation relation $R$ on $X$ is an AM-bisimulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We prove that $\sigma$ also is a witness for $R$ to be an AM-bisimulation. For every $(x_1,x_2)\in R$ we have
|
|
% \begin{enumerate}
|
|
% \item $\alpha(x_1)=(p_1+1)\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
|
|
% \item $\alpha(x_2)\sappr (p_2+1)\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
|
|
% % \end{enumerate}
|
|
% and since $R$ is symmetric, we have
|
|
% % \begin{enumerate}
|
|
% \item $\alpha(x_2)=(p_1+1)\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
|
|
% \item $\alpha(x_1)\sappr (p_2+1)\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
|
|
% \end{enumerate}
|
|
% Followed by $\alpha(x_2)\sappr (p_2+1)\comp\sigma(x_1,x_2)$ we have the following cases:
|
|
% \begin{itemize}
|
|
% \item $\alpha(x_2)=(p_2+1)\comp\sigma(x_1,x_2)$: In this case, we already have $\sigma$ as a witness for $R$ to be a bisimulation.
|
|
% \item $p_2+1\comp\sigma(x_1,x_2)=\bot$: In this case, we have
|
|
% \begin{align*}
|
|
% (p_2+1)\comp\sigma(x_1,x_2)=\bot,&\\
|
|
% \Rightarrow&\sigma(x_1,x_2)=\bot,\\
|
|
% \Rightarrow&(p_1+1)\comp\sigma(x_1,x_2)=\bot,\\
|
|
% \Rightarrow&\alpha(x_1)=\bot,&\eqref{eq:maybe-func-set-1}\\
|
|
% \Rightarrow&(p_2+1)\comp\sigma(x_2,x_1)=\bot,&\eqref{eq:maybe-func-set-3}\\
|
|
% \Rightarrow&\sigma(x_2,x_1)=\bot,\\
|
|
% \Rightarrow&(p_1+1)\comp\sigma(x_2,x_1)=\bot,\\
|
|
% \Rightarrow&\alpha(x_2)=\bot,&\eqref{eq:maybe-func-set-2}\\
|
|
% \Rightarrow&\alpha(x_2)=(p_1+1)\comp\sigma(x_1,x_2).
|
|
% \end{align*}
|
|
% \end{itemize}\qed
|
|
\begin{enumerate}
|
|
\item $\alpha\comp p_1=(p_1+1)\comp\sigma$\label{eq:maybe-func-set-1},
|
|
\item $\alpha\comp p_2\sappr (p_2+1)\comp\sigma$\label{eq:maybe-func-set-2},\\
|
|
and since $R$ is symmetric, we have
|
|
\end{enumerate}
|
|
Followed by~\eqref{eq:maybe-func-set-2} for every $(x_1,x_2)\in R$ we have $\alpha(x_2)=(p_2+1)\comp\sigma(x_1,x_2)$ and $\Dom((p_2+1)\comp\sigma)\subseteq\Dom(\alpha\comp p_2)$. We have
|
|
\begin{align*}
|
|
\Dom((p_2+1)\comp\sigma)&\\
|
|
=&\Dom(\sigma)\\
|
|
=&\Dom((p_1+1)\comp\sigma)\\
|
|
=&\Dom(\alpha\comp p_1)&\by{\eqref{eq:maybe-func-set-1}}\\
|
|
=&\Dom(\alpha\comp p_1\comp s)\\
|
|
=&\Dom(\alpha\comp p_2)
|
|
\end{align*}
|
|
So, we have $\alpha\comp p_2= (p_2+1)\comp\sigma$ as well that means that $\sigma$ also serves as a witness for $R$ to be a bisimulation.
|
|
\end{proof}
|
|
|
|
Now, we want to abstract the given proof for an extensive category that has terminal objects (so that we have the maybe functor). We assume a natural order structure $\appr$ for the maybe functor. We use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$.%For every objects $X$ and $Y$, we define $\appr$ on each $\Hom(X,Y+1)$ by saying that for $f,g\in\Hom(X,Y+1)$ we have $f\appr g$ whenever either $f=g$ or $f=\bot$.\sgnote{This abstract order is coarser than (not the concretisation of) the pointwise Set order at \autoref{lem:set-ord-str}: here the \emph{whole} map must be $\bot$, whereas pointwise a map may be $\bot$ on some points and agree elsewhere. Please check the two agree on the hom-sets you actually use, or justify why the coarser order suffices.}
|
|
%Although even in the context that we are at the moment, it does not seem plausible to prove that a symmetric simulation is a bisimulation without having an operator like $\join$ in~\autoref{prop:sym-sim-bis-maybe} that takes two morphisms of the same type and gives one.
|
|
The following lemma is an abstraction of saying that in $\Set$ assuming $f\c X\to Y+1$ and $g\c Y\to Z$, we have $\Dom((g+1)\comp f)=\Dom(f)$ that we have already used multiple times in the concrete proofs.
|
|
\begin{lemma}
|
|
Assuming that $(X_Y,f_Y,i)$ and $(X_Y,p,i)$ are pullbacks in the following diagram, then $p=g\comp f_Y$.
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{X_Y} \& X \\
|
|
Y \& {Y+Q} \\
|
|
Z \& {Z+Q'}
|
|
\arrow["i", tail, from=1-1, to=1-2]
|
|
\arrow["{f_Y}"', from=1-1, to=2-1]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
|
\arrow["p"', bend right=50, from=1-1, to=3-1]
|
|
\arrow["f", from=1-2, to=2-2]
|
|
\arrow["{\inl_Y}"', tail, from=2-1, to=2-2]
|
|
\arrow["{g}"', from=2-1, to=3-1]
|
|
\arrow["g+q", from=2-2, to=3-2]
|
|
\arrow["{\inl_Z}"', tail, from=3-1, to=3-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
IT simply relies on the fact that injections in an extensive category are assumed to be monic, and that the pullback of a monomorphism gives a monomorphism leg in the pullback's span.
|
|
\end{proof}
|
|
Assuming $f,g\in\Hom(X,Y+1)$, and that $f'\c X_f\rightarrowtail X$ and $g'\c X_g\rightarrowtail X$, and $X_f$ and $X_g$ are pullbacks of $f$ and $\inl$, and $g$ and $\inl$ accordingly, then $f\appr g$ if there exists $h\c X_f\rightarrowtail X_g$ that commutes in the following diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{X_g} \& X \\
|
|
{X_f}
|
|
\arrow["{g'}", tail, from=1-1, to=1-2]
|
|
\arrow["h", dashed, tail, from=2-1, to=1-1]
|
|
\arrow["{f'}"', tail, from=2-1, to=1-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\begin{lemma}\label{prop:maybe-lif-abs}
|
|
The order structure on the functor $F\c\BC\to\BC$ defined as $FX=X+1$ is a liftable order.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
\& Z \& \\
|
|
{X_h} \&\& {Z+1} \\
|
|
{X_k} \& X \\
|
|
Y \& {Y+1} \\
|
|
Z \& {Z+1}
|
|
\arrow["{\inl_Z}", tail, from=1-2, to=2-3]
|
|
\arrow["q", from=2-1, to=1-2]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125, rotate=45}, draw=none, from=2-1, to=2-3]
|
|
\arrow["e"', dashed, from=2-1, to=3-1]
|
|
\arrow["{h_d}", tail, from=2-1, to=3-2]
|
|
\arrow["{k_d}", tail, from=3-1, to=3-2]
|
|
\arrow["{k_Y}"', from=3-1, to=4-1]
|
|
\arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-1, to=4-2]
|
|
\arrow["p"', bend right=50, from=3-1, to=5-1]
|
|
\arrow["h"', from=3-2, to=2-3]
|
|
\arrow["k"', from=3-2, to=4-2]
|
|
\arrow["{\inl_Y}"', tail, from=4-1, to=4-2]
|
|
\arrow["g"', from=4-1, to=5-1]
|
|
\arrow["{g+1}"', from=4-2, to=5-2]
|
|
\arrow["{\inl_Z}"', tail, from=5-1, to=5-2]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
We define $k'=k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}$. We have:
|
|
\begin{align*}
|
|
(g+1)\comp k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}\\
|
|
(g+1)\comp k\comp\mathsf{iso}\comp(h_d+1)\comp\mathsf{iso}
|
|
\end{align*}
|
|
\todo{Finish! Perhaps, $k'=k\comp\mathsf{iso}\comp(k_d+1)\comp(e+1)\comp\mathsf{iso}$.}
|
|
\end{proof}
|
|
%\begin{proof}
|
|
% For morphisms $h\c X\to Z+1$, $g\c Y\to Z$, and $k\c X\to Y+1$, we assume $h\appr Fg\comp k$ that means we have two cases:
|
|
% \begin{itemize}
|
|
% \item $h=\bot$: In this case we take $k'=\bot$. Now, $k'\appr k$ and $Fg\comp k'=h$.
