vault backup: 2025-05-05 09:54:15

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Marco Realacci 2025-05-05 09:54:15 +02:00
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@ -71,3 +71,20 @@ It is not very effective for concretely proving equivalences:
- so L(P2) is infinite because so is the action set
#### Sub-Logics
Let's consider the sub-logic without negation: $$\phi := TT | \phi \land \phi|◇ a \phi \quad where a \in Action$$
*Remark:* the formula ☐aFF is not expressible anymore.
Hence, we can only express through formulae what a process is able to do.
Let's call L(P) the set of negation-free formulae satisfied by process P.
##### Theorem
P simulates Q if and only if $L(Q) \subseteq L(P)$
*Proof:* The proof is similar to the one for the previous Theorem. The main difference is in the (⇐) implication, when φ = ⋄aφ, because here we do not prove that the inclusion cannot be proper (indeed, in general it is not).
An easy corollary of this result is that there is a double simulation between P and Q if and only if L¬(P) = L¬(Q).
*Example:* it can be checked that L(P2) = L(P1)
indeed, P1 can simulate P2 and viceversa, but the simulations are not bisimulations.
![300](../../Pasted%20image%2020250505095249.png)