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?
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 $v_2 \not\models \Box p \rightarrow p$.