vault backup: 2025-05-05 09:07:04
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.obsidian/workspace.json
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.obsidian/workspace.json
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@ -213,6 +213,7 @@
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},
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"active": "cdcc59f1bf6d4ae1",
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"lastOpenFiles": [
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"Pasted image 20250505090603.png",
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"Concurrent Systems/slides/class 15.pdf",
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"Concurrent Systems/notes/15 - A formal language for LTSs.md",
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"Pasted image 20250505085454.png",
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@ -34,4 +34,14 @@ hence, P |= ☐aFF holds true only if P cannot perform any action a
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Let us now define the set of formulae satisfied by a process as $$L(P) = \{ \phi \in Form:P \models \phi \}$$
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To simplify the proof, let us modify the set of formulae
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To simplify the proof, let us modify the set of formulae by allowing conjunctions over a numerable set of formulae
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##### Theorem $$P \sim Q \space iif \space L(P)=L(Q)$$
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(=>) By induction on the syntax tree of the formula, let's show that $$φ∈L(P) \space iff \space φ∈L(Q)$$
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*Base case:* The only possible case is with φ = TT.
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- by the satisfiability relation, $\phi$ belongs to the set of formulae of every process
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- so also to L(P) and L(Q)
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*Inductive step:* Let's assume the thesis for every tree of height at most h. Let h+1 be the height of $\phi$. Let's distinguish on the outmost operator in $\phi$
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