|
|
% \item $h=Fg\comp k$: In this case we take $k'=k$. Now, $k'\appr k$ and $Fg\comp k'=h$.\qed
|
|
% \end{itemize}
|
|
%\end{proof}
|
|
%To have an abstraction of~\autoref{lem:maybe-func-set} recalling that our category is extensive, we use the fact that for an arbitrary morphism $f\c X\to Y+Z$, we can have morphisms $f_Y\c X_Y\to Y$ and $f_Z\c X_Z\to Z$, such that $X_Y+X_Z\iso X$, so we follow with $f_Y+f_Z$. We take $\brks{\alpha\comp p_1,\alpha\comp p_2}\c R\to (X+1)\times(X+1)$, where $(X+1)\times (X+1)\iso X^2+(2\times X)+1$, and we assume the following pullbacks exist:
|
|
%\begin{equation*}
|
|
% \begin{tikzcd}[ampersand replacement=\&]
|
|
% {R_{X^2}} \& R \& {R_{2\times X}} \& R \\
|
|
% {X^2} \& {X^2+(2\times X)+1} \& {(2\times X)} \& {X^2+(2\times X)+1} \\
|
|
% \& {R_1} \& R \\
|
|
% \& 1 \& {X^2+(2\times X)+1}
|
|
% \arrow["{q_1}", from=1-1, to=1-2]
|
|
% \arrow["{q_2}"', from=1-1, to=2-1]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-1, to=2-2]
|
|
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-2, to=2-2]
|
|
% \arrow["{r_1}", from=1-3, to=1-4]
|
|
% \arrow["{r_2}"', from=1-3, to=2-3]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=1-3, to=2-4]
|
|
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=1-4, to=2-4]
|
|
% \arrow["{\mathsf{in}_1}"', from=2-1, to=2-2]
|
|
% \arrow["{\mathsf{in}_2}"', from=2-3, to=2-4]
|
|
% \arrow["{s_1}", from=3-2, to=3-3]
|
|
% \arrow["{s_2}"', from=3-2, to=4-2]
|
|
% \arrow["\lrcorner"{anchor=center, pos=0.125}, draw=none, from=3-2, to=4-3]
|
|
% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}", from=3-3, to=4-3]
|
|
% \arrow["{\mathsf{in}_3}"', from=4-2, to=4-3]
|
|
% \end{tikzcd}
|
|
%\end{equation*}
|
|
%%\begin{equation*}
|
|
%% \begin{tikzcd}[ampersand replacement=\&]
|
|
%% R \& I \& {X^2+(2\times X)+1}
|
|
%% \arrow["e", two heads, from=1-1, to=1-2]
|
|
%% \arrow["{\brks{\alpha\comp p_1,\alpha\comp p_2}}"', bend right=20, from=1-1, to=1-3]
|
|
%% \arrow["m", tail, from=1-2, to=1-3]
|
|
%% \end{tikzcd}
|
|
%%\end{equation*}
|
|
%So, we have the following:
|
|
%\begin{gather*}
|
|
% R\iso R_{X^2}+R_{2\times X}+R_{1}
|
|
%\end{gather*}
|
|
%And now we can use $q_2+r_2+s_2\c R_{X^2}+R_{2\times X}+R_{1}\to X^2+(2\times X)+1$ instead of $\brks{\alpha\comp p_1,\alpha\comp p_2}$. To prove~\autoref{lem:maybe-func-set} abstractly is to prove that $q_2+r_2+s_2$ factors through $X^2+1$. To achieve this, we need to show that $R_{2\times X}\iso 0$.
|
|
%\sgnote{This abstraction is unfinished: the crux $R_{2\times X}\iso 0$ (the ``no mixed pairs'' content of \autoref{lem:maybe-func-set}) is stated but not proved, and the subsection ends here. Either complete the argument or mark it clearly as work in progress.}
|
|
\section{Relators}
|
|
\subsection{Two-way similarity in Hughes-Jacobs}
|
|
Hughes and Jacobs define two-way similarity as $\leq\cap\leq^\op$. They give a sufficient condition for the two-way similarity to be the bisimilarity. We discuss that this condition does not allow us to say that a symmetric simularity is a bisimilarity. The condition is:
|
|
\begin{gather*}
|
|
\appr;(FR_1)^\dagger;\appr\quad\cap\quad\appr^\op;(FR_2)^\dagger;\appr^\op\qquad\subseteq\qquad(F(R_1\cap R_2))^\dagger
|
|
\end{gather*}
|
|
We need the case that $R_1=R_2$, and we refer to it with $R$. So, we need to have
|
|
\begin{gather*}
|
|
\appr;(FR)^\dagger;\appr\quad\cap\quad\appr^\op;(FR)^\dagger;\appr^\op\qquad\subseteq\qquad(FR)^\dagger.
|
|
\end{gather*}
|
|
Assuming $(x_1,x_2)\quad\in\quad \appr;(FR)^\dagger;\appr\quad\cap\quad\appr^\op;(FR)^\dagger;\appr^\op$ means that there exist $(u,v),(u',v')\in (FR)^\dagger$, such that $x_1\appr u$, $u'\appr x_1$, $v\appr x_2$, and $x_2\appr v'$. We can also use the symmetry of $R$, and then from $(x_1,x_2)\in R$ derive that there exist $(w,z),(w',z')$ such that $x_1\appr z$, $z'\appr x_1$, $w\appr x_2$, and $x_2\appr w'$. The situation can be illustrated as the following:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
{u'} \&\& u \& v \&\& {v'} \\
|
|
\& {x_1} \&\&\& {x_2} \\
|
|
{z'} \&\& z \& w \&\& {w'}
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=1-4, to=2-5]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=3-3]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-5, to=1-6]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=2-5, to=3-6]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=3-1, to=2-2]
|
|
\arrow["\appr"{marking, allow upside down}, draw=none, from=3-4, to=2-5]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
|
|
But then how can we derive $(x_1,x_2)\in (FR)^\dagger$?
|
|
|
|
\todo{Investigate more! Can it be doable really?!}
|
|
\subsection{Uniqueness of the witness in Hughes-Jacobs definition}
|
|
In this section we set $\rel$ to be the category that has sets as objects and binary relations as morphisms.
|
|
We answer the question that why there can be multiple simulation witnesses based on~\autoref{def:sim}, while for the same relation, there is only one witness according to Hughes-Jacobs simulation.
|
|
\begin{definition}[Hughes-Jacobs Simulation]
|
|
For a functor $F$, and a poset $\appr$ over $F$ a HuJ-simulation is a relation $r$ for which there exists a morphism $\sigma\c r\to (Fr)^\dagger$ called \emph{witness} such that the following diagram commutes ($;$ is the relation composition):
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y \\
|
|
{FX} \& {\appr;(FR)^\dagger;\appr} \& {FY}
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["{{Fp_1}_\appr}", from=2-2, to=2-1]
|
|
\arrow["{{Fp_2}_\appr}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
\end{definition}
|
|
At the moment we have limited the discussion to the category of sets and we are talking about the powerset functor. We know that $\sigma$ is unique in the following diagram:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y\\
|
|
{\powf X} \& {\subseteq;(\powf R)^\dagger;\subseteq} \& {\powf Y}
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", dashed, from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["{{\powf p_1}_\subseteq}", from=2-2, to=2-1]
|
|
\arrow["{{\powf p_2}_\subseteq}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
It is defined as $\sigma(x_1,x_2)=(\alpha(x_1),\beta(x_2))$.
|
|
But $\sigma'$ in the following diagram is not unique:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y \\
|
|
{\powf X} \& {(\powf R)^\dagger} \& {\powf Y}
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["\subseteq"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["{\sigma'}", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["\subseteq"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
|
|
\arrow["{{\powf p_1^\dagger}}", from=2-2, to=2-1]
|
|
\arrow["{{\powf p_2^\dagger}}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
Because assuming we have $\sigma$, for every given $\sigma'$ we can define a $\delta\c(\powf R)^\dagger\to\subseteq;(\powf R)^\dagger;\subseteq$ that $\sigma=\delta\comp\sigma'$, i.e., the following diagram commutes:
|
|
\begin{equation*}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& R \& Y \\
|
|
{\powf X} \& {(\powf R)^\dagger} \& {\powf Y} \\
|
|
{\powf X} \& {\subseteq;(\powf R)^\dagger;\subseteq} \& {\powf Y}
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["{\sigma'}", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["id"', from=2-1, to=3-1]
|
|
\arrow["\delta", from=2-2, to=3-2]
|
|
\arrow["id", from=2-3, to=3-3]
|
|
\arrow["{{\powf p_1}_\subseteq}", from=3-2, to=3-1]
|
|
\arrow["{{\powf p_2}_\subseteq}"', from=3-2, to=3-3]
|
|
\end{tikzcd}
|
|
\end{equation*}
|
|
To define $\delta$, we define $c\c(\powf R^\dagger)\to((\powf R^\dagger)\times R)+(\powf R)^\dagger$ and $u\c((\powf R^\dagger)\times R)+(\powf R)^\dagger\to\subseteq;(\powf R)^\dagger;\subseteq$ and then we define $\delta=u\comp c$. Here are the definitions for $c$ and $u$:
|
|
\begin{gather*}
|
|
c(w)=
|
|
\begin{cases}
|
|
\inl(w,(x_1,x_2)) & \exists x_1,x_2, \sigma'(x_1,x_2)=w \\
|
|
\inr w & \mathsf{o.w}
|
|
\end{cases}\\
|
|
u(\inl w,(x_1,x_2))=(\alpha(x_1),\alpha(x_2))\\
|
|
u(\inr w)=w
|
|
\end{gather*}
|
|
\subsection{Symmetric simulation}
|
|
\begin{notation}
|
|
From now on, we denote relations with small letters, and for two relations $r_1$ and $r_2$ by $r_1\leq r_2$ we mean $r_1\subseteq r_2$. Also, we denote the category of relations over set that we represent by spans with $\spa$, and $\rel$ is the category of sets and binary relations between them.
