Normal modal Logic with finite proposition letters

Question

Assume our modal language $L$ has only diamonds, and the set of proposition letters $Prop$ is finite. The deduction rules are the same as normal modal logic. Now consider $M$ is a finite model of this language $L$. We first define an equivalence relation $\approx$ on $M$ : $u\approx v$ $\Leftrightarrow$ $M,u\models\varphi$ iff $M,v\models\varphi$ for every $\varphi\in L$. Since M is finite, we then have finite $\approx$-equivalence classes $M_1$, $M_2$,..., $M_n$.

Now the question is: how to construct formulas $\varphi_1$, $\varphi_2$,..., $\varphi_n$ such that for every $i=1,2,...,n$, for $u\in M_i$, we have :

$M,u\models\psi$ iff $(\varphi_i\rightarrow\psi)$ is a $L$-theory, for every $\psi\in L$.

Answer 1

Assuming that by “normal modal logic” you mean the smallest normal logic K, this is not possible in general.

For example, let $M$ be the one-element reflexive model (with whatever valuation of variables). Then assuming for contradiction that $\phi$ is a formula that axiomatizes over K the set of formulas true in the unique element of $M$ as in the question, we see that K proves $\phi\to\Box\phi$ (that is, $\Diamond\neg\phi\to\neg\phi$, if you insist on a language without $\Box$). By Theorem 3.6 in my paper Blending margins: The modal logic K has nullary unification type, this is only possible if K either proves $\phi$, or it proves $\phi\to\Box^n\bot$ for some $n$. But these are impossible: the former would imply that only K-tautologies can be true in $M$, whence $M$ satisfies neither $p$ nor $\neg p$ for any propositional variable $p$; on the other hand, the latter would falsely imply that $\Box^n\bot$ is true in $M$.

In general, over K, such formulas $\phi_i$ exist for a given finite model $M$ if and only if $M$ is acyclic (i.e., iff $M$ validates $\Box^n\bot$ for some $n$).