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I'm a bit confused regarding the relation between trace equivalence and bisimilarity. These lecture notes I found and a few others documents I've read state that "if an LTS is deterministic then two states are bisimilar if they are trace equivalent".

When reading around the topic I found this page, which shows the following image: enter image description here

These LTS' are trace equivalent and deterministic(?), why does the rule not hold that they are then bisimilar?

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The answer is that the LTS on the left isn't deterministic, as the label (or action) open_door doesn't go to a single state and hence that action is non-deterministic.

This example shows that determinism is indeed required for trace equivalence and bisimilarity to be equivalent.

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  • $\begingroup$ Wow, I stupidly mixed up determinism of a graph as being a cyclic or not. Thanks. $\endgroup$ – UnhingedCS Apr 17 '18 at 10:45

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