So, I know that I can construct a frame {W, R, I} which is not transitive and in which schema 4 is not true (more specifically, Axiom Schema K and Axiom Schema 4 are not both true). I also know that I can do the same for a frame that is irreflexive in which schema T is not true (Axiom Schema K + Axiom Schema T). Can we make a frame where K4 (or, more weakly, KT4) is true but R is not transitive? Or similarly for KT and irreflexivity?

For the non-transitive case, my thoughts have been along these lines: take a frame with two worlds w and u where the only relation is wRu. Then, if □p is true in w, □□p can be true in w so long as □p is also true in u (which it is, vacuously, since u doesn't have access to any worlds). I'm not sure if the vacuously true part works, though (because then it seems that ~□p should also be true in u, which is a contradiction). If this sort of argument isn't possible, then it seems like we should never be able to construct such a non-transitive frame?

For the irreflexive case, my hunch is that there is no such frame. I'm not super clear on how to prove this, but I'm wondering if an argument like the following works: if R isn't reflexive, then for every world where □p is true, there is an interpretation in which ~p is true in that same world, which means that □p -> p is not true in that world. (It may be obvious that I am a bit confused about how I works in a frame: is there a fixed truth value in this frame for every sentence in every world in W, or is it that there are (2^a)^|W| possible interpretations within I where a is the number of sentences in the frame?)

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  • I recommend you to glance over the section "Frame Definability" of the splendid text (complete build) openly accessible at openlogicproject.org and if it still remains unsatisfactory, please rephrase your question clearly. – Tankut Beygu Nov 1 '20 at 20:32

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