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Let $L$ be a finite set of labels, and let $\mathcal{C}$ be the set of finitely-branching transition systems labeled by $L$ and with a countable set of states. Let $\sim$ denote the bisimulation relation, and let $\mathcal{C} / \sim$ denote the set of bisimulation equivalence classes of LTS's in $\mathcal{C}$.

Q1: Does every bisimilarity equivalence class $\alpha \in \mathcal{C} / \sim$ have a tree model?

Let $L\mu$ denote the set of closed formulas in modal $\mu$-calculus.

Q2: True or false: $\forall \alpha \in \mathcal{C} / \sim$, $\exists F \in L\mu$ such that models$(F) \cap \mathcal{C} = \alpha$. Or in plain English, every bisimilarity class of countable and finitely-branching LTS's over $L$ is exactly the set of countable and finitely-branching models of some modal $\mu$-calculus formula.

An answer to either would be helpful, and answers to both would be sublime :)

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Q1: Yes, every LTS is bisimilar to its unfolding, which is a tree.

Q2: No, by a cardinality argument. For instance take infinite binary trees with $L=\{a,b\}$. Each tree has countable set of states and is finitely-branching, but you have uncountably many such trees, even up to bisimilarity. However you have only countably many $\mu$-calculus formulas, so some bisimilarity classes are not captured by formulas.

Actually in this setting, a bisimilarity class can be exactly captured by a $L_\mu$ formula if and only if it is the bisimilarity class of a finite structure. The unfolding of a finite structure is called a regular tree. Any formula accepting a non-regular tree must also accept a regular tree, which is not bisimilar to it. See for instance [1] for more details.

[1] The mu-calculus and model checking. J Bradfield, I Walukiewicz. Handbook of Model Checking, 871-919

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  • $\begingroup$ Wow, way shorter answers than I expected! I think I dissuaded myself from thinking Q1 was easy because I first considered countably branching LTS's and then was thinking that unfolding might give me cardinality problems that would take the result outside of $\mathcal{C}$ by creating too many states, although now that I rethink about it, even with countable branching I would have each level of the unfolding as a countable union of countable sets and the entire state space would be the countable union of levels, so still countable. Q2 answer makes perfect sense as well, although I'm trying... $\endgroup$ – wanderingmathematician Oct 16 at 23:23
  • $\begingroup$ ... to wrap my head around your second comment. Weird to try to think up an example of a finite LTS which is bisimilar to a non-regular tree! $\endgroup$ – wanderingmathematician Oct 16 at 23:27
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    $\begingroup$ There is probably a little misunderstanding: a non-regular infinite binary tree cannot be bisimilar to a finite LTS. What my last comment means is that if a formula accepts such a non-regular tree, it must accept a union of several $\sim$-classes, actually uncountably many, one of which contains a regular tree. $\endgroup$ – Denis Oct 16 at 23:51
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    $\begingroup$ Maybe the confusion comes from the distinction between "captures" and "accept". I used "$\varphi$ captures $\alpha$" to say that $\textit{models}(\varphi)\cap\mathcal C=\alpha$ , and "$\varphi$ accepts $M$" for $M\in\textit{models}(\varphi)$. $\endgroup$ – Denis Oct 17 at 0:00

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