|
|
\end{notation}
|
|
\begin{lemma}\label{lem:rel-span-equiv}
|
|
$r\c X\rto Y$ is a morphism in $\rel$ iff there is an object $(r,p_1,p_2)$ in $\spa$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
($\Rightarrow$): $r\c X\rto Y$ being a morphism in $\rel$ means that in $\Set$ there exist an object $r$ with a unique mono of type $r\to X\times Y$ that is a pairing that we denote with $\brks{p_1,p_2}$. So, $(r,p_1,p_2)$ form an object in $\spa$.
|
|
|
|
($\Leftarrow$): If $(r,p_1,p_2)$ is an object in $\spa$, then $r$ is a binary relation from $X$ to $Y$ so, it is a morphism of type $X\rto Y$ in $\rel$.\qed
|
|
\end{proof}
|
|
The above translation seems to be true in a more general case, where $\spa$ and $\rel$ are defined on an arbitrary category (the latter is called an allegory then).
|
|
\begin{definition}[Relator]
|
|
Assuming $F$ is a functor on $\Set$, a $F$-relator or simply a relator $\relar$ is a monotone map that sends a morphism of $\Rel$ that is a relation $X\rto Y$ to $FX\rto FY$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Hermida-Jacobs Simulation]\label{def:hej-sim}
|
|
For a relator $\relar$ on a functor $F$ a HJ-simulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ is a relation $r$ for which there exists a morphism $\sigma\c r\to\relar r$ called \emph{witness} such that the following diagram commutes ($;$ is the relation composition):
|
|
\begin{equation}\label{eq:hej-sim}
|
|
\begin{tikzcd}[ampersand replacement=\&]
|
|
X \& r \& Y \\
|
|
{FX} \& {\relar r} \& {FY}
|
|
\arrow["\alpha"', from=1-1, to=2-1]
|
|
\arrow["{p_1}"', from=1-2, to=1-1]
|
|
\arrow["{p_2}", from=1-2, to=1-3]
|
|
\arrow["\sigma", from=1-2, to=2-2]
|
|
\arrow["\beta", from=1-3, to=2-3]
|
|
\arrow["{{(Fp_1)}^\relar}", from=2-2, to=2-1]
|
|
\arrow["{{(Fp_2)}^\relar}"', from=2-2, to=2-3]
|
|
\end{tikzcd}
|
|
\end{equation}
|
|
\end{definition}
|
|
|
|
\begin{definition}[Relator-based Simulation]
|
|
Given a relator $\relar$, a relation $r\c X\rto Y$ is a $\relar$-simulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ if $r\leq \alpha;\relar r;\beta^\op$, i.e, if $(x,y)\in r$ entails $(\alpha(x),\beta(y))\in\relar r$, for all $x\in X$ and $y\in Y$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Symmetric Relator]
|
|
A relator $\relar$ is symmetric if and only if for every relation $r$ we have $\relar(r^\op)=(\relar r)^\op$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Relator-based Bisimulation]
|
|
Given a relator $\relar$, a relation $r\c X\rto Y$ is a $\relar$-bisimulation from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ if $r$ is a $\relar$-simulation, and $\relar$ is a symmetric relator.
|
|
\end{definition}
|
|
|
|
\begin{notation}
|
|
From now on we refer to relator-based simulations of a relator $\relar$ with $\relar$-simulation. If we talk about a HJ-simulation we specify the witness.
|
|
\end{notation}
|
|
|
|
\begin{lemma}\label{lem:sim-simp}
|
|
$r$ being a $\relar$-simulation means that assuming $x\;r\;y$ we have $\alpha(x)\;\relar r\;\beta(y)$.
|
|
\end{lemma}
|
|
\begin{proof}
|
|
$r$ being a $\relar$-simulation means $r\leq\beta^\op\comp\relar r\comp\alpha$, meaning that if $x\;r\;y$ then $x \;\alpha;\relar r;\beta^\op\; y$, and it means that there exist $w\in\relar r$ that its first element is equal with $\alpha(x)$ and its second element is equal with $\beta(y)$, enabling us to say $\alpha(x)\;\relar r\; \beta(y)$.\qed
|
|
\end{proof}
|
|
|
|
\begin{prop}
|
|
For a relator $\relar$ of a functor $F$, a relation $r$ is a $\relar$-simulation from $\alpha\c X\to FX$ to $\beta\c Y\to FY$ iff it is a HJ-simulation with the witness $\sigma\c r\to\relar r$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
($\Rightarrow$): $r$ being a $\relar$-simulation means that $x\;r\;y$ gives $\alpha(x)\;\relar r\;\beta(y)$. Since $r$ and $\relar r$ are both relations, by~\autoref{lem:rel-span-equiv} there exist objects $(r,p_1,p_2)$ and $(\relar r,(Fp_1)^\relar,(Fp_2)^\relar)$ in $\spa$. We define $\sigma\c r\to\relar r$ to be $\sigma(x,y)=(\alpha(x),\beta(y))$. $\sigma$ commutes in~\eqref{eq:hej-sim}, so we have a HJ-simulation.
|
|
|
|
($\Leftarrow$): Assuming we have a $\sigma$ that commutes in~\eqref{eq:hej-sim}, we want to prove that if $x\;r\;y$ we have $\alpha(x)\;\relar r\;\beta(y)$. By~\eqref{eq:hej-sim} we have $\alpha\comp p_1=(Fp_1)^\relar\comp\sigma$ and $\alpha\comp p_2=(Fp_2)^\relar\comp\sigma$. It means that since $x\;r\;y$ then exists $w\in\relar r$, such that $\sigma(x,y)=w$, where $(Fp_1)^\relar(w)=\alpha(x)$ and $(Fp_2)^\relar(w)=\beta(y)$. So, we have $\alpha(x) \;\relar r \;\beta(y)$.\qed
|
|
\end{proof}
|
|
\begin{prop}
|
|
For an arbitrary relator $\relar$ on a functor $F$, if a relation $r$ is a HJ-simulation, the witness is unique.
|
|
\end{prop}
|
|
\begin{proof}
|
|
It only relies on the fact that ${(Fp_1)}^\relar$ and ${(Fp_2)}^\relar$ in~\eqref{def:hej-sim} are jointly monic.\qed
|
|
\end{proof}
|
|
\begin{definition}
|
|
We call $\hat{\relar}$ a symmetrization of a relator $\relar$ iff for a relation $r$ it is defined as follows:
|
|
\begin{gather*}
|
|
\hat{\relar}r=\relar r\cap (\relar(r^\op))^\op
|
|
\end{gather*}
|
|
\end{definition}
|
|
\begin{prop}
|
|
For every relator $\relar$, $\hat{\relar}$ is a relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Almost obvious!\qed
|
|
\end{proof}
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a relator, and $r$ and $r^\op$ are both $\relar$-simulations from a coalgebra $\alpha\c X\to FX$ to a coalgebra $\beta\c Y\to FY$ and vice-versa respectively, then $r$ is also a $\hat{\relar}$-simulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We need to prove that $x\; r\;y$ gives $\alpha(x)\;\hat{\relar} r\;\beta(y)$.
|
|
$r$ being a $\relar$-simulation means that assuming $x\;r\;y$ we have $\alpha(x)\;\relar r\;\beta(y)$. So, we are left to prove $\alpha(x)\;(\relar (r^\op))^\op\;\beta(y)$. $x\;r\;y$ gives $y\;r^\op\;x$, and $r^\op$ being a $\relar$-simulation from $\beta$ to $\alpha$ gives $\beta(y)\;\relar r^\op\; \alpha(x)$. So, we have $\alpha(x) \;(\relar(r^\op))^\op \;\beta(y)$. \qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Assuming that $\relar$ is a relator, and $r$ is a symmetric $\relar$-simulation from a coalgebra $\alpha\c X\to FX$ to itself, then $r$ is also a $\hat{\relar}$-simulation.
