The only way I could think of doing this is to show that p → □p with the K axioms would imply a contradiction, but I don't think that's true. Not sure how to get started on this. Do I just have to construct a possible countermodel?

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    No. A statement not being theorem does not mean that it is a contradiction. You would merely have to show that it is a contingency. – Graham Kemp Oct 21 '20 at 1:48
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    Constructing a countermodel using Kripke semantics would seem to be the best way to go. – Bumble Oct 21 '20 at 3:15
  • I recommend you to ponder over this question further; it would lead you to an illuminating debate on the availability of a deduction theorem in modal logic. See R. Hakli and S. Negri's 2012 paper "Does the Deduction Theorem Fail for Modal Logic?" Synthese 187, pp. 849-867. – Tankut Beygu Oct 23 '20 at 11:32

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