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 $OO_1 \equiv_S OO_2$


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.

I'd really appreciate any help you could offer!

  • $\begingroup$ You need to explain the role of $\phi$ in your definition of $\equiv_{S}$. $\endgroup$ – Rob Arthan Jan 21 at 21:54
  • $\begingroup$ I answered that here $\endgroup$ – Jishin Noben Jan 22 at 15:14

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