Hi Brent,

You didn't answer my last post where I explain that Bp is different from Bp & p. I hope you were not too much disturbed by my "teacher's" tone (which can be enervating I imagine). Or is it because you don't recognize the modal form of Godel's theorem:


                                                ~Bf -> ~B(~Bf),

which is equivalent to B(Bf -> f) -> Bf, by simple contraposition "p -> q" is equivalent with "~p -> ~q", and using also that "~p" is equivalent to "p -> f", where f is put for "false".

This shows that for a consistent (~Bf) machine, although Bf -> f is true, it cannot be proved by the machine. Now (Bf & f) -> f trivially. So Bf and Bf & f are not equivalent for the machine
(although they are for the "guardian angel").


Bruno


http://iridia.ulb.ac.be/~marchal/

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