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You can see : James Garson, Modal Logic for Philosophers (2006), Ch.10 : Axioms and Their Corresponding Conditions on $R$, page 209-on. we will prove a theorem that may be used to determine conditions on $R$ from axioms (and vice versa) for a wide range of axioms (Lemmon and Scott, 1977). (For a more general result of this kind see Sahlqvist, 1975 ...

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I think the rationale here is correct, but I do think the solution is wrong. I think our island critters can deduce that they should leave after 2 days. All blue-eyed people see 99 people with blue eyes, and all brown eyed people see 100 people with blue eyes. They will reason as in the solution, based on common knowledge. However, they don't have to wait ...

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Regarding : $∃x \square (x=^{\vee} j)$, you are right; in order to compute its truth-value, we have to "unwind" the semantical specifications of Definition 4, page 123. We have that : $[[∃x \square (x=^\lor j)]]=1$ iff for some $d \in D, [[\square (x=^{\vee} j)]]_{M,w,g(x/d)} = 1$, where $x$ is a variable of type $e$ and $j$ a constant of the same ...

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I'll refer to : L.T.F. Gamut, Logic, Language, and Meaning, Volume 2 : Intensional Logic and Logical Grammar (1991), page 54 : Definition 2 Let $M$ a model and $w \in W$; then the truth conditions are : [...] (iii) $V_{M,w} (\phi \rightarrow \psi) = 1$ iff $V_{M,w} (\phi)=0$ or $V_{M,w} (\psi)=1$ (iv) $V_{M,w} (\square \phi)=1$ iff for ...

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The idea of an independence proof (in general) is : (i) define a "sort of" interpretation (in this case : $*$) for formulae, starting from atomic ones : $\bot, p, \ldots$ (ii) show that all axioms have a certain "property" : in this case, are tautologies, i.e. they are equivalent to $\top$ (iii) show that the rules of inference preserve that "property" : ...

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$\square p\land p\to p$ is a theorem; therefore $\square(\square p\land p\to p)$ by necessitation, and so $$\square(\square p\land p)\to \square p$$ by $\mathbf K$ (and modus ponens). Now apply $p\land\cdot$ to both sides and use the equivalence of $a\land b\to c$ and $a\to(b\to c)$.

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