# Equivalence of strings in modal logic

I'm trying to solve a question which asks me to show that for any two finite strings $$O_1$$ and $$O_2$$ of $$\square$$s and $$\lozenge$$s, (e.g. $$\square\lozenge\lozenge\square\lozenge\square)$$, that

i) if $$O_1\equiv_S O_2$$ then $$OO_1 \equiv_S OO_2$$

and

ii) if $$O_1\equiv_S O_2$$ then $$O_1O \equiv_S O_2O$$

where $$O$$ is any other string and we say $$O_1\equiv_S O_2$$ (they express the same modality) if $$\models_S (O_1\phi \leftrightarrow O_2\phi)$$, where $$S$$ is just some modal logic system.

I've managed to solve the first part quite easily using the fact that if $$\models \phi_1 \leftrightarrow \phi_2$$ then $$\models \chi(\phi_1)\leftrightarrow\chi(\phi_2)$$ where $$\chi(\phi)$$ is just the result of replacing $$P$$ with $$\phi$$ in the formula $$\chi(P)$$.

I'm not sure how to go about solving the second though, because in this case the new string is inserted between the old one and the sentence $$\phi$$, and I can't see an obvious way to do that with this formula.

I'm also trying to find a way to show, using these, that under the modal system $$B$$ (where the accessibility relation is reflexive and symmetric), that there are infinitely many modalities, using some sort of informal semantic argument.

I've done the same for $$S_4$$ and $$S_5$$, but those are finite numbers of modalities so the strategy was different.

• You need to explain the role of $\phi$ in your definition of $\equiv_{S}$. – Rob Arthan Jan 21 at 21:54