Merge branch 'master' of git.wlog.site:pouya/coalgebraic-simulation
This commit is contained in:
commit
9b511e74b3
178
draft/draft.tex
178
draft/draft.tex
@ -1025,12 +1025,12 @@ We show this relation with $F_\rel(R,X)$.
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X \& R \& X \\
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FX \& (FR)^\dagger \& FX
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\arrow["\alpha"', from=1-1, to=2-1]
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\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
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\arrow["\appr"{marking, allow upside down}, draw=none, from=1-1, to=2-2]
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\arrow["{p_1}"', from=1-2, to=1-1]
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\arrow["{p_2}", from=1-2, to=1-3]
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\arrow["\sigma", from=1-2, to=2-2]
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\arrow["\alpha", from=1-3, to=2-3]
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\arrow["\sqsubseteq"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
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\arrow["\appr"{marking, allow upside down}, draw=none, from=2-2, to=1-3]
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\arrow["{{(Fp_1)^\dagger}}", from=2-2, to=2-1]
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\arrow["{{(Fp_2)^\dagger}}"', from=2-2, to=2-3]
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\end{tikzcd}
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@ -1123,7 +1123,7 @@ We show this relation with $F_\rel(R,X)$.
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\end{proof}
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%
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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.
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We set $R=\{(A,B),(B,A),(C_1,C_2),(C_2,C_1),(C'_2,C_2),(C_2,C'_2),(C_2,C_2)\}$, $F=\mathbf{Id}$, $\sqsubseteq=\Delta\cup\{(C_1,C_2),(C_2,C'_2)\}$, and the coalgebra $\alpha$ is defined with the following set of reductions:
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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:
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\begin{gather*}
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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
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\end{gather*}
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@ -2194,13 +2194,28 @@ For every relation $r\rto X\to Y$ $\emre r=\subseteq\;\emre r=\emre r\;\subseteq
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Barr relator is a generalization of the Egli-Milner relator, where the functor is generalized.
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\begin{definition}
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\begin{definition}[Barr relator]
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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:
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\begin{gather*}
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\bar{F}r=F\pi_2\comp(F\pi_1)^\op
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\end{gather*}
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\end{definition}
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\begin{prop}
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For every set-functor $F$, the barr relator $\bar{F}$ is symmetric.
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\end{prop}
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\begin{proof}
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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:
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\begin{align*}
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\hat{\bar{F}}r&\\
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=&\bar{F}r\cap(\bar{F}r^\op)^\op\\
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=&\bar{F}(\pi_2\comp\pi_1^\op)\cap(\bar{F}(\pi_2\comp\pi_1^\op)^\op)^\op\\
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=&\bar{F}(\pi_2\comp\pi_1^\op)\cap(\bar{F}(\pi_1\comp\pi_2^\op))^\op\\
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=&F\pi_2\comp(F\pi_1)^\op\cap(F\pi_1\comp (F\pi_2)^\op)^\op\\
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=&F\pi_2\comp(F\pi_1)^\op\cap F\pi_2\comp (F\pi_1)^\op\\
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=&F\pi_2\comp(F\pi_1)^\op\\
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=&\bar{F}r
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\end{align*}\qed
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\end{proof}
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\begin{prop}
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$\hat{L}$ is a Barr relator.
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\end{prop}
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@ -2208,66 +2223,67 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
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\todo{Finish.}
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\end{proof}
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\begin{definition}[One-sided Barr relator]
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Given a relation $r$, and take a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$. Assuming that $\sqsubseteq$ is a partial order over a functor $F$, then the relator over $F$ and shown with $\overrightarrow{F}$ is a \emph{one-sided Barr relator} iff we have:
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\begin{definition}[Mid-lax Barr relator]
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Given a relation $r$, and take a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ that $r=\pi_2\comp\pi_1^\op$. Assuming that $\appr$ is a partial order over a functor $F$, then the relator over $F$ and shown with $\overrightarrow{F}$ is a \emph{mid-lax Barr relator} if we have:
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% 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:
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\begin{gather*}
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\overrightarrow{F}r=F\pi_2\comp\sappr\comp(F\pi_1)^\op
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\overrightarrow{F}r=F\pi_2\comp\appr\comp(F\pi_1)^\op
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\end{gather*}
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\end{definition}
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\begin{prop}
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For every functor $F\c\Set\to\Set$, the symmetrization of the one-sided Bar relator is equal with the Barr relator.
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For every functor $F\c\Set\to\Set$, the symmetrization of the mid-lax Bar relator is equal with the Barr relator.