|
|
\end{cor}
|
|
\begin{remark}
|
|
If we want to have the previous corollary for two different coalgebras $\alpha\c X\to FX$ and $\beta\c X\to FX$, we need to assume that $r^\op$ is also a $\relar$-simulation from $\beta$ to $\alpha$.
|
|
\end{remark}
|
|
|
|
\begin{prop}
|
|
$\hat{\relar}$ is a symmetric relator, i.e., every $\hat{\relar}$-simulation is actually a $\hat{\relar}$-bisimulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\begin{align*}
|
|
\hat{\relar}(r^\op)&\\
|
|
&=\relar(r^\op)\cap(\relar (r^\op)^\op)^\op\\
|
|
&=\relar(r^\op)\cap(\relar r)^\op\\
|
|
&=(\relar r)^\op\cap\relar(r^\op)\\
|
|
&=(((\relar r)^\op\cap\relar(r^\op))^\op)^\op\\
|
|
&=(\relar r\cap(\relar(r^\op))^\op)^\op\\
|
|
&=(\hat{\relar}r)^\op
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a symmetric relator, and $r$ is a $\relar$-simulation from a coalgebra $(X,\alpha)$ to itself, then $r^\op_{s}$ is a $\relar$-simulation as well.
|
|
\end{prop}
|
|
\begin{proof}
|
|
It is easy to directly show that $r^\op$ is a $\relar$-simulation.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Assuming that $\relar$ is a symmetric relator, then the $\relar$-simularity from a coalgebra $(X,\alpha)$ to itself is a symmetric relation.
|
|
\end{cor}
|
|
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a symmetric relator, then for every $r$ that is a $\relar$-simulation, $r^\op$ is also a $\relar$-simulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming $y\; r^\op x$ we have $x\; r\; y$. Since $r$ is $\relar$-simulation we have $\alpha(x)\;\relar r\;\beta (y)$. So, we have $\beta(y)\;(\relar r)^\op\;\alpha(x)$, and since $\relar$ is symmetric we have $\beta(y)\;\relar r^\op\;\alpha(x)$.\qed
|
|
\end{proof}
|
|
|
|
\begin{definition}[Behavioural Equivalence]
|
|
Two states $x$ and $y$ of two coalgebras $(X,\alpha)$ and $(Y,\beta)$ are behaviourally equivalent iff there exist a coalgebra $(Z,\gamma)$ and coalgebra morphisms $f\c(X,\alpha)\to(Z,\gamma)$ and $g\c(Y,\beta)\to(Z,\gamma)$ such that $f(x)=g(y)$. The relation $r$ consisting of all behaviourally equivalent states of these two coalgebras is called behavioural equivalence.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Difunctional Relation]
|
|
A relation $r\c X\rto Y$ is difunctional iff there are functions $f\c X\to Z$ and $g\c Y\to Z$ such that for every $(x,y)\in R$ we have $f(x)=g(y)$.
|
|
\end{definition}
|
|
|
|
\begin{definition}[Soundness and Completeness of $\relar$-similarity]
|
|
For a relator $\relar$ the $\relar$-similarity from a coalgebra $(X,\alpha)$ to a coalgebra $(Y,\beta)$ is sound iff it is less than or equal to their behavioural equivalence, and it is complete iff it is greater than or equal to their behavioural equivalence.
|
|
\end{definition}
|
|
|
|
\begin{thm}
|
|
Let $\relar$ be a relator for a functor $F$:
|
|
\begin{enumerate}
|
|
\item If for all functions $f\c X\to A$ and $g\c Y\to A$, $\relar(g^\op\comp f)\geq (Fg)^\op\comp Ff$, then $\relar$-similarity is complete.
|
|
\item If $\relar$ preserves difunctional relations and for every epi-cospan $(f\c X\to A,g\c Y\to A)\in\Set$, $\relar(g^\op\comp f)\leq(Fg)^\op\comp Ff$, then $\relar$-similarity is sound.
|
|
\end{enumerate}\qed
|
|
\end{thm}
|
|
|
|
\begin{prop}\label{prop:difunc-preser}
|
|
Assuming that $\relar$ is a $F$-relator that preserves difunctional relations, then $\hat{\relar}$ does the same.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming $r$ is difunctional, then there exist $f_1,f_2\c FX\to FZ$ and $g_1,g_2\c FY\to FZ$ such that $p\;\relar r\;q$ iff $f_1(p)=g_1(q)$ and $q \;\relar r^\op \;p$ iff $f_2(p)=g_2(q)$. By the definition of $\hat{\relar}$ we have $p\;\hat{\relar}r\;q$ iff $p\;\relar r\;q$ and $p\;(\relar r^\op)^\op\;q$ that is equivalent to say that $\brks{f_1,f_2}(p)=\brks{g_1,g_2}(q)$.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Assuming that $\relar$ is a $F$-relator that preserves difunctional relations, for every symmetric relation $r$ we have $\hat{\relar}r=\relar r$.
|
|
\end{cor}
|
|
\begin{proof}
|
|
$\relar$-similarity being symmetric means that for the $f_1$, $f_2$, $g_1$ and $g_2$ in the proof of~\autoref{prop:difunc-preser}, we have $f_1=f_2$ and $g_1=g_2$.\qed
|
|
\end{proof}
|
|
|
|
Assuming that $\relar$-similarity is complete, does not guarantee that $\hat{\relar}$-similarity is sound and complete. We give a counter-example. Assuming that $\relar$ is a $\powf$-relator that takes $r\c X\rto Y$ to $\powf X\times \powf Y$, then $\hat{\relar} r= \powf X\times\powf Y$ as well. It means that for every coalgebras $(X,\alpha)$, $(Y,\beta)$, $\hat{\relar}$-similarity is equal to $X\times Y$, which is rare to be equal to behavioural equivalence. For example, if we take $X=\{x_1,x_2\}$ and $Y=\{y_1,y_2,y_3\}$, and we define $\alpha$ and $\beta$ as
|
|
\begin{gather*}
|
|
\alpha(x)=
|
|
\begin{cases}
|
|
\{x_1,x_2\} & x=x_1 \\
|
|
\{x_2\} & x=x_2
|
|
\end{cases}
|
|
\intertext{and}
|
|
\beta(y)=
|
|
\begin{cases}
|
|
\{y_3\} & y=y_1 \\
|
|
Y & y=y_2\\
|
|
\emptyset & y=y_3
|
|
\end{cases}
|
|
\end{gather*}
|
|
then at least $(x_1,y_3)$ is not in the behavioural equivalence, while it is in $\hat{\relar}$-similarity.
|
|
|
|
\begin{prop}
|
|
Assuming that $\relar$-similarity is symmetric and complete, then $\hat{\relar}$-similarity from a coalgebra $\alpha\c X\to FX$ to itself is sound and complete.
|
|
\end{prop}
|
|
\begin{proof}
|
|
(Completeness): We show $\relar$-similarity with $r_s$ and $\hat{\relar}$-similarity with $r_{\hat{s}}$. Also, we show the behavioural equivalence with $r_b$. Since
|
|
\end{proof}
|
|
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a $F$-relator ($F$ is a set functor), that for every functions $f\c X\to Z$ and $g\c Y\to Z$, we have $\relar(g^\op\comp f)\geq (Fg)^\op\comp Ff$, then $\hat{\relar}$-similarity is complete.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We need to prove that for every functions $f\c X\to Z$ and $g\c Y\to Z$ we have $\hat{\relar}(g^\op\comp f)\geq (Fg)^\op\comp Ff$. By the assumption we have $\relar(g^\op\comp f)\geq(Fg)^\op\comp Ff$. Also, again from the assumption we have $\relar(f^\op\comp g)\geq(Ff)^\op\comp Fg$ that gives $(\relar(f^\op\comp g))^\op\geq(Fg)^\op\comp Ff$. So, we have $\hat{\relar}(g^\op\comp f)\geq F(g)^\op\comp Ff$.\qed
|
|
\end{proof}
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a symmetric relator for a functor $F\c\Set\to\Set$, then the $\relar$-bisimilarity from a coalgebra $\alpha\c X\to FX$ to itself is sound, using the axiom of choice.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We call the bisimilarity relation $r$, and we assume $x_1\;r\;x_2$, now we need to prove that $x_1$ and $x_2$ are behaviourally equivalent. We take $Z=X/r$, where $X/r=\{[x]\mid [x]=\{y\mid x\;r\;y\}\}$. Now, we define the coalgebra homomorphism $f\c X\to X/r$ as $f(x)=[x]$. So, assuming $x_1\;r\;x_2$ gives $f(x_1)=f(x_2)$. Now, assuming that exists a choice function $c\c X/r\to X$ that $c\comp f=\id_X$, we define $\gamma\c X/r\to F(X/r)$, as $\gamma([x])=Ff\comp\alpha\comp c([x])$. Now, we have
|
|
\begin{align*}
|
|
\gamma\comp f&\\
|
|
&=Ff\comp\alpha\comp c\comp f\\
|
|
&=Ff\comp\alpha.
