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

vault backup: 2025-05-05 09:32:04

Browse source
This commit is contained in:
Marco Realacci 2025-05-05 09:32:04 +02:00
parent 41b1be1927
commit 830289beeb
1 changed files with 1 additions and 1 deletions
Download patch file Download diff file Expand all files Collapse all files

2
Concurrent Systems/notes/15 - A formal language for LTSs.md
View file

@ -48,7 +48,7 @@ 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). (<=) 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 $P \xrightarrow{a} P'$. Let's consider $$\phi \triangleq \phi' \space where \space \phi'\triangleq \bigwedge_{\phi'' \in L(P')}\phi''$$ Let $(P, Q) \in R$ and $P \xrightarrow{a} P'$. Let's consider $$\phi \triangleq \lozenge a \phi' \space where \space \phi'\triangleq \bigwedge_{\phi'' \in L(P')}\phi''$$
By construction, $P \models \phi$, hence, $Q \models \phi$, because $L(Q) = L(P)$; then, $\exists Q' : Q \xrightarrow{a} Q'$ such that $Q'\models \phi'$. By construction, $P \models \phi$, hence, $Q \models \phi$, because $L(Q) = L(P)$; then, $\exists Q' : Q \xrightarrow{a} Q'$ such that $Q'\models \phi'$.
We proven that the formulas satisfied by P' are all satisfied by Q' We proven that the formulas satisfied by P' are all satisfied by Q'