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\end{prop}
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\begin{proof}
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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:
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\begin{align*}
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s \;\hat{\overrightarrow{F}}r\; t&\\
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&\iff s\;\overrightarrow{F}r\;t\qquad\&\qquad s \;(\overrightarrow{F}r^\op)^\op\; t\\
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&\iff s\;F\pi_2\comp\sappr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;(F\pi_1\comp\sappr\comp(F\pi_2)^\op)^\op\;t\\
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&\iff s\;F\pi_2\comp\sappr\comp(F\pi_1)^\op\;t \qquad\&\qquad s\;F\pi_2\comp\appr\comp(F\pi_1)^\op\;t
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\end{align*}
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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:
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\begin{gather*}
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w\;F\pi_2\comp\sappr\; t \qquad\&\qquad w\;F\pi_2\comp\appr\; t
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\end{gather*}
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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:
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\begin{align*}
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&w\;\sappr\; v \qquad\&\qquad w\;\appr\; v\\
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\iff&(F\pi_1)^\op(s)\;\sappr\; (F\pi_2^\op)(t) \qquad\&\qquad (F\pi_1)^\op(s)\;\appr\; (F\pi_2^\op)(t)\\
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\iff&(F\pi_1)^\op(s)\;=\; (F\pi_2^\op)(t)\\
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\iff&s\; F\pi_2\comp(F\pi_1)^\op\;t\\
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\iff&s\;\bar{F}r\;t
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\end{align*}
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So we have $\hat{\overrightarrow{F}}\leq \bar{F}$.
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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.
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% 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:
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% \begin{align*}
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% s \;\hat{\overrightarrow{F}}r\; t&\\
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% &\iff s\;\overrightarrow{F}r\;t\qquad\&\qquad s \;(\overrightarrow{F}r^\op)^\op\; t\\
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% &\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\\
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% &\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
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% \end{align*}
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% 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:
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% \begin{gather*}
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% w\;F\pi_2\comp\appr\; t \qquad\&\qquad w\;F\pi_2\comp\appr\; t
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% \end{gather*}
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% 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:
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% \begin{align*}
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% &w\;\appr\; v \qquad\&\qquad w\;\appr\; v\\
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% \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)\\
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% \iff&(F\pi_1)^\op(s)\;=\; (F\pi_2^\op)(t)\\
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% \iff&s\; F\pi_2\comp(F\pi_1)^\op\;t\\
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% \iff&s\;\bar{F}r\;t
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% \end{align*}
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% So we have $\hat{\overrightarrow{F}}\leq \bar{F}$.
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% 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.
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\todo{Finish.}
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\end{proof}
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\begin{lemma}
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The following propositions hold:
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\begin{prop}
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For a span $(\pi_1\c A\to X,\pi_2\c A\to Y)$ the following propositions hold:
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\begin{enumerate}
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\item $\powf\pi_2\comp\supseteq\comp(\powf\pi_1)^\op\quad=\quad\supseteq\comp \powf\pi_2\comp(\powf\pi_1)^\op$
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\item $\powf\pi_1\comp\supseteq\comp(\powf\pi_2)^\op\quad=\quad\supseteq\comp \powf\pi_1\comp(\powf\pi_2)^\op$
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\item $\powf\pi_2\comp\subseteq\comp(\powf\pi_1)^\op\quad=\quad\subseteq\comp \powf\pi_2\comp(\powf\pi_1)^\op$
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\item $\powf\pi_1\comp\subseteq\comp(\powf\pi_2)^\op\quad=\quad\subseteq\comp \powf\pi_1\comp(\powf\pi_2)^\op$
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\end{enumerate}
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\end{lemma}
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\end{prop}
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\begin{proof}
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Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $\powf\pi_j\comp\supseteq\comp(\powf\pi_i)^\op\quad=\quad\supseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op$.
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Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $\powf\pi_j\comp\subseteq\comp(\powf\pi_i)^\op\quad=\quad\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op$.
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Assuming $x \mathrel{\powf\pi_j\comp\supseteq\comp(\powf\pi_i)^\op} y$, then exist $z$ and $z'$ such that
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Assuming $x \mathrel{\powf\pi_j\comp\subseteq\comp(\powf\pi_i)^\op} y$, then exist $z$ and $z'$ such that
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\begin{gather*}
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z \mathrel{(\powf\pi_i)} x,\\
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z \mathrel{\subseteq} z',\\
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z'\mathrel{(\powf\pi_j)} y.