|
|
\end{align*}
|
|
So, $x_1$ and $x_2$ are behaviourally equivalent. So, the $\relar$-bisimilarity is sound.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Assuming that a relator $\relar$ over a functor $F\c\Set\to\Set$ satisfies $\relar(g^\op\comp f)\geq (Fg)^\op\comp Ff$ for every functions $f\c X\to Z$ and $g\c Y\to Z$, then $\hat{\relar}$-bisimilarity from a coalgebra $\alpha\c X\to FX$ to itself is sound and complete, using the axiom of choice.
|
|
\end{cor}
|
|
\subsection{Egli-Milner relator and Barr relators}
|
|
\begin{definition}
|
|
We call the map $\emre\c\rel\to\rel$ the Egli-Milner $\powf$-relator, whenever for every relation $r\c X\rto Y$ it is defined as follows:
|
|
\begin{gather*}
|
|
\emre r=\{(S,T)\mid x\in S\Rightarrow \exists y\in T, x\;r\;y\}
|
|
\end{gather*}
|
|
\end{definition}
|
|
Egli-Milner relator is not sound or complete, although its symmetrization is sound and complete.
|
|
\begin{prop}
|
|
$\hat{\emre}$-similarity from a coalgebra $(\alpha,X)$ to $(\beta,Y)$ is sound and complete.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We need to prove that for every functions $f\c X\to Z$ and $g\c Y\to Z$, $\hat{\emre}(g^\op\comp f)=(\powf g)^\op\comp\powf f$.
|
|
We have $S\;\hat{\emre}(g^\op\comp f)\;T$ iff $S\;\emre(g^\op\comp f)\;T$ and $T\;\emre(f^\op\comp g)\;S$. Then we have
|
|
\begin{align*}
|
|
S\;\emre(g^\op\comp f)\;T&\\
|
|
&\iff\forall x\in S,\exists y\in T, x\;g^\op\comp f\; y\\
|
|
&\iff \forall x\in S,\exists y\in T,z\in Z, x\;f\;z\; , \; y\;g\;z,
|
|
\end{align*}
|
|
and
|
|
\begin{align*}
|
|
T\;\emre(f^\op\comp g)\;S&\\
|
|
&\iff\forall y\in T,\exists x\in S, y\;f^\op\comp g\; x\\
|
|
&\iff \forall y\in T,\exists x\in S, z\in Z, x\;f\;z\; , \; y\;g\;z.
|
|
\end{align*}
|
|
It is equivalent with the following:
|
|
\begin{gather*}
|
|
\forall x\in S,\exists y\in T, f(x)=g(y),\\
|
|
\forall y\in T,\exists x\in S, f(x)=g(y).
|
|
\end{gather*}
|
|
Equivalently, $Im(f\mid_S)=Im(g\mid_T)$, and we call images $U$ that is in $\powf Z$. So, we equivalently have
|
|
\begin{align*}
|
|
S\;\powf f\;U,\; T\;\powf g\;U&\\
|
|
&\iff S\;\powf f\;U,\; U\;(\powf g)^\op\;T\\
|
|
&\iff S\;(\powf g)^\op\comp\powf f\;T
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
For every relation $r\rto X\to Y$ $\emre r=\subseteq\;\emre r=\emre r\;\subseteq=\subseteq;\emre r;\subseteq$.
|
|
%\begin{prop}
|
|
% Assuming that $r\c X\rto Y$, then $\subseteq;\hat{\emre}r;\subseteq=\subseteq;\hat{\emre}r$ and $\subseteq;\hat{\emre}r=\hat{\emre}r;\subseteq$.
|
|
%\end{prop}
|
|
|
|
|
|
Barr relator is a generalization of the Egli-Milner relator, where the functor is generalized.
|
|
|
|
\begin{definition}[Barr relator]
|
|
A relator over a functor $F$ is a Barr relator, shown by $\bar{F}$, iff for a relation $r\c X\rto Y$, and a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$ we have:
|
|
\begin{gather*}
|
|
\bar{F}r=F\pi_2\comp(F\pi_1)^\op
|
|
\end{gather*}
|
|
\end{definition}
|
|
\begin{prop}
|
|
For every set-functor $F$, the barr relator $\bar{F}$ is symmetric.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming that for a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ we have $r=\pi_2\comp\pi_1^\op$, then we have:
|
|
\begin{align*}
|
|
\hat{\bar{F}}r&\\
|
|
=&\bar{F}r\cap(\bar{F}r^\op)^\op\\
|
|
=&\bar{F}(\pi_2\comp\pi_1^\op)\cap(\bar{F}(\pi_2\comp\pi_1^\op)^\op)^\op\\
|
|
=&\bar{F}(\pi_2\comp\pi_1^\op)\cap(\bar{F}(\pi_1\comp\pi_2^\op))^\op\\
|
|
=&F\pi_2\comp(F\pi_1)^\op\cap(F\pi_1\comp (F\pi_2)^\op)^\op\\
|
|
=&F\pi_2\comp(F\pi_1)^\op\cap F\pi_2\comp (F\pi_1)^\op\\
|
|
=&F\pi_2\comp(F\pi_1)^\op\\
|
|
=&\bar{F}r
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
\begin{prop}
|
|
$\hat{L}$ is a Barr relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We have
|
|
\begin{gather*}
|
|
\hat{\emre}r=\emre r\cap (\emre r^\op)^\op,\\
|
|
\hat{\emre}r=\{(S,T)\mid x\in S\Rightarrow \exists y\in T, x\;r\;y\}\cap\{(T,S)\mid x\in S\Rightarrow \exists y\in T, y\;r\;x\}.
|
|
\end{gather*}
|
|
Assuming that $r=\pi_2\comp(\pi_1)^\op$ we have to prove that $\hat{L}r=\powf\pi_2\comp(\powf\pi_1)^\op$.
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
|
|
\begin{definition}[Mid-lax Barr relator]
|
|
Given a relation $r$, and take a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$, and $\pi_1$ and $\pi_2$ are surjective, assuming that $\appr$ is a partial order over a functor $F$, then the relator over $F$ and shown with $\relar$ is a \emph{mid-lax Barr relator} if we have:
|
|
% A relator over a functor $F$ is a one-sided Barr relator, shown by $\overrightarrow{F}$, iff for a partial order $\appr$ over $F$, a relation $r\c X\rto Y$, and a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$ we have:
|
|
\begin{gather*}
|
|
\relar=F\pi_2\comp\appr\comp(F\pi_1)^\op
|
|
\end{gather*}
|
|
\end{definition}
|
|
|
|
\begin{prop}
|
|
For every functor $F\c\Set\to\Set$, the symmetrization of the mid-lax Bar relator is equal with the Barr relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
% Where there exist $\pi_1\c A\to X$ and $\pi_2\c A\to Y$ such that $r=\pi_2\comp\pi_1^\op$, we assume that $s \;\hat{\overrightarrow{F}}r\; t$, and we need to show that $s\;F\pi_2\comp(F\pi_1)^\op\;t$. Considering that $r^\op=\pi_1\comp\pi_2^\op$, we have:
|
|
% \begin{align*}
|
|
% s \;\hat{\overrightarrow{F}}r\; t&\\
|
|
% &\iff s\;\overrightarrow{F}r\;t\qquad\&\qquad s \;(\overrightarrow{F}r^\op)^\op\; t\\
|
|
% &\iff s\;F\pi_2\comp\appr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;(F\pi_1\comp\appr\comp(F\pi_2)^\op)^\op\;t\\
|
|
% &\iff s\;F\pi_2\comp\appr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;F\pi_2\comp\appr\comp(F\pi_1)^\op\;t
|
|
% \end{align*}
|
|
% Since $F\pi_1$ is a surjective function, then exists at least one $w\in FA$ such that $(F\pi_1)^\op(s)=w$, and:
|
|
% \begin{gather*}
|
|
% w\;F\pi_2\comp\appr\; t \qquad\&\qquad w\;F\pi_2\comp\appr\; t
|
|
% \end{gather*}
|
|
% And similarly, since $F\pi_2$ is also a surjective function we have at least one $v\in FA$ such that $(F\pi_2^\op)(t)=v$, and:
|
|
% \begin{align*}
|
|
% &w\;\appr\; v \qquad\&\qquad w\;\appr\; v\\
|
|
% \iff&(F\pi_1)^\op(s)\;\appr\; (F\pi_2^\op)(t) \qquad\&\qquad (F\pi_1)^\op(s)\;\appr\; (F\pi_2^\op)(t)\\
|
|
% \iff&(F\pi_1)^\op(s)\;=\; (F\pi_2^\op)(t)\\
|
|
% \iff&s\; F\pi_2\comp(F\pi_1)^\op\;t\\
|
|
% \iff&s\;\bar{F}r\;t
|
|
% \end{align*}
|
|
% So we have $\hat{\overrightarrow{F}}\leq \bar{F}$.
|
|
% Now, we are left to show that $\bar{F}\leq\hat{\overrightarrow{F}}$. For that, reading the given proof from the end to the starting point is sufficient.
