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@@ -2205,15 +2205,15 @@ Now, we prove that the given order on the maybe functor is a liftable one.
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\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
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\end{itemize}
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\end{proof}
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\begin{remark}
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The mentioned order on the maybe functor is not coliftable.
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\end{remark}
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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 $(X,\alpha)$ such 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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