Matthew Milano edited subsection_b_bf_Knowledge_generalization__.tex  about 9 years ago

Commit id: 47308c7515875f1c98d308bfa1e56b1c405eda5b

deletions | additions      

       

\\  (b) I will argue by contradiction. Assume that we have some $M \vDash \varphi \Rightarrow E_S(\psi \wedge \varphi)$, but there is some $(M,s) \vDash \phi \wedge \neg C_S \psi$. As $(M,s) \vDash \varphi \wedge \varphi \Rightarrow E_S(\psi \wedge \varphi)$ we conclude $(M,s) \vDash E_S(\psi \wedge \varphi)$. Again by our non-empty assumption, part (a), and knowledge-axiom analog, we get $(M,s) \vDash \varphi \wedge \psi$. But by these same