Proof for Transitivity of Weak Bisimulation ≈ I have already posted this question on the other page, but I was suggested to post it here as it might be more relevant. 

I am currently following a course in concurrency theory and I am currently trying to prove that the Weak Bisimulation relation ≈ is an equivalence relation.

I have managed to prove cases for reflexivity and symmetry, however I can't seem to make the connection for transitivity. Does any one have any clues how can I go about this ?

I have tried the relation {(P,R) | ∃Q.P≈Q,and Q≈R} as my witness relation and then at one point I get stuck.

I assumed that P->a P' which gives me that there is Q =>a Q' and that P' ≈ Q' Now the thing is that since Q =>a Q' then by the definition of => Q ->tau* Q1 ->a Q2 ->tau* Q' Q1 ->a Q2 will cause some R1=>a R2 such that Q2 ≈ R2 Then I am not sure how to proceed to show that somehow there is a transition R =>a R' such that (P', R') are in R. May be I believe I have to use numerical induction on the number of tau ?

Thanks !
You are on the right track. It can be useful to prove the following lemma: if $\sim$ is a weak bisimulation, if $P\sim Q$ and if you have a sequence of transitions $P\xrightarrow{w}P'$ for $w\in (Act\cup\{\tau\})^*$ then there must be a sequence $Q\xrightarrow{w'}Q'$ with $P'\sim Q'$ and $w =_{\tau} w'$, i.e., $w$ and $w'$ are "the same modulo $\tau$". This can be done by induction on $w$, and it will come in handy when you deal with transitivity. Check that $u=_\tau u'$ and $v=_\tau v'$ imply $uv=_\tau u'v'$, i.e., that $=_\tau$ is a congruence.