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\end{gather*}
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Then from $z \mathrel{\subseteq} z'$ we get $\powf\pi_j(z)\mathrel{\subseteq}y$. So, we have $z\mathrel{\supseteq\comp \powf\pi_j} y$, thus $x\mathrel{\supseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$.
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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 $x\mathrel{\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$.
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Now, assuming $x \mathrel{\supseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
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Now, assuming $x \mathrel{\subseteq\comp \powf\pi_j\comp(\powf\pi_i)^\op} y$, then there exist $z$ and $y'$ such that
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\begin{gather*}
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z\mathrel{(\powf\pi_i)} x,\\
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z \mathrel{(\powf\pi_j)} y',\\
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y' \mathrel{\subseteq} y.
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\end{gather*}
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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 $x \mathrel{\powf\pi_j\comp\supseteq\comp(\powf\pi_i)^\op} y$.\qed
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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
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\end{proof}
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%\begin{lemma}
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@ -2295,9 +2311,87 @@ Barr relator is a generalization of the Egli-Milner relator, where the functor i
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% y' \mathrel{\sqsubseteq} y.
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% \end{gather*}
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%\end{proof}
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\begin{prop}
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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 relator that takes $r\c X\rto Y$ to $\appr_{X};\relar r;\appr_{Y}$ is a Barr relator.
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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:
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\begin{enumerate}
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\item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad\appr\comp F\pi_2\comp(F\pi_1)^\op$
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\item $F\pi_1\comp\appr\comp(F\pi_2)^\op\quad=\quad\appr\comp F\pi_1\comp(F\pi_2)^\op$
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\end{enumerate}
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\end{prop}
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\begin{proof}
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\todo{Finish.}
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\end{proof}
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\begin{definition}[Cogood Order Structure]\label{def:cogood}
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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:
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\begin{enumerate}[label=(\Roman*), ref=(\Roman*)]
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\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}
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\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}
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\end{enumerate}
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\end{definition}
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\begin{prop}
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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:
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\begin{enumerate}
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\item $F\pi_2\comp\appr\comp(F\pi_1)^\op\quad=\quad\appr\comp F\pi_2\comp(F\pi_1)^\op$
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\item $F\pi_1\comp\appr\comp(F\pi_2)^\op\quad=\quad\appr\comp F\pi_1\comp(F\pi_2)^\op$
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\end{enumerate}
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\end{prop}
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\begin{proof}
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Without loss of generality, we assume $i,j\in\{1,2\}$, and $i\neq j$, and prove $F\pi_j\comp\appr\comp(F\pi_i)^\op\quad=\quad\appr\comp F\pi_j\comp(F\pi_i)^\op$.
|
||||
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||||
Assuming $x \mathrel{F\pi_j\comp\appr\comp(F\pi_i)^\op} y$, then exist $z$ and $z'$ such that
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\begin{gather*}
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z \mathrel{(F\pi_i)} x,\\
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z \mathrel{\appr} z',\\
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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 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 $x\mathrel{\appr\comp F\pi_j\comp(F\pi_i)^\op} y$.
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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
|
||||
\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 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
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\end{proof}
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\begin{prop}
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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:
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\begin{enumerate}
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\item $F\pi_2\comp\sappr\comp(F\pi_1)^\op\quad=\quad F\pi_2\comp(F\pi_1)^\op\comp\appr$
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\item $F\pi_1\comp\sappr\comp(F\pi_2)^\op\quad=\quad F\pi_1\comp(F\pi_2)^\op\comp\appr$
|
||||
\end{enumerate}
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||||
\end{prop}
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\begin{proof}
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\todo{Finish.}
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||||
\end{proof}
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||||
\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 $F\pi_2\comp\appr\comp\sappr\comp(F\pi_1)^\op\quad=\quad \appr\comp F\pi_2\comp(F\pi_1)^\op\comp\appr$
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||||
\item $F\pi_1\comp\appr\comp\sappr\comp(F\pi_2)^\op\quad=\quad \appr\comp F\pi_1\comp(F\pi_2)^\op\comp\appr$
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||||
\end{enumerate}
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||||
\end{prop}
|
||||
\begin{proof}
|
||||
\todo{Finish.}
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||||
\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:
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||||
\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.}
|
||||
\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.
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\end{prop}
|
||||
\begin{proof}
|
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\todo{Finish.}
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||||
\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.
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||||
\end{prop}
|
||||
\begin{proof}
|
||||
\todo{Finish.}
|
||||
|
||||
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