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?


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.

  • $\begingroup$ 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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