diff --git a/draft/draft.tex b/draft/draft.tex index b2c1b62..004aab2 100644 --- a/draft/draft.tex +++ b/draft/draft.tex @@ -2207,14 +2207,11 @@ Now, we prove that the given order on the maybe functor is a liftable one. \end{proof} 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 $\alpha\comp p_1=(p_1+1)\comp\sigma$. \begin{lemma}\label{lem:maybe-func-set} - Assuming that $R$ is a symmetric AM-simulation over an $F$-coalgebra $(X,\alpha)$ that $FX=X+1$, then for every $(x_1,x_2)\in R$ either + Assuming that $R$ is a symmetric AM-simulation over an $F$-coalgebra $(X,\alpha)$ that $FX=X+1$, then for every $(x_1,x_2)\in R$ \begin{gather*} - \alpha(x_1),\alpha(x_2)\in X, + \alpha(x_1)\in X\Rightarrow \alpha(x_2)\in X\\ + \alpha(x_1)=\bot \Rightarrow \alpha(x_2)=\bot \end{gather*} - or - \begin{gather*} - \alpha(x_1)=\alpha(x_2)=\bot. - \end{gather*} \end{lemma} \begin{proof} 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 $p_1+1\comp\sigma=\alpha\comp p_1$. Since $R$ is symmetric, for every $(x_1,x_2)\in $ we have the following: @@ -2317,7 +2314,7 @@ 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 $R_{2\times X}\iso 0$. +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 $R_{2\times X}\iso 0$ that is indeed the abstraction of~\autoref{lem:maybe-func-set}. \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: