vault backup: 2025-05-05 09:49:15
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@ -56,4 +56,18 @@ Hence, $\forall \phi'' \in L(P'), \space \phi'' \in L(Q'), \space i.e. \space L(
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By contradiction, let us assume that the inclusion is a proper, i.e. $L(P') \subset L(Q')$. Thus, $\exists \hat{\phi}:\hat{\phi}\in L(Q') \land \hat{\phi} \not \in L(P')$.
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Then, $\lnot \hat{\phi} \in L(P')$ and this would imply that $\lnot \hat{\phi} \in L(Q')$. This is a contradiction because $L(Q')$ cannot contain a formula and its negation.
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Then, $\lnot \hat{\phi} \in L(P')$ and this would imply that $\lnot \hat{\phi} \in L(Q')$. This is a contradiction because $L(Q')$ cannot contain a formula and its negation.
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### Proving unequivalences
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The Logic approach presented so far is very natural for proving unequivalences:
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- show one formula that is satisfied by a proc but not by the other
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It is not very effective for concretely proving equivalences:
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- e.g. to show that $P \sim Q$, we should check that every formula in L(P) belongs to L(Q) and conversely
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- the problem is that L(P) is infinite, for every P, it contains TT, TT^TT, TT^TT^TT, ...
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- even if we restrict to logical equivalence class, the situation does not change
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- EXAMPLE: consider process P2 
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- it satisfies ☐bFF, ☐cFF, ☐dFF, ...
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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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