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
|
|
\begin{prop}
|
|
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ the following propositions hold:
|
|
\begin{enumerate}
|
|
\item $\powf\pi_2\comp\subseteq\comp(\powf\pi_1)^\op\quad=\quad\subseteq\comp \powf\pi_2\comp(\powf\pi_1)^\op$
|
|
\item $\powf\pi_1\comp\subseteq\comp(\powf\pi_2)^\op\quad=\quad\subseteq\comp \powf\pi_1\comp(\powf\pi_2)^\op$
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $\powf\pi_j\comp\subseteq\comp(\powf\pi_i)^\op\quad=\quad\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op$.
|
|
|
|
Assuming $x \mathrel{\powf\pi_j\comp\subseteq\comp(\powf\pi_i)^\op} y$, then exist $z$ and $z'$ such that
|
|
\begin{gather*}
|
|
z \mathrel{(\powf\pi_i)} x,\\
|
|
z \mathrel{\subseteq} z',\\
|
|
z'\mathrel{(\powf\pi_j)} y.
|
|
\end{gather*}
|
|
Then from $z \mathrel{\subseteq} z'$ we get $\powf\pi_j(z)\mathrel{\subseteq}y$. So, we have $z\mathrel{\subseteq\comp \powf\pi_j} y$, thus $x\mathrel{\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$.
|
|
|
|
Now, assuming $x \mathrel{\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
|
|
\begin{gather*}
|
|
z\mathrel{(\powf\pi_i)} x,\\
|
|
z \mathrel{(\powf\pi_j)} y',\\
|
|
y' \mathrel{\subseteq} y.
|
|
\end{gather*}
|
|
We take the set $w=z\cup (\powf\pi_i(z)\times y)$ for which we have $z\subseteq w$ and $\powf\pi_j(w)=y$. So, we have $w \mathrel{(\powf\pi_j)} y$, $z\mathrel{\subseteq} w$, and $z\mathrel{(\powf\pi_i)} x$ that gives $x \mathrel{\powf\pi_j\comp\subseteq\comp(\powf\pi_i)^\op} y$.\qed
|
|
\end{proof}
|
|
|
|
%\begin{lemma}
|
|
% Assuming that $F\pi_1$ and $F\pi_2$ are monotone with respect to $\sqsubseteq$ that is an order over $F$, then the following propositions hold:
|
|
% \begin{enumerate}
|
|
% \item $F\pi_2\comp\sqsupseteq\comp(F\pi_1)^\op\quad=\quad\sqsupseteq\comp F\pi_2\comp(F\pi_1)^\op$
|
|
% \item $F\pi_1\comp\sqsupseteq\comp(F\pi_2)^\op\quad=\quad\sqsupseteq\comp F\pi_1\comp(F\pi_2)^\op$
|
|
% \end{enumerate}
|
|
%\end{lemma}
|
|
%\begin{proof}
|
|
% Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $F\pi_j\comp\sqsupseteq\comp(F\pi_i)^\op\quad=\quad\sqsupseteq\comp F\pi_j\comp(F\pi_i)^\op$.
|
|
%
|
|
% Assuming $x \mathrel{F\pi_j\comp\sqsupseteq\comp(F\pi_i)^\op} y$, then exist $z$ and $z'$ such that
|
|
% \begin{gather*}
|
|
% z \mathrel{(F\pi_i)} x,\\
|
|
% z \mathrel{\sqsubseteq} z',\\
|
|
% z'\mathrel{(F\pi_j)} y.
|
|
% \end{gather*}
|
|
% Then from $z \mathrel{\sqsubseteq} z'$ we get $F\pi_j(z)\mathrel{\sqsubseteq}y$. So, we have $z\mathrel{\sqsupseteq\comp F\pi_j} y$, thus $x\mathrel{\sqsupseteq\comp F\pi_j\comp(F\pi_i)^\op} y$.
|
|
%
|
|
% Now, assuming $x \mathrel{\sqsupseteq\comp F\pi_j\comp(F\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
|
|
% \begin{gather*}
|
|
% z\mathrel{(F\pi_i)} x,\\
|
|
% z \mathrel{(F\pi_j)} y',\\
|
|
% y' \mathrel{\sqsubseteq} y.
|
|
% \end{gather*}
|
|
%\end{proof}
|
|
\begin{prop}
|
|
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ is either the maybe monad, the subdistribution monad or $FX=\powf(X\times A)$, where $A$ is a set of labels, then the following propositions hold:
|
|
\begin{enumerate}
|
|
\item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad\appr\comp F\pi_2\comp(F\pi_1)^\op$
|
|
\item $F\pi_1\comp\appr\comp(F\pi_2)^\op\quad=\quad\appr\comp F\pi_1\comp(F\pi_2)^\op$
|
|
\end{enumerate}
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
%\begin{definition}[Natural Order Structure]\label{def:nat-ord}
|
|
% A \emph{natural order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ such that if $\alpha\appr\beta$ in $\Hom(X,FY)$, $f\c X'\to X$, $g\c Y\to Y'$, then:
|
|
% \begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
|
|
% \item $\alpha\comp f\appr\beta\comp f$ in $\Hom(X',FY)$. \label{item:nat-ord:I}
|
|
% \item $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,FY')$. \label{item:nat-ord:II}
|
|
% \end{enumerate}
|
|
%\end{definition}
|
|
|
|
|
|
%\begin{prop}
|
|
% For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a coliftable order structure $\appr$, the following propositions hold:
|
|
% \begin{enumerate}
|
|
% \item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad\appr\comp F\pi_2\comp(F\pi_1)^\op$
|
|
% \item $F\pi_1\comp\appr\comp(F\pi_2)^\op\quad=\quad\appr\comp F\pi_1\comp(F\pi_2)^\op$
|
|
% \end{enumerate}
|
|
%\end{prop}
|
|
%\begin{proof}
|
|
% Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $F\pi_j\comp\appr\comp(F\pi_i)^\op\quad=\quad\appr\comp F\pi_j\comp(F\pi_i)^\op$.
|
|
%
|
|
% Assuming $x \mathrel{F\pi_j\comp\appr\comp(F\pi_i)^\op} y$, then exist $z$ and $z'$ such that
|
|
% \begin{gather*}
|
|
% z \mathrel{(F\pi_i)} x,\\
|
|
% z \mathrel{\appr} z',\\
|
|
% z'\mathrel{(F\pi_j)} y.
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|
% \end{gather*}
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% Then from $z \mathrel{\appr} z'$ since $\appr$ is a coliftable order structure by~\ref{item:nat-ord:I} we get $F\pi_j(z)\mathrel{\appr}y$. So, we have $z\mathrel{\appr\comp F\pi_j} y$, thus $x\mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$.
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%
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% Now, assuming $x \mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
|
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% \begin{gather*}
|
|
% z\mathrel{(F\pi_i)} x,\\
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% z \mathrel{(F\pi_j)} y',\\
|
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% y' \mathrel{\appr} y.
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|
% \end{gather*}
|
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% Since $\appr$ is a coliftable order structure by~\autoref{def:coliftable-ord} there exists a $w$ such that $z\appr w$ and $F\pi_j(w)=y$. So, we have $w \mathrel{(F\pi_j)} y$, $z\mathrel{\appr} w$, and $z\mathrel{(F\pi_i)} x$ that gives $x \mathrel{F\pi_j\comp\appr\comp(F\pi_i)^\op} y$.\qed
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%\end{proof}
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\begin{prop}\label{prop:all-rel-compa}
|
|
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $\pi_1$ and $\pi_2$ are surjective, assuming that a set-functor $F$ has an order structure $\appr$, the following propositions hold:
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\begin{enumerate}
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\item If $\appr$ is liftable, we have $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad F\pi_2\comp(F\pi_1)^\op\comp\appr$
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\item If $\appr$ is coliftable, we have $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad \appr\comp F\pi_2\comp(F\pi_1)^\op$
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\item If $\appr$ is both liftable and coliftable, all the following are equal:
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\begin{itemize}
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|
\item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad$
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\item $F\pi_2\comp(F\pi_1)^\op\comp\appr$
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|
\item $\appr\comp F\pi_2\comp(F\pi_1)^\op$
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\item $\appr\comp F\pi_2\comp(F\pi_1)^\op\comp\appr$
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|
\end{itemize}
|
|
\end{enumerate}
|
|
\end{prop}
|
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\begin{proof}
|
|
They all follow in an obvious way from~\autoref{lem:liftable} and~\autoref{lem:coliftable}. The last one needs $\appr\comp\appr=\appr$ that comes from transitivity of $\appr$. \qed
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\end{proof}
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\begin{notation}
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|
From now on we denote relators $\bar{F}-\comp\appr$ with $F^\rightarrow$, $\appr\comp\bar{F}-$ with $F^\leftarrow$, and $\appr\comp\bar{F}-\comp\appr$ with $F^\leftrightarrow$.
