vault backup: 2025-05-05 09:54:15
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@ -71,3 +71,20 @@ It is not very effective for concretely proving equivalences:
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- so L(P2) is infinite because so is the action set
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#### Sub-Logics
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Let's consider the sub-logic without negation: $$\phi := TT | \phi \land \phi|◇ a \phi \quad where a \in Action$$
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*Remark:* the formula ☐aFF is not expressible anymore.
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Hence, we can only express through formulae what a process is able to do.
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Let's call L(P) the set of negation-free formulae satisfied by process P.
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##### Theorem
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P simulates Q if and only if $L(Q) \subseteq L(P)$
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*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).
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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).
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*Example:* it can be checked that L(P2) = L(P1)
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indeed, P1 can simulate P2 and viceversa, but the simulations are not bisimulations.
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