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

@@ -49,10 +49,10 @@ 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 $P \xrightarrow{a} P'$. 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 $Q'\models \phi'$. *(quindi "deve esistere un Q' che fa almeno tutto quello che fa P'")*
 We proven that the formulas satisfied by P' are all satisfied by Q'
 
-Hence, $\forall \phi'' \in L(P').\phi'' \in L(Q'), \space i.e. \space L(P') \subseteq L(Q')$
+Hence, $\forall \phi'' \in L(P'), \space \phi'' \in L(Q'), \space i.e. \space L(P') \subseteq L(Q')$
 
 By contradiction, let us assume that the inclusion is a proper, i.e. $L(P') \subset L(Q')$. Thus, $\exists \hat{\phi}:\hat{\phi}\in L(Q') \land \hat{\phi} \not \in L(P')$.