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I have the two LTS (labeled transition system) as seen in the following picture: enter image description here And the book is telling me that between those two LTS, their $1$ and $1'$ are non-bisimilar.

So I tried to get a bisimilation starting from the pair $\{\{1,1'\}\}$ by continuously extending it whenever I found a conflict, ending up with: $$\{ \{1,1'\} ,\{2,2'\},\{3,3'\},\{2,4'\},\{4,3'\},\{3,5'\},\{4,5'\} \}$$

Finally, to check whether it truly was a bisimilation, I checked for each pair each node, and asserted that all possible pairs of derivatives were part of the set.

I am arriving that they are a bisimilation - is the crux in the $a$'s and $b$'s?
(They didn't really explain what it means)

Sudix
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1 Answers1

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The left diagram has a b and the right diagram has a c. Thus, the pair $(2,4')$ does not satisfy the conditions required to be a bisimulation. In particular, the book is correct that those two systems are not bisimilar.

D.W.
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  • So, in conclusion, it is irrelevant that we have parallel branches with the same action (e.g. $a$), but it is relevant for a bisimulation that ... the actions taken, i.e. $a,b,c$, are the same? – Sudix Sep 13 '18 at 19:22
  • @Sudix, I didn't try to work that out; I am merely answering the question, for the specific example you gave. If you are wondering about what would happen if you replaced a c with a b in the right-hand picture, I suggest you spend some time working through the details. Checking whether the conditions of the definition hold is a mechanical process that you should be able to do as well as i can. – D.W. Sep 13 '18 at 19:51