Prove that $\Sigma\Vdash^g_M \phi$ iff $\{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$

Question

(Modal Logic, Blackburn, Ex 1.5.3): Let $\Sigma$ be a set of formulas in the basic modal language, and let $M$ denote the class of all models. Prove that $$\Sigma\Vdash^g_M \phi \Longleftrightarrow \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$$

My work:

1. For the $[\Leftarrow]$ direction: Assume $\{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \Vdash_M\phi$, i.e. $\forall \mathfrak M\in M,\forall w\in W, \mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \implies \mathfrak M,w\Vdash\phi$. We want to show that $\forall \mathfrak M\in M, \mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. Consider arbitrary $\mathfrak M$, and suppose $\mathfrak M\Vdash\Sigma$. So, $\mathfrak M,w\Vdash\Sigma$ for all $w\in W$ (some would like to write $\mathfrak M,w\Vdash\sigma$ for all $w\in W,\sigma\in\Sigma$ - but that notation doesn't add any value). To show $\mathfrak M\Vdash\phi$, we take arbitrary $w\in W$ and show $\mathfrak M,w\Vdash\phi$. Since we have $\forall w\in W, \mathfrak M,w\Vdash\Sigma, $ we conclude $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$. By assumption, $\mathfrak M,w\Vdash \phi$.

2. I'm stuck with the $[\Rightarrow]$ direction. Assume $\Sigma\Vdash^g_M \phi$ holds, i.e. $\forall \mathfrak M\in M, \mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. We want to show that $\forall \mathfrak M\in M, \mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\} \implies \mathfrak M,w\Vdash\phi$. Let us argue by contradiction. Suppose, there exists $\mathfrak M\in M$ and some $w\in W$, for which $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ but $\mathfrak M,w\nVdash\phi$. What do I do next?

I really need to be able to use the $\mathfrak M$ and $w$ above to construct some model $\mathfrak M'$ wherein $\mathfrak M'\Vdash\Sigma$ holds so I can use the assumption. I don't know how exactly that would help though.

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Definitions if you need them:

1. $\Sigma\Vdash^g_M \phi$ - For all $\mathfrak M \in M$ (class of all models), $\mathfrak M\Vdash\Sigma \implies \mathfrak M\Vdash\phi$. This is the global semantic consequence relation.
2. $\Sigma\Vdash_M \phi$ - For all $\mathfrak M \in M$ (class of all models), $\forall w\in W$, $\mathfrak M,w\Vdash\Sigma \implies \mathfrak M,w\Vdash\phi$. This is the local semantic consequence relation.

Please help, thanks a lot! Hints are okay, and solutions are good too!

Answer 1

Here is a proof $[\Rightarrow]$ with some details omitted. Let me know if this helps.

Assume you have a countermodel to the right hand side of the equivalence, i.e. a pair $(\mathfrak M,w)$ such that $\mathfrak M,w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ and $\mathfrak M,w\nVdash\phi$.

In the next step, you should consider the model $\mathfrak M'$ which arises by removing from $\mathfrak M$ all worlds which cannot be reached from $w$ via a directed path using the accessibility relation. Note that $\mathfrak M'\in M$ as $M$ is the class of all models.

(Maybe you want to do the proof from here for yourself. Otherwise, the rest goes as follows:)

Convince yourself that removing such unreachable worlds does not have an effect on which modal formulas are true at $w$. So we still have (i) $\mathfrak M',w\Vdash \{\square^n\sigma: \sigma\in \Sigma,n\in\omega\}$ and (ii) $\mathfrak M',w\nVdash\phi$. Let $w'$ be any world in $\mathfrak{M}'$. By construction $w'$ is reachable from $w$ in $\mathfrak{M}'$ and so from (i) it follows (check!) that $\mathfrak{M}',w'\Vdash \sigma$ for all $\sigma\in\Sigma$. Therefore $\mathfrak M'\Vdash\Sigma$, and from this and (ii) you get a countermodel to the left hand side of the equivalence.