From 7dd5d662fc762a9c8acaa48bd85fbc7ba2f7a481 Mon Sep 17 00:00:00 2001 From: Marco Realacci Date: Mon, 5 May 2025 09:22:04 +0200 Subject: [PATCH] vault backup: 2025-05-05 09:22:04 --- .../notes/15 - A formal language for LTSs.md | 7 ++++++- 1 file changed, 6 insertions(+), 1 deletion(-) diff --git a/Concurrent Systems/notes/15 - A formal language for LTSs.md b/Concurrent Systems/notes/15 - A formal language for LTSs.md index d9abe4a..c1d2c4e 100644 --- a/Concurrent Systems/notes/15 - A formal language for LTSs.md +++ b/Concurrent Systems/notes/15 - A formal language for LTSs.md @@ -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 $$ \ No newline at end of file +Let $(P, Q) \in R$ and $P \xrightarrow{a} P'$. Let's consider $$\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. $L(P') \subset L(Q')$. \ No newline at end of file