On 5/2/2019 5:56 PM, bradklee@gmail.com wrote:
Let x=y be a doubtable proposition P and introduce function B(P)=1 for belief in P, and B(P)=0 for doubt. These are again doubtable propositions.
So does B(P)=1 => B(~P)=-1 ?
Let x’=y’ be an undoubtable proposition P’, such that B(P’)=0 by definition.
? If P' is undoubtable then B(P')=1, or I don't understand what "belief in P" means. Brent
This is again a doubtable proposition, but an honest person will say B(B(P’)=0)=1 and B(B(P’)=1)=0.
An abuse victim is made to say that B(P’)=1. The abuse suffered is so bad that victim reports B(B(P’)=1)=1, and positive belief for all recursions.
An oracle machine is sensitive to abuse, and can tell that the victim has been coerced into lying at every juncture. The operator of the machine then confronts the victim with this opinion.
The victim does not attain enlightenment, but suffers more from abuse at the hands of the epistemologist, end scene 2.
***
Smullyan is very fun about birds and Combinators, but on this sort of thing, I also lose patience.
Nagarjuna’s four propositions are more to my taste, and usually shorter to read.
Cheers,
Brad
On May 2, 2019, at 1:17 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I sometimes lose patience ...
math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun