The concrete proof for maybe revisited
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partowp
+93
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@@ -2206,7 +2206,7 @@ Now, we prove that the given order on the maybe functor is a liftable one.
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\end{itemize}
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\end{proof}
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\begin{lemma}\label{lem:maybe-colif}
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The order structure on the set-functor $FX=X+1$ is a liftable order.
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The order structure on the set-functor $FX=X+1$ is a coliftable order.
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\end{lemma}
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\begin{proof}
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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:
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@@ -2217,64 +2217,106 @@ Now, we prove that the given order on the maybe functor is a liftable one.
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\end{itemize}
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\end{proof}
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So, proven by Dubut, for every symmetric AM-simulation relation over a coalgebra $(X,\alpha)$ of the maybe functor, we have a witness $\sigma\c R\to R+1$ such that $\alpha\comp p_1=Fp_1\comp\sigma$.
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\begin{lemma}\label{lem:maybe-func-set}
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Assuming that $R$ is a symmetric AM-simulation over an $F$-coalgebra $(X,\alpha)$ that $FX=X+1$, then for every $(x_1,x_2)\in R$
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\begin{gather*}
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\alpha(x_1)\in X\Rightarrow \alpha(x_2)\in X\\
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\alpha(x_1)=\bot \Rightarrow \alpha(x_2)=\bot
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\end{gather*}
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\end{lemma}
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\begin{proof}
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By~\autoref{prop:alph-prod-dubut} and~\autoref{lem: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.}:
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\begin{enumerate}
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\item $\alpha(x_1)=Fp_1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1}
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\item $\alpha(x_2)=Fp_1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2}
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\item $\alpha(x_1)\sappr Fp_2\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3}
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\item $\alpha(x_2)\sappr Fp_2\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4}
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\end{enumerate}
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Now, we have two cases:
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\begin{itemize}
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\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 $\alpha(x_2)\in X$.
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\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
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\end{itemize}
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\end{proof}
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%\begin{proof}
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% 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
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%\begin{lemma}\label{lem:maybe-func-set}
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% Assuming that $R$ is a symmetric AM-simulation over an $F$-coalgebra $(X,\alpha)$ that $FX=X+1$, then for every $(x_1,x_2)\in R$
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% \begin{gather*}
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% \alpha(x_1)\in X\quad\&\quad \alpha(x_2)=\bot,
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% \alpha(x_1)\in X\Rightarrow \alpha(x_2)\in X\\
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% \alpha(x_1)=\bot \Rightarrow \alpha(x_2)=\bot
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% \end{gather*}
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%\end{lemma}
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%\begin{proof}
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% By~\autoref{prop:alph-prod-dubut} and~\autoref{lem: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.}:
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%% \begin{enumerate}
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%% \item $\alpha(x_1)=Fp_1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1}
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%% \item $\alpha(x_2)=Fp_1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2}
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%% \item $\alpha(x_1)\sappr Fp_2\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3}
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%% \item $\alpha(x_2)\sappr Fp_2\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4}
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%% \end{enumerate}
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% For every $(x_1,x_2)$ we have
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% \begin{enumerate}
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% \item $\alpha(x_1)=Fp_1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
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% \item $\alpha(x_2)\sappr Fp_2\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
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%% \end{enumerate}
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% and since $R$ is symmetric, we have
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%% \begin{enumerate}
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% \item $\alpha(x_2)=Fp_1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
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% \item $\alpha(x_1)\sappr Fp_2\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
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% \end{enumerate}
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% Now, we have two cases:
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% \begin{itemize}
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% \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 $\alpha(x_2)\in X$.
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% \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
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% \end{itemize}
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%\end{proof}
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%%\begin{proof}
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%% 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
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%% \begin{gather*}
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%% \alpha(x_1)\in X\quad\&\quad \alpha(x_2)=\bot,
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%% \end{gather*}
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%% or
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%% \begin{gather*}
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%% \alpha(x_1)=\bot\quad\&\quad \alpha(x_2)\in X.
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%% \end{gather*}
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%% 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
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%%
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%% 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
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%%\end{proof}
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%
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%\begin{prop}\label{prop:sym-sim-bis-maybe}
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% 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.
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%\end{prop}
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%\begin{proof}
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% By~\autoref{lem:maybe-func-set} for every $(x_1,x_2)\in R$, we either have
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% \begin{gather*}
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% \alpha(x_1)=\bot\quad\&\quad \alpha(x_2)=\bot,
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% \end{gather*}
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% or
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% \begin{gather*}
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% \alpha(x_1)=\bot\quad\&\quad \alpha(x_2)\in X.
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% \alpha(x_1)\in X\quad\&\quad \alpha(x_2)\in X.
