Re: Fitch's paradox and OWA

On 2 jan, 22:52, stevendaryl3..._at_yahoo.com (Daryl McCullough) wrote:
> Jan Hidders says...
>
> >But I'm afraid I don't think that will work. The reason is that in
> >your logic you can still express the same things that could be
> >expressed in the old logic. Take for example the following proposition
> >in the old model theory:
>
> >(1)  K(p & ~K(p))
>
> >You can still express this in your logic.
>
> Yes, but with the correct axiomatization of knowability
> predicate, the corresponding proposition will not be true.
>
> >You can do this by using a predicate CW(w) that expresses
> >that w is (equivalent to) the current world. You can express
> >this as follows:
>
> >(2)  CW(w)  =def=  For all p, ( t(w,p) <-> p )
>
> >With that you can write (1) in your logic as:
>
> >(3)  Forall w : W, ( CW(w) -> k(w, (p & ~k(w, p))) )
>
> >This can be done for all for all formulas in the old logic and so it
> >seems to me that you will still have the same paradox but written down
> >differently.
>
> I don't see how it is a paradox. Your proposition (3) will
> (with the appropriate axiomatization of the knowability
> predicate) be provably false.

The purpose of (3) was only to illustrate the translation of formulas in the original logic to your logic. You are right that by itself it does not show the paradox. But if this translation exists then all formulas used in the proof of the paradox will have their equivalents in your logic. If your logic is complete it will also have the equivalents of all the used axioms and principles, and so the proof will still proceed but will just be phrased in a different syntax.

For example, on the Stanford page the formulas (4) and (5) both have their equivalents in your logic. You should also have the principle (A) in your logic, but of course translated to your syntax, so in your logic we should be able to derive the equivalent of (5) from the equivalent of (4). Et cetera.

• Jan Hidders