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I have a question about how I should read the formula: $\square(\square\varphi\rightarrow \varphi)\rightarrow \square \varphi$.

I do understand why this formula is not universally valid based on the example of the book(van benthem: modal logic for open minds), but I am still misreading the part $\square(\square\varphi\rightarrow \varphi)$ of the formula for $\square\square\varphi\rightarrow \square\varphi$.

According to the book $\square \square \varphi \to \square \varphi$ and $\square (\square \varphi\to \varphi)$ are not the same why is that?

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  • $\begingroup$ I don't understand what you're asking. The part of the formula in question is written as $\square (\square \varphi\to \varphi).$ In what sense are you reading it as $\square \square \varphi \to \square \varphi$ and why? $\endgroup$ – spaceisdarkgreen Feb 9 at 7:23
  • $\begingroup$ @spaceisdarkgreen $\square\square\varphi\rightarrow \square\varphi$ is wrong, but how should the formula actually be read? $\endgroup$ – jennifer ruurs Feb 9 at 7:32
  • $\begingroup$ I don't know what you mean by "read". Like you think these two formulas should be equivalent intuitively, and want to know why they aren't? If so, it would help to know why you think that. $\endgroup$ – spaceisdarkgreen Feb 9 at 7:33
  • $\begingroup$ @spaceisdarkgreen according to the book $\square \square \varphi \to \square \varphi$ and $\square (\square \varphi\to \varphi)$ are not the same why is that? $\endgroup$ – jennifer ruurs Feb 9 at 7:38
  • $\begingroup$ Why would they be? $\endgroup$ – spaceisdarkgreen Feb 9 at 7:40
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The formulas \begin{equation} P_1 = \Box(\Box \varphi \rightarrow \varphi) \quad \text{and} \quad P_2 = \Box\Box \varphi \rightarrow \Box \varphi \end{equation} are not equivalent. It is clear from the usual axiom \begin{equation} K: \Box(\varphi_1 \rightarrow \varphi_2) \rightarrow (\Box\varphi_1 \rightarrow \Box\varphi_2) \end{equation} that $P_1$ implies $P_2$. However, $P_2$ does not imply $P_1$, which we can show by constructing a model in which $P_2$ is true but $P_1$ is not.

Let $\varphi = p$ be a literal and consider the model with five worlds $u$, $v_1$, $v_2$, $w_1$, and $w_2$ where $w_2$ is the only world satisfying the literal $p$. Let the accessibility relation given by \begin{equation} R = \{(u,v_1), (u,v_2), (v_1,w_1), (v_2,w_2)\}. \end{equation} Then $u \models P_2$ because $u \models \neg \Box\Box p$, since $w_1 \not\models p$. On the other hand, $u \not\models P_1$ because $v_2 \not\models \Box p \rightarrow p$.

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