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% \end{gather*}
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% 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
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%
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% 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
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% We define $\beta$ as a witness for $R$ to be an AM-bisimulation as follows:
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% \begin{gather*}
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% \beta(x_1,x_2)=
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% \begin{cases}
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% \bot & \alpha(x_1)=\bot\;\&\; \alpha(x_2)=\bot \\
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% (\alpha(x_1),\alpha(x_2)) & \alpha(x_1)\in X\;\&\;\alpha(x_2)\in X
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% \end{cases}
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% \end{gather*}
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% 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 $Fp_i\comp\beta(x_1,x_2)=\alpha(x_i)\in X$.
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% 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 $\beta(x_1,x_2)\in R$.
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% \qed
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%\end{proof}
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\begin{prop}\label{prop:sym-sim-bis-maybe}
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\begin{prop}\label{prop:sym-sim-bis-may}
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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.
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\end{prop}
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\begin{proof}
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By~\autoref{lem:maybe-func-set} for every $(x_1,x_2)\in R$, we either have
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\begin{gather*}
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\alpha(x_1)=\bot\quad\&\quad \alpha(x_2)=\bot,
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\end{gather*}
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or
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\begin{gather*}
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\alpha(x_1)\in X\quad\&\quad \alpha(x_2)\in X.
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\end{gather*}
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We define $\beta$ as a witness for $R$ to be an AM-bisimulation as follows:
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\begin{gather*}
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\beta(x_1,x_2)=
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\begin{cases}
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\bot & \alpha(x_1)=\bot\;\&\; \alpha(x_2)=\bot \\
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(\alpha(x_1),\alpha(x_2)) & \alpha(x_1)\in X\;\&\;\alpha(x_2)\in X
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\end{cases}
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\end{gather*}
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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 $Fp_i\comp\beta(x_1,x_2)=\alpha(x_i)\in X$.
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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 $\beta(x_1,x_2)\in R$.
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\qed
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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
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\begin{enumerate}
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\item $\alpha(x_1)=p_1+1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-1},
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\item $\alpha(x_2)\sappr p_2+1\comp\sigma(x_1,x_2)$\label{eq:maybe-func-set-4},\\
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% \end{enumerate}
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and since $R$ is symmetric, we have
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% \begin{enumerate}
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\item $\alpha(x_2)=p_1+1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-2},
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\item $\alpha(x_1)\sappr p_2+1\comp\sigma(x_2,x_1)$\label{eq:maybe-func-set-3},
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\end{enumerate}
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Followed by $\alpha(x_2)\sappr p_2+1\comp\sigma(x_1,x_2)$ we have the following cases:
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\begin{itemize}
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\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.
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\item $Fp_2\comp\sigma(x_1,x_2)=\bot$: In this case, we have
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\begin{align*}
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p_2+1\comp\sigma(x_1,x_2)=\bot,&\\
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\Rightarrow&\sigma(x_1,x_2)=\bot,\\
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\Rightarrow&p_1+1\comp\sigma(x_1,x_2)=\bot,\\
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\Rightarrow&\alpha(x_1)=\bot,&\eqref{eq:maybe-func-set-1}\\
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\Rightarrow&p_2+1\comp\sigma(x_2,x_1)=\bot,&\eqref{eq:maybe-func-set-3}\\
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\Rightarrow&\sigma(x_2,x_1)=\bot,\\
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\Rightarrow&p_1+1\comp\sigma(x_2,x_1)=\bot,\\
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\Rightarrow&\alpha(x_2)=\bot,&\eqref{eq:maybe-func-set-2}\\
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\Rightarrow&\alpha(x_2)=p_1+1\comp\sigma(x_1,x_2)\bot.
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\end{align*}
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\end{itemize}\qed
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\end{proof}
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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.}
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@@ -3049,9 +3091,7 @@ Perhaps if we can relax the definition of liftable by allowing $g$ to be a relat
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\begin{proof}
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\todo{Finish.}
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\end{proof}
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\begin{lemma}
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The non-trivial order on the maybe functor $FX=X+1$ is a liftable order.
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\end{lemma}
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\subsection{Symmetrization of a Barr relator}
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\begin{prop}
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Assuming $F$ is the maybe functor $FX=X+1$, then the syemmetrization of the left-lax Barr relator of $F$ is a Barr-relator.
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\end{prop}
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@@ -3073,6 +3113,7 @@ Perhaps if we can relax the definition of liftable by allowing $g$ to be a relat
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\begin{remark}
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With a similar argument we can prove that the symmetrization of a right-lax Barr-relator of $F$ is a Barr-relator.
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\end{remark}
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\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.}
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\end{document}
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