problem set 5 (landscape of normal modal logics)

You may use the soundness and completeness results from the slides: each system in the landscape is sound and complete with respect to the corresponding class of frames.

A rule \(\varphi/\psi\) is admissible in a system \(\Sigma\) when the theorems of \(\Sigma\) are closed under it: if \(\vdash_\Sigma \varphi\), then \(\vdash_\Sigma \psi\).

1. Admissibility of \(\Box \varphi / \varphi\).

   1. \(\Box \varphi/\varphi\) is admissible in \(\textsf{K}\). That is, for each modal formula \(\varphi\), if \(\vdash_\textsf{K} \Box \varphi\), then \(\vdash_\textsf{K} \varphi\).

   2. Explain why your construction from 1.1 is not available for \(\textsf{K5}\).

   For 1.1., given a countermodel for \(\varphi\) at a world \(w\), add a new world \(w_0\) that sees only \(w\). For 1.2., give a euclidean frame and a world \(w\) such that adding a new world \(w_0\) that sees only \(w\) yields a frame that is not euclidean.

2. \(\textsf{KB4}\) is a proper extension of \(\textsf{K5}\):

   1. \(\textsf{K5} \sqsubseteq \textsf{KB4}\). You may give a semantic argument: use the fact that every symmetric and transitive frame is euclidean, and explain why that suffices.

   2. \(\textsf{KB4} \not\sqsubseteq \textsf{K5}\).

3. \(\textsf{K5}\) proves the necessitation of every axiom of \(\textsf{KB4}\). That is:

   1. \(\vdash_{\textsf{K5}} \Box (\varphi \to \Box \Diamond \varphi)\).

   2. \(\vdash_{\textsf{K5}} \Box (\Box \varphi \to \Box \Box \varphi)\).

   You may use the fact that \(\textsf{K5}\) is sound and complete with respect to the class of euclidean frames.

4. Use your answers to the previous problems to show:

   1. \(\Box \varphi / \varphi\) is not admissible in \(\textsf{K5}\).

   2. \(\Box \varphi/\varphi\) is not admissible in \(\textsf{KB4}\).