From 2c4736dd54dbafb2a06b49ffa2d900f01eeb7731 Mon Sep 17 00:00:00 2001 From: partowp Date: Sat, 18 Jul 2026 01:47:10 +0100 Subject: [PATCH] no maybe anymore --- draft/draft.tex | 216 ++++++++++++++++++++++++++++++++---------------- 1 file changed, 147 insertions(+), 69 deletions(-) diff --git a/draft/draft.tex b/draft/draft.tex index 271cbad..6eaec3a 100644 --- a/draft/draft.tex +++ b/draft/draft.tex @@ -2193,8 +2193,8 @@ 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 sets $X$ and $Y$, and functions $f,g\c X\to Y+1$ we have $f\appr g$ whenever $\Dom(f)\subseteq\Dom(g)$ ($\Dom(f)$ is the domain of a function $f$). -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} @@ -2204,7 +2204,15 @@ Now, we prove that the given order on the maybe functor is a liftable one. % \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 $h\appr (g+1)\comp k$ +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. @@ -2287,41 +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 $p_2+1\comp\sigma(x_1,x_2)=\bot$: 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 $\alpha(x_1)=(p_1+1)\comp\sigma(x_1,x_2)$\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. %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=\&] @@ -2336,50 +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} + \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} -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{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: