lemma
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@@ -2207,13 +2207,10 @@ Now, we prove that the given order on the maybe functor is a liftable one.
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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 $\alpha\comp p_1=(p_1+1)\comp\sigma$.
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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 $(x_1,x_2)\in R$ either
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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),\alpha(x_2)\in X,
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\end{gather*}
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or
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\begin{gather*}
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\alpha(x_1)=\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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@@ -2317,7 +2314,7 @@ So, we have the following:
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\begin{gather*}
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R\iso R_{X^2}+R_{2\times X}+R_{1}
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\end{gather*}
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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 $R_{2\times X}\iso 0$.
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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}.
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\section{Relators}
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\subsection{Two-way similarity in Hughes-Jacobs}
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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:
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