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Matthew Milano edited subsection_b_bf_Knowledge_generalization__.tex
about 9 years ago
Commit id: 47308c7515875f1c98d308bfa1e56b1c405eda5b
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(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