Compare commits
2 Commits
81da58d639
...
2c4736dd54
| Author | SHA1 | Date | |
|---|---|---|---|
 partowp
|
2c4736dd54 | ||
|
partowp
|
38d506650f |
+151
-72
@@ -293,6 +293,7 @@
|
||||
\newcommand{\rto}{\mathrel{\tikz{\draw[-{Stealth}] (0.04,0) -- (0.4,0); \draw (0.17,0.07) -- (0.17,-0.07);\draw (0.04,0) -- (0,-0.07);\draw (0.04,0) -- (0,0.07);}}}
|
||||
\newcommand{\powf}{\mathcal{P}}
|
||||
\newcommand{\sappr}{\sqsupseteq}
|
||||
\newcommand{\Dom}{\mathsf{Dom}}
|
||||
|
||||
\newcommand{\simeet}{%
|
||||
\mathbin{%
|
||||
@@ -2192,18 +2193,26 @@ Now, we prove our main statement.
|
||||
\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.
|
||||
\subsection{Maybe Functor}
|
||||
We prove that symmetric simulation is a bisimulation for the case that $FX=X+1$. First, we prove it for $\Set$. The order structure that we can define for this functor is that for a set $X$, the order is $\id_{X+1}\cup\{(\bot,x)\mid x\in X\}$.\\
|
||||
Now, we prove that the given order on the maybe functor is a liftable one.
|
||||
We prove that symmetric simulation is a bisimulation for the case that $FX=X+1$. First, we prove it for $\Set$. 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 $f$).
|
||||
Now, we prove that the given order on the maybe functor is liftable and coliftable.
|
||||
\begin{lemma}\label{lem:maybe-lif}
|
||||
The order structure on the set-functor $FX=X+1$ is a liftable order.
|
||||
\end{lemma}
|
||||
\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 $h\appr Fg(k)$, 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}
|
||||
% 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}
|
||||
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{lemma}\label{lem:maybe-colif}
|
||||
The order structure on the set-functor $FX=X+1$ is a coliftable order.
|
||||
@@ -2286,41 +2295,79 @@ So, proven by Dubut, for every AM-simulation relation over a coalgebra $(X,\alph
|
||||
% 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 $Fp_2\comp\sigma(x_1,x_2)=\bot$: In this case, 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*}
|
||||
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).
|
||||
\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*}
|
||||
\end{itemize}\qed
|
||||
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. %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.}
|
||||
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=\&]
|
||||
@@ -2335,50 +2382,82 @@ Assuming $f,g\in\Hom(X,Y+1)$, and that $f'\c X_f\rightarrowtail X$ and $g'\c X_g
|
||||
The order structure on the functor $F\c\BC\to\BC$ defined as $FX=X+1$ is a liftable order.
|
||||
\end{lemma}
|
||||
\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{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]
|
||||
\& 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*}
|
||||
\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 \& 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]
|
||||
% {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*}
|
||||
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.}
|
||||
%%\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:
|
||||
@@ -3101,7 +3180,7 @@ Perhaps if we can relax the definition of liftable by allowing $g$ to be a relat
|
||||
\begin{proof}
|
||||
\todo{Finish.}
|
||||
\end{proof}
|
||||
\subsection{Symmetrization of a Barr relator}
|
||||
\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}
|
||||
|
||||
Reference in New Issue
Block a user