Modal Logic: Proof that a Maximally Consistent Set is Complete

Question

To understand the completeness proof for modal logic, I need to show that Maximally Consistent Sets are complete. A set $\Sigma$ is consistent if $\Sigma \not \vdash \bot$. A consistent set is maximal if any set that has $\Sigma$ as a proper subset is inconsistent. The properity I have trouble proving is that a maximally consistent set is complete, i.e. that for every formula $\varphi$, one of $\varphi \in \Sigma$ or $\neg\varphi\in \Sigma$.

Many people online claim that $\Sigma\cup\{\varphi\}\vdash \bot$ implies that $\Sigma\vdash\neg\varphi$. And if that is the case, then the proof is easy. But it sounds a lot like the deduction theorem, which I was told is not true for modal logic. Where it the case, one could prove that $\vdash \alpha \to \square\alpha$, which is clearly not true.

So, how does one prove the completeness of maximally ocnsistent sets?

Answer 1

In your case, $\Sigma \cup \{\varphi\} \vdash \bot$ indeed implies that $\Sigma \vdash \lnot \varphi$. And it is indeed a version of the deduction theorem. However there are caveats:

1. Deduction theorem fails in modal logic if we allow the unrestricted use of the necessitation rule. In the classic modal logic $\mathbf{K}$, necessitation is only applied to theorems: From $\vdash \varphi$ infer $\vdash \square \varphi$. So we cannot prove $\vdash \varphi \rightarrow \square \varphi$. (The obligatory link to the corresponding paper)
2. The notion of '$\varphi$ provable from $\Sigma$' in modal logic is a bit different from the one in propositional logic. When we write $\Sigma \vdash \varphi$, what we really mean is that $\exists \psi_1, ..., \psi_n \in \Sigma$: $\vdash \bigwedge_i \psi_i \rightarrow \varphi$, where $i \in \{1, ..., n\}$. Compare this to the definition for propositional logic: $\exists \psi_1, ..., \psi_n \in \Sigma$: $\bigwedge_i \psi_i \vdash \varphi$, where $i \in \{1, ..., n\}$.