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\end{notation}
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\begin{prop}\label{prop:lax-relator-full-comm}
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|
For a functor $F$ with a liftable order we have:
|
|
\begin{gather*}
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|
F^\rightarrow r\leq F^\leftarrow r
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|
\end{gather*}
|
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\end{prop}
|
|
\begin{proof}
|
|
Assuming $x \mathrel{(F^\rightarrow r)} y$ and $\bar{F}r=F\pi_2\comp(F\pi_1)^\op$, then there exist $x'$ and $p$ such that $x\appr x'$, $p\mathrel{F\pi_1}x'$, and $p\mathrel{F\pi_2}y$. So, we have $x\appr F\pi_1(p)$ then by liftability there exists $p'$ such that $p'\appr p$, and $F\pi_1(p')=x$. Then from $p'\appr p$ we get $F\pi_2(p')\appr y$ that is equivalent with $p' \mathrel{(\appr\comp F\pi_2)} y$, and then we have $x \mathrel{(\appr\comp F\pi_2\comp(F\pi_1)^\op)} y$ that is $x\mathrel{(F^\leftarrow r)}y$.\qed
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\end{proof}
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|
\begin{cor}
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|
Assuming that $r$ is an $F^\rightarrow$-simulation, if $\appr$ is liftable, then it is an $F^\leftarrow$-simulation as well.
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\end{cor}
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%\begin{example}
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% In the category of sets, subset over the powerset functor is an example of a liftable order. Using~\autoref{lem:set-ord-str} we only prove the case for every $h\in\Hom(1,\powf Z)$, $k\in\Hom(1,\powf Y)$. Additionally, for every $g\c Y\to Z$, such that $h\subseteq\powf g(k)$, we define $k'\in\powf Y$ that that $k'=\{y\mid g(y)\in h\}$. We show that $\powf g(k')=h$.
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%
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% Assuming $z\in\powf g(k')$, then there exists $y'\in \{y\mid g(y)\in h\}$ that $z=g(y')$, so $z\in h$ and $\powf g(k')\subseteq h$.
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%
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% Assuming $z\in h$, since $h\subseteq \powf g(k)$, then $z\in\powf g(k)$. So, there exists $y'\in k$ such that $g(y')=z$. So, by the definition of $k'$ we have $y'\in k'$ that means $z\in\powf g(k')$.\qed
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%\end{example}
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|
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|
\begin{definition}[Natural Relator]
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|
An $F$-relator $\relar$ is called \emph{natural}, whenever for every relation $r\c X\rto Y$, and all functions $f\c A\to X$, and $g\c B\to Y$, we have $\relar (g^\op\comp r\comp f)=(Fg)^\op\comp\relar r\comp Ff$.
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\end{definition}
|
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\begin{definition}[Relaional Connector]
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A natural $F$-relator $\relar$ is a \emph{relational connector}, whenever for every set $X$, $\id_{FX}\subseteq\relar\id_X$.
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\end{definition}
|
|
\begin{definition}[Normal Relator]
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|
An $F$-relator $\relar$ is \emph{normal}, whenever for every set $X$, $\id_{FX}=\relar\id_X$.
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\end{definition}
|
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\begin{prop}
|
|
Every normal relational connector is difunctionally functorial.\qed
|
|
\end{prop}
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|
|
|
\begin{prop}
|
|
Assuming that a set-functor $F$ has a liftable order structure $\appr$, then the relator $F^\leftarrow$ is natural.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\begin{align*}
|
|
F^\leftarrow (g^\op\comp r\comp f)&\\
|
|
=&\appr\comp F(g^\op\comp r\comp f)\\
|
|
=&\appr\comp(Fg)^\op\comp Fr\comp Ff\\
|
|
=&(Fg)^\op\comp\appr\comp Fr\comp Ff&\by{\autoref{lem:liftable}}\\
|
|
=&(Fg)^\op\comp F^\leftarrow r\comp Ff
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
\begin{remark}
|
|
One may wonder if the above proposition is true for relators $F^\rightarrow$ or $F^\leftrightarrow$, where the order structure $\appr$ over $F$ is coliftable. But it may not be the case because the surjectivity condition on the $g$ in~\autoref{def:coliftable-ord} prevents the same reasoning.
|
|
\end{remark}
|
|
\begin{prop}
|
|
Symmetrization of a natural relator, is natural.
|
|
\end{prop}
|
|
\begin{proof}
|
|
Assuming $\relar$ is a natural $F$-relator, we prove that $\hat{\relar}$ is natural as well. Assuming $r\c X\rto Y$, $f\c A\to X$, and $g\c B\to Y$, by naturality of $\relar$ we have $\relar (g^\op\comp r\comp f)=Fg^\op\comp \relar r\comp Ff$. Then we have:
|
|
\begin{align*}
|
|
\hat{\relar}(g^\op\comp r\comp f)&\\
|
|
=&\relar(g^\op\comp r\comp f)\cap(\relar(f^\op\comp r^\op\comp g))^\op\\
|
|
=&(Fg)^\op\comp\relar r\comp Ff\cap((Ff)^\op\comp\relar r^\op\comp Fg)^\op\\
|
|
=&(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff
|
|
\end{align*}
|
|
So we are left to prove
|
|
\begin{gather*}
|
|
(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff=(Fg)^\op\comp(\relar r\cap(\relar r^\op)^\op)\comp Ff
|
|
\end{gather*}
|
|
because then we will have $\hat{\relar}(g^\op\comp r\comp f)=(Fg)^\op\comp\hat{\relar}\comp Ff$. We have
|
|
\begin{align*}
|
|
(t,s)\in &(Fg)^\op\comp\relar r\comp Ff\cap (Fg)^\op\comp(\relar r^\op)^\op\comp Ff,\\
|
|
&\iff(t,s)\in (Fg)^\op\comp\relar r\comp Ff\quad\&\quad (t,s)\in (Fg)^\op\comp(\relar r^\op)^\op\comp Ff,\\
|
|
&\iff Ff(t) \mathrel{(\relar r)} Fg(s) \quad\&\quad Ff(t) \mathrel{(\relar r^\op)^\op} Fg(s),\\
|
|
&\iff Ff(t) \mathrel{(\relar\cap(\relar r^\op)^\op)} Fg(s),\\
|
|
&\iff t\mathrel{((Fg)^\op\comp(\relar\cap(\relar r^\op)^\op)\comp Ff)} s.
|
|
\end{align*}
|
|
It worth noting that when we have $(t,s)\in (Fg)^\op\comp\relar r\comp Ff$, it mean that there exist $t'$ and $s'$ that $(t,t')\in Ff$, $(t',s')\in \relar r$, and $(s',s)\in Fg^\op$. Since $Ff$ and $Fg$ are functions, then $t'=Ff(t)$ and $s'=Fg(s)$, and these are unique elements. The uniqueness enables us in the above reasoning to go from the second line to the third line.\qed
|
|
\end{proof}
|
|
\begin{prop}
|
|
If $\appr$ is an order structure on $F$ that for every sets $X$ and $Y$, $(\Hom(X,FY),\appr)$ is antisymmetric as well (making the posets), then the symmetrization of the left-lax Barr relator of $F$ and $\appr$ is normal.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We denote the left-lax Barr relator with $\relar$. We have:
|
|
\begin{align*}
|
|
\hat{\relar}\id&\\
|
|
=&\relar\id\cap(\relar\id^\op)^\op\\
|
|
=&\appr\comp F\id\cap (\appr\comp F\id^\op)^\op\\
|
|
=&\appr\cap\sappr\\
|
|
=&\id
|
|
\end{align*}\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Since the symmetrization of left-lax Barr relator with a lifatble order structure is natural, and normal, it is a normal relational connector. So, it is a sound and complete relator.
|
|
\end{cor}
|
|
\begin{cor}
|
|
By~\autoref{prop:all-rel-compa}.(1), if $\appr$ is liftable, then the mid-lax Barr relator is a normal relation connector, and thus a sound and complete relator as well.
|
|
\end{cor}
|
|
Perhaps if we can relax the definition of liftable by allowing $g$ to be a relation rather than a function, we can have a lax Barr relator that its symmetrization is a normal lax extension. Well, this idea actually does not work because if $g$ is supposed to be a relation, then $F$ can not be a set-functor, but it should be a functor on $\rel$, and it is already a big assumption.
