As I understand it from Modal Logic 2.1 - the systems M, B, S4 & S5, I should not be able to show "⬜A ➡A" in modal logic K. The following truth tree seems to confirm this, if I did it correctly:
There is no contradiction in w0 and I have no possibly operator allowing me to continue the tree into a new world. Because the accessibility relation is not reflexive in K, I cannot derive A in w0 to reach a contradiction that I would be able to in modal system M.
The truth tree remains open, implying that I cannot show "⬜A ➡A" in K. However, when I try to find a counterexample in K, I am stuck.
Let v be the valuation function. There are two possibilities:
- v(A) = 1. If that is the case then "⬜A" is 1 and so is "A" which means the conditional is true. So that valuation does not lead to a contradiction.
- v(A) = 0. If that is the case then "⬜A" is 0 and so is "A" which means the conditional is again true.
I don't see how I can construct a counterexample in K. That counterexample should also work as a counterexample in M where this can be derived. Perhaps being invalid does not mean I can always create a counterexample, but without the counterexample is it really invalid?
Kane B channel, Modal logic 2.1 - the systems M, B, S4 & S5 https://youtu.be/VRVX7B5Iw14