marcorealacci/master-degree-notes
1
0
Fork 0
Code Issues Pull requests Projects Releases Packages Wiki Activity Actions

vault backup: 2025-05-05 09:22:04

Browse source
This commit is contained in:
Marco Realacci 2025-05-05 09:22:04 +02:00
parent 65d4d02315
commit 7dd5d662fc
1 changed files with 6 additions and 1 deletions
Download patch file Download diff file Expand all files Collapse all files

7
Concurrent Systems/notes/15 - A formal language for LTSs.md
View file

@ -48,4 +48,9 @@ To simplify the proof, let us modify the set of formulae by allowing conjunction
(<=) We prove that $$R \triangleq \{ (P, Q) : L(P)=L(Q) \}$$ is a simulation; this suffices, since the relation just defined is trivially symmetric (so is a bisimulation too).
Let $(P, Q) \in R$ and $$
Let $(P, Q) \in R$ and $P \xrightarrow{a} P'$. Let's consider $$\phi \triangleq \phi' \space where \space \phi'\triangleq \bigwedge_{\phi'' \in L(P')}\phi''$$
By construction, $P \models \phi$, hence, $Q \models \phi$, because $L(Q) = L(P)$; then, $\exists Q' : Q \xrightarrow{a} Q'$ such that $Q'\models \phi'$.
Hence, $\forall \phi'' \in L(P').\phi'' \in L(Q'), \space i.e. \space L(P') \subseteq L(Q')$
By contradiction, let us assume that the inclusion is a proper, i.e. $L(P') \subset L(Q')$.