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I'm trying to summarize the issue without too many details. We have to start from the language of modal logic, with the "operator" $\Box$, and its semantical interpretation, following Kripke. Of course, we want that the calculus for modal logic turns to be sound and complete with respect to semantics; and so it is. The first step is to prove that every ...

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i haven't read Chellas but I refer to the new edition of 'Hughes and Cresswell's Introduction to Modal Logic' for an extensive treatment of your question. Completeness of a modal logic also depends onto the axioms chosen.

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Definition 2.5. [page 35] states the conditions for a formula $A$ to be true at the possible world $\alpha$ in the model $\mathcal M$ starting with : (1) $\vDash_{\alpha}^{\mathcal M} \mathbb{P}_n \ \text {iff} \ \alpha \in \mathbb{P}_n$, for $n = 0,1,2, \ldots$ where the $\mathbb{P}_i$s are the atomic sentences, and so on. The conditon for a ...

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For a countermodel consider any 2-element K-model, where the first element accesses the second and where p is true in the first, but not in the second. $F \rightarrow \Box F$ is valid in every S4-model satisfying the hereditary condition: If $p$ is true in $w$, then $p$ is true in $v$ for all $v$ with $wRv$.

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The property : if $\vDash \varphi$, then $\vDash \square \varphi$ is the ground for the Rule of Necessitation. It says that $\square \varphi$ is a theorem of a normal modal logic whenever $\varphi$ is a theorem of the logic. This does not contradict the fact that $\varphi \to \square \varphi$ is not valid. A "tricky" but simple counterexample ...

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Suppose that $S$ and $T$ are both spheres, and that neither is a subset of the other. Then $S' = S\setminus T$ and $T'=T\setminus S$ are both nonempty. Setting $A=S' \cup T'$ in the second condition for being a sphere we learn that $f(S'\cup T')\subseteq S$ and $f(S'\cup T') \subseteq T$. Together with axiom (2) this means that $$f(S'\cup T') \subseteq S ... 2 Mauro Allegranza's answer shows that your condition is not only sufficient but also necessary, but the actual proof is left as an exercise. Here's a solution to that exercise. I'll prove the contrapositive; I assume your condition is not satisfied, so some w_1 is R-related to two distinct elements w_2 and w_3, and I'll show that that \Diamond ... 1 It seems to me that the condition you have stated is an iff condition. See : Brian Chellas, Modal Logic : An Introduction (1980), page 85-on for the G^{k,l,m,n} schema :$$\Diamond^k \Box^l A \to \Box^m \Diamond^l A.$$Thus, your formula is the case with 1,0,1,0. Then, see EXERCISE 3.38 [page 90] : Prove that 1,0,1,0-incestuality is the ... 1 Some terminology ... A frame is a pair \mathfrak F = \langle W, R \rangle, where W is a non-empty set (the set of "possible worlds") and R is a partial order on W. A valuation V is a map associating to each sentential variable p of the language a subset V(p) \subseteq W. A model is a pair \mathfrak M = \langle \mathfrak F, V \rangle. We ... 2 I think that the problem you're having here is similar to the following apparent "paradox" we often see in more general modal logics: if we can infer \phi, then by necessitation we can infer \square \phi, yet the sentence \phi \rightarrow \square \phi is not in general valid. The point is that if we can really infer \phi, starting from no axioms, ... 1 Contingency can be somehow defined in terms of necessity:$$\operatorname{C}p\leftrightarrow(\lnot\operatorname{N}p\,\land\,\lnot\operatorname{N}\lnot p) You can add the above (defining) axiom to a system which you use to formalize necessity and have system for both necessity and contingency. Note that the "meaning" of the above axiom is just "Contingency ...

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