# Normal modal Logic with finite proposition letters

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 $$\psi\in L$$.

• There are many normal modal logics. Do you mean K? Sep 24 at 10:59
• Yes! I mean deduction rules are (K), (MP) and (Gen) Sep 25 at 2:47

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 $$n$$).