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See generated subframe: A frame $\mathbf G=\langle G,S,W\rangle$ is a generated subframe of a frame $\mathbf F=\langle F,R,V\rangle$, if the Kripke frame $\langle G,S\rangle$ is a generated subframe of the Kripke frame $\langle F,R\rangle$ (i.e., $G$ is a subset of $F$ closed upwards under $R$, and $S$ is the restriction of $R$ to $G$), and $W ... 1$p \land \lnot q$is$\lnot (p \to q)$. Thus, we can rewrite$\varphi := □(□p∧¬◊p)$as$◻¬(◻p \to ◊p)$. 1 Hint There is an algorithm described into: Patrick Blackburn & Maarten de Rijke & Yde Venema, Modal Logic (2001), Ch3.6 Sahlqvist Formulas; see Example 3.43, page 159. The formula:$St_x(\varphi) := ∀P∀x \ [∃y(R(x,y) ∧ ∃z(R(y,z) ∧ P(z))) → ∃w(R(x,w) ∧ P(w))]$must be converted by the algorithm, with the instantiation$\sigma(P) := \lambda ...

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It is true more general fact. Proposition. If $U$ is open and $A$ an arbitrary subset of $X$, then $IC(U)\cap IC(A)\subseteq IC(U\cap A)$. Proof. Suppose $$x\in IC(U)\cap IC(A)$$. $x\in IC(U)\cap IC(A)$ iff there exists $U'$ an open neighborhood of $x$ such that $U'\subseteq C(U)\cap C(A)$. Let us denote $$U'\cap U=V_1$$ and $$U'\cap A=V_2.$$ ...

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Let $(X,\tau)$ be a topological space with interior operator $I$ and closure operator $C$. Let $A$ be a subset of $X$. For convenience lets make the following notations: $C(A)=^{def}A^{-}$ and $X\backslash A=^{def}A'$ There is conjugation $I(A)=X\backslash (C(X\backslash A))=(( A') ^{-})'$. We will need two following lemas. Lemma 1: Let $A$ be a subset ...

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No, not necessarily. When changing the connective, the modality has to interact with the negation ($p \wedge q \equiv \neg (\neg p \vee \neg q)$), so it may behave differently. The negation may change the modality, $L$ may be changed to a dual modality $K$, with each modality having its own syntactic rules. For example, the exponential in linear logic ...

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