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ACV-abstract-2026/sym-sim.tex
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@ -241,7 +241,7 @@ We ask:
\begin{enumerate}
\item How to construct relators for established notions of simulation?
\item How to generally prove that ensuing symmetric simulations are bisimulations?
\item How alternative notions of simulations align with the relator-based one?
\item How alternative notions of simulation align with the relator-based one?
\end{enumerate}
\paragraph{Relaxing Barr Relators} %\label{sec:}

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@ -2321,68 +2321,135 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
\begin{proof}
\todo{Finish.}
\end{proof}
\begin{definition}[Cogood Order Structure]\label{def:cogood}
A \emph{cogood order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ such that:
\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 If $\alpha\appr\beta$ in $\Hom(X,FY)$, $g\c Y\to Y'$, then $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,Y')$.\label{item:cogood:I}
\item If $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $Fg\comp k\appr h $ in $\Hom(X,FZ)$, then there is $k'\c X\to FY$ such that $k\appr k'$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.\label{item:cogood:II}
\item $\alpha\comp f\appr\beta\comp f$ in $\Hom(X',Y)$. \label{item:nat-ord:I}
\item $Fg\comp\alpha\appr Fg\comp\beta$ in $\Hom(X,Y')$. \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 cogood 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 $F\pi_j\comp\appr\comp(F\pi_i)^\op\quad=\quad\appr\comp F\pi_j\comp(F\pi_i)^\op$.
\begin{definition}[Good Order Structure]\label{def:good-ord}
A \emph{good order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $h\appr Fg\comp k$ in $\Hom(X,FZ)$, then there is $k'\c X\to FY$ such that $k'\appr k$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
\end{definition}
\begin{definition}[Cogood Order Structure]\label{def:cogood-ord}
A \emph{cogood order structure} on a functor $F$ is a preorder $\appr$ on each Hom-set of the form $\Hom(X,FY)$ that is a natural order structure, and if $h\c X\to FZ$, $k\c X\to FY$, $g\c Y\to Z$, $Fg\comp k\appr h $ in $\Hom(X,FZ)$, then there is $k'\c X\to FY$ such that $k\appr k'$ in $\Hom(X,FY)$ and $h=Fg\comp k'$.
\end{definition}
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.
\end{gather*}
Then from $z \mathrel{\appr} z'$ since $\appr$ is a cogood order structure by~\autoref{def:cogood}.\ref{item:cogood:I} we get $F\pi_j(z)\mathrel{\appr}y$. So, we have $z\mathrel{\appr\comp F\pi_j} y$, thus $x\mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$.
\begin{lemma}\label{lem:good}
Assuming that a a functor $F$ has an order structure $\appr$ that is good, then for every $f\in\Hom(X,FY)$ 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 good, and thus natural, by~\autoref{def:nat-ord}, from $s\appr t$ we get $x\appr Ff(t)$ that is $t\mathrel{\sappr\comp Ff} x$.
Now, assuming $x \mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
Assuming $t\mathrel{\sappr\comp Ff} x$, there exists $y$ such that $Ff(t)=y$ and $y\sappr x$. By~\autoref{def:good-ord} since $Ff(t)\sappr x$ there exists $s$ that $t\sappr s$ 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*}
z\mathrel{(F\pi_i)} x,\\
z \mathrel{(F\pi_j)} y',\\
y' \mathrel{\appr} y.
(Ff\comp\appr)^\op=\sappr\comp(Ff)^\op,\\
(\appr\comp Ff)^\op=(Ff)^\op\comp\sappr.
\end{gather*}
Since $\appr$ is a cogood order structure by~\autoref{def:cogood}.\ref{item:cogood:II} 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
So it follows directly from applying $\op$ on both sides of $(I)$.\qed
\end{proof}
\begin{lemma}\label{lem:cogood}
Assuming that a a functor $F$ has an order structure $\appr$ that is cogood, then for every $f\in\Hom(X,FY)$ 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 cogood, 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:cogood-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}
% For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a cogood 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.
% \end{gather*}
% Then from $z \mathrel{\appr} z'$ since $\appr$ is a cogood 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 $x\mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$.
%
% Now, assuming $x \mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
% \begin{gather*}
% z\mathrel{(F\pi_i)} x,\\
% z \mathrel{(F\pi_j)} y',\\
% y' \mathrel{\appr} y.
