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Math Forum / Mathematics / Mathematical Logic / July 2009S4, Clavius and Triv
Dear All I am currently echewing on a little hypothesis I am not sure about. Lets assume we have S4, and now we add the following axiom schema: L(A -> c) v A (Clavius) Where c is a modal constant, not a propositional constant. Thus c has some truth value in each world, but it cannot be instantiated, like propositional variables would do. Could it be that we then have: S4 + (Clavius) = Triv Or to expell it in another way, does the above axiom when added to S4 imply the following? B -> L B If yes, how can it be shown? Best Regards > .... Lets assume we have S4, and now we > add the following axiom schema: [quoted text clipped - 9 lines] > S4 + (Clavius) = Triv > .... Surely it depends upon your c. In the extreme case where c is the constant true proposition, your "Clavius" is already in S4. In the opposite extreme case where c is the constant false proposition, it does indeed give the trivial system. If you wish to allow every conceivable c, then in particular it could be the constant false proposition so you have the trivial system. Or am I misunderstanding something? Ken Pledger. >> .... Lets assume we have S4, and now we >> add the following axiom schema: [quoted text clipped - 14 lines] > opposite extreme case where c is the constant false proposition, it does > indeed give the trivial system. Yes, if it happens that c=0, then we have L~A v A. Ok, I see, if I take A to be ~B, then I get L~~B v ~B, which is then B -> L B. > If you wish to allow every conceivable c, then in > particular it could be the constant false proposition > so you have the trivial system. Yes, if it happens that c=1, then we have L 1 v A, and hence 1. But this does not imply B -> L B, so the conclusion is not valid in general. Viewed from this angel Triv does not result in general. Or am I wrong here? Bye >> If you wish to allow every conceivable c, then in particular it could >> be the constant false proposition so you have the trivial system. [quoted text clipped - 7 lines] > > Bye Situation might be different when L1 <> 1. Bye > Dear All > [quoted text clipped - 3 lines] > > L(A -> c) v A (Clavius) Why Clavius? The Clavius that I know(*) is (~p -> p) -> p. (* I should own up and say that I know very little but this comes from Lukasiewicz's Aristotle's Syllogistic.)  SignatureWhich of the seven heavens / Was responsible her smile / Wouldn't be sure but attested / That, whoever it was, a god / Worth kneeling-to for a while / Had tabernacled and rested. Frederick Williams schrieb: >> Dear All >> [quoted text clipped - 10 lines] > (* I should own up and say that I know very little but this comes from > Lukasiewicz's Aristotle's Syllogistic.) Yes, thats it, now observe, this can be written as: ((p > f) > p) > p And this is in modal logic: L(L(L(p -> c) -> p) -> p) (See other thread about Neighborhood II) I say this is interderivable in S4 with: (L(p->c) -> p) -> p) Hence the same as: L(p -> c) v p But meanwhile I have doubt that my interderivability hypothesis is true in S4. Probably I was a little bit too sloppy in eliminating the second L. So maybe lets rephrase the question, take (Clavius') to be: L(L(p -> c) -> p) -> p What is the result of S4 + (Clavius'). The assumption that c is a constant still holds. Bye |
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