|
|
\begin{prop}
|
|
For a natural order structure $\appr$ on a set-functor $F$, if for every $f\in\Hom(X,FY)$ we have $Ff\comp\appr=\appr\comp Ff$, then $\appr$ is coliftable.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Investigate if it's true! If it is, is naturality necessary? It is really weird! If having a liftable order structure forces every witness for AM-simulation to have the left cell of the lax diagram as equality, and subset on powerset functor is a liftable order structure, then what is the counter-example number 4 that you have?!}
|
|
\end{proof}
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a difunctionally functorial relator, then the symmetrization of the relator that takes $r\c X\rto Y$ to $\relar r\comp\appr$ is a sound relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
\begin{prop}
|
|
Assuming that $\relar$ is a relator over $F\c\Set\to\Set$, and $\appr_{X}$ and $\appr_{Y}$ are posets over $FX$ and $FY$ respectively, then the symmetrization of the relator that takes $r\c X\rto Y$ to $\appr_{X};\relar r;\appr_{Y}$ is a Barr relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
\todo{Finish.}
|
|
\end{proof}
|
|
\subsection{Symmetrization of a lax Barr relator}
|
|
\begin{prop}
|
|
Assuming $F$ is the maybe functor $FX=X+1$, then the syemmetrization of the left-lax Barr relator of $F$ is a Barr-relator.
|
|
\end{prop}
|
|
\begin{proof}
|
|
We have the following:
|
|
\begin{align*}
|
|
\appr\comp\bar{F}r\cap(\appr\comp\bar{F}r^\op)^\op&\\
|
|
=&\appr\comp\bar{F}r\cap(\appr\comp F\pi_1\comp(F\pi_2)^\op)^\op\\
|
|
=&\appr\comp\bar{F}r\cap F\pi_2\comp(F\pi_1)^\op\comp\sappr\\
|
|
=&\appr\comp\bar{F}r\cap \bar{F}r\comp\sappr
|
|
\end{align*}
|
|
We prove that $\appr\comp\bar{F}r\cap \bar{F}r\comp\sappr=\bar{F}r$. It is obvious that $\bar{F}r\subseteq(\appr\comp\bar{F}r)$, and $\bar{F}r\subseteq(\bar{F}r\comp\sappr)$. Assuming $x\mathrel{(\appr\comp\bar{F}r\cap \bar{F}r\comp\sappr)} y$ then we have $x\mathrel{(\appr\comp\bar{F}r)} y$ and $x\mathrel{(\bar{F}r\comp\sappr)} y$, so by $x\mathrel{(\appr\comp\bar{F}r)} y$ there exists $y'$ such that $x\mathrel{\bar{F}r}y'$ and $y'\appr y$. In fact, $y'\appr y$ gives either of the following cases:
|
|
\begin{itemize}
|
|
\item $y=y'$: In this case we directly have $x\mathrel{\bar{F}r}y$.
|
|
\item $y'=\bot$: In this case we have $x\mathrel{\bar{F}r}\bot$ that means $x\mathrel{(\pi_2+1\comp(\pi_1+1)^\op)}\bot$ so there exists $z$ such that $\pi_2+1(z)=\bot$ that entails $z=\bot$ meaning that $x=\bot$. Recalling $x\mathrel{(\bar{F}r\comp\sappr)} y$, there exists $x'$ such that $x'\appr x$ and $x' \mathrel{\bar{F}r}y$. Since $x=\bot$ then $x'\appr x$ gives $x'=\bot$ that entails $x=x'$. So, from $x'\mathrel{\bar{F}r}y$ we have $x\mathrel{\bar{F}r}y$.
|
|
\end{itemize}
|
|
So, we have $(\appr\comp\bar{F}r)\subseteq\bar{F}r$ that entails $(\appr\comp\bar{F}r\cap \bar{F}r\comp\sappr)\subseteq\bar{F}r$.\qed
|
|
\end{proof}
|
|
\begin{remark}
|
|
With a similar argument we can prove that the symmetrization of a right-lax Barr-relator of $F$ is a Barr-relator.
|
|
\end{remark}
|
|
\begin{lemma}\label{lem:lax-relator-str}
|
|
For a relation $r$, for functions $k_s\c F\pi_1(\bar{F}r)\to\powf FX$ and $k_b\c F\pi_2(\bar{F}r)\to\powf FX$, defined as
|
|
\begin{gather*}
|
|
k_s(x)=\{(x',y)\mid(x,y)\in\bar{F}r,x'\appr x, (x',y)\notin\bar{F}r\},\\
|
|
k_b(x)=\{(x',y)\mid(x,y)\in\bar{F}r,x\appr x', (x',y)\notin\bar{F}r\},
|
|
\end{gather*}
|
|
we have
|
|
\begin{gather*}
|
|
(\bar{F}r)\comp\appr=\bar{F}r\cup(\bigcup_{x\in F\pi_1(\bar{F}r)} k_s(x)),\\
|
|
(\bar{F}r)\comp\sappr=\bar{F}r\cup(\bigcup_{x\in F\pi_1(\bar{F}r)} k_b(x)).
|
|
\end{gather*}
|
|
\end{lemma}
|
|
\begin{proof}
|
|
\todo{Write it down. You have it in your notes.}
|
|
\end{proof}
|
|
\todo{Try $FX=\powf(X^2)$ to see if the symmetrization of its lax Barr relator is a Barr relator. The order is just the set inclusion. See if~\autoref{prop:lax-relator-full-comm} or~\autoref{lem:lax-relator-str} can help!}
|
|
%
|
|
\begin{prop}\label{prop:left-lax-inc-triv}
|
|
For every functor $F\c\Set\to\Set$, we have $\bar{F}\leq F^\leftarrow$.
|
|
\end{prop}
|
|
\begin{proof}
|
|
For a relation $r$, assuming $(x,y)\in \bar{F}r$, since $\appr$ is reflexive we have $y\mathrel{\appr}y$, so we have $x\mathrel{(\bar{F}\comp\appr)} y$ that is $(x,y)\in F^\leftarrow r$.\qed
|
|
\end{proof}
|
|
%
|
|
\subsection{Symmetric relation}
|
|
\begin{prop}
|
|
Assuming that $r$ is a symmetric relation, and it is an $\relar$-simulation on a coalgebra $(X,\alpha)$, then $r$ is an $\hat{\relar}$-bisimulation.
|
|
\end{prop}
|
|
\begin{proof}
|
|
$r$ being an $\relar$-simulation means that $r\leq \alpha^\op\comp\relar r\comp\alpha$, and $r$ being symmetric means that $r^\op=r$. So, we need to prove that $r\leq\alpha^\op\comp\hat{\relar} r\comp\alpha$. We have:
|
|
\begin{align*}
|
|
x \mathrel{(\alpha^\op\comp\hat{\relar} r\comp\alpha)} y&\\
|
|
\iff &x\mathrel{(\alpha^\op\comp(\relar r\cap (\relar r)^\op)\comp\alpha)} y\\
|
|
\iff &\exists x',y',\quad x\mathrel{\alpha}x'\quad\&\quad y\mathrel{\alpha}y'\quad\&\quad x'\mathrel{(\relar r\cap (\relar r)^\op)}y'\\
|
|
\iff &\exists x',y',\quad x\mathrel{\alpha}x'\quad\&\quad y\mathrel{\alpha}y'\quad\&\quad x'\mathrel{\relar r}y'\quad\&\quad x'\mathrel{(\relar r)^\op}y'\\
|
|
\iff &\exists x',y',\quad x\mathrel{\alpha}x'\quad\&\quad y\mathrel{\alpha}y'\quad\&\quad x'\mathrel{\relar r}y'\quad\&\quad y'\mathrel{\relar r}x'
|
|
\end{align*}
|
|
Now, assuming $x\mathrel{r}y$ gives us $x\mathrel{(\alpha^\op\comp\relar r\comp \alpha)} y$ that is equivalent with saying that exist $x'$ and $y'$ such that $x\mathrel{\alpha}x'$, $y\mathrel{\alpha}y'$, and $x'\mathrel{\relar r}y'$. Since $r$ is symmetric, we have $y\mathrel{r} x$ that means that exist $x''$ and $y''$ such that $x\mathrel{\alpha}x''$, $y\mathrel{\alpha}y''$, and $y''\mathrel{\relar r}x''$. On the other hand since $\alpha$ is a function, we have $x''=x'$ and $y''=y'$, so we have $y'\mathrel{\relar r}x'$ that ultimately gives $x\mathrel{(\alpha^\op\comp\hat{\relar}r\comp\alpha)}y$. So, $r$ is an $\hat{\relar}$-bisimulation as well.\qed
|
|
\end{proof}
|
|
\begin{cor}
|
|
Recalling~\autoref{prop:left-lax-inc-triv}, for a functor $F\c\Set\to\Set$, assuming that $F^\leftarrow\leq\bar{F}$, we get $F^\leftarrow=\bar{F}$. So, if $r$ is symmetric, and it is a $F^\leftarrow$-simulation, then it is a $\bar{F}$-bisimulation.
|
|
\end{cor}
|
|
\end{document}
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