% \end{gather*}
% Since $\appr$ is a cogood order structure by~\autoref{def:cogood-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
%\end{proof}
\begin{prop}
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a cogood order structure $\appr$, the following propositions hold:
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has an order structure $\appr$, the following propositions hold:
\begin{enumerate}
\item $F\pi_2\comp\sappr\comp(F\pi_1)^\op\quad=\quad F\pi_2\comp(F\pi_1)^\op\comp\appr$
\item $F\pi_1\comp\sappr\comp(F\pi_2)^\op\quad=\quad F\pi_1\comp(F\pi_2)^\op\comp\appr$
\item If $\appr$ is good, we have $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad F\pi_2\comp(F\pi_1)^\op\comp\appr$
\item If $\appr$ is cogood, we have $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad \appr\comp F\pi_2\comp(F\pi_1)^\op$
\item If $\appr$ is both good and cogood, all the following are equal:
\begin{itemize}
\item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad$
\item $F\pi_2\comp(F\pi_1)^\op\comp\appr$
\item $\appr\comp F\pi_2\comp(F\pi_1)^\op$
\item $\appr\comp F\pi_2\comp(F\pi_1)^\op\comp\appr$
\end{itemize}
\end{enumerate}
\end{prop}
\begin{proof}
\todo{Finish.}
They all follow in an obvious way from~\autoref{lem:good} and~\autoref{lem:cogood}. The last one needs $\appr\comp\appr=\appr$ that comes from transitivity of $\appr$. \qed
\end{proof}
\begin{example}
In the category of sets, subset over the powerset functor is an example of a good order. For every $h\c X\to\powf Z$, $k\c X\to \powf Y$, and $g\c Y\to Z$, such that $h\subseteq\powf g\comp k$, there exists $k'\c X\to\powf Y$ that for every $x\in X$, $k'(x)=k(x)\setminus\{y\mid g(y)\notin h(x)\}$. We show that for every $x\in X$, we have $\powf g\comp k'(x)=h(x)$.
Assuming $z\in\powf g(k'(x))$, then there exist $y'\in k(x)\setminus \{y\mid g(y)\notin h(x)\}$ that $g(y')=z$, so $y'\in k(x)$, $y'\notin\{y\mid g(y)\notin h(x)\}$. $y'\notin\{y\mid g(y)\notin h(x)\}$ means that $g(y')\in h(x)$, thus $\powf g\comp k'\subseteq h$.
Assuming $z\in h(x)$, since $h\subseteq \powf g\comp k$, then $z\in\powf g\comp k(x)$. So, there exists $y'$ such that $y'\in k(x)$ and $g(y')=z$. Since $g(y')\in h(x)$ then $y'\notin\{y\mid g(y)\notin h(x)\}$ meaning that $y'\in k'(x)$ that means $z\in \powf g(k'(x))$, thus $h\subseteq \powf g\comp k'$.\qed
\end{example}
\begin{example}
In the category of sets, subset over the powerset functor is NOT an example of a cogood 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{example}
The previous example suggests that cogoodness is a strong condition. Perhaps it can be limited to be meaningful. Maybe the morphism $g$ can be limited to legs of a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, where for a relation $r$, $r=\pi_2\comp\pi_1^\op$.
\begin{prop}
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a cogood order structure $\appr$, the following propositions hold:
\begin{enumerate}
\item $F\pi_2\comp\appr\comp\sappr\comp(F\pi_1)^\op\quad=\quad \appr\comp F\pi_2\comp(F\pi_1)^\op\comp\appr$
\item $F\pi_1\comp\appr\comp\sappr\comp(F\pi_2)^\op\quad=\quad \appr\comp F\pi_1\comp(F\pi_2)^\op\comp\appr$
\end{enumerate}
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 cogood.
\end{prop}
\begin{proof}
\todo{Finish.}
\end{proof}
\begin{prop}
For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$, assuming that $F$ has a good order structure $\appr$, the following propositions hold:
\begin{enumerate}
\item $F\pi_2\comp\sappr\comp(F\pi_1)^\op\quad=\quad\appr\comp F\pi_2\comp(F\pi_1)^\op$
\item $F\pi_1\comp\sappr\comp(F\pi_2)^\op\quad=\quad\appr\comp F\pi_1\comp(F\pi_2)^\op$
\end{enumerate}
\end{prop}
\begin{proof}
\todo{Finish.}
\todo{Investigate if it's true! If it is, is naturality necessary? It is really weird! If having a good 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 good 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.
@ -2396,7 +2463,6 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
\begin{proof}
\todo{Finish.}
\end{proof}
\end{document}