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The exercise is from Modal logic by Patrick Blackburn et al. The approach is the same as in the Example 2.4, i.e. you assume, that the backward looking modality is definable in the basic modal language (that is using $\diamond$). Then, according to the Proposition 2.6, some formula $\varphi$, which might include the new modality, is true in some state of a ...

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See What is Modal Logic?: Narrowly construed, modal logic studies reasoning that involves the use of the expressions ‘necessarily’ and ‘possibly’. However, the term ‘modal logic’ is used more broadly to cover a family of logics with similar rules and a variety of different symbols [including logics for belief, for tense and other temporal expressions, ...

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Mathematical truths are generally held to be necessary truths, so there shouldn't actually be any examples. I notice that Carl Mummert's answer claims to have an example, so let me say what I think is wrong with it. The question "is the parallel postulate true?" is not a question at all until you specify the context -- are we talking about Euclidean ...

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Semantically, the axiom says, that $(M,s) \models \langle ?\psi \rangle \varphi$ iff $(M,t) \models \psi$ and $(M,t) \models \varphi$. The result of an execution of a test $?\psi$ is a set $\{(s,s) \mid (M,s) \models \psi\}$, i.e. the set of states of the model $M$ in which $\psi$ holds, and the relation is reflexive. So, according to the axiom, $\langle ... 0 Your Kripke model is$M = (\{u,v,w\}, \{(u,v), (u,w). (v,w), (w,v)\}, \{u,v\})$. The first set are states in the model. The second one defines an accessibility relation (i.e. for each program (agent) a set of states that can be reached from the current one), and the last set is a valuation set for a propositional variable$p$(i.e. where$p\$ is true). So, ...

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