Hi Aditya,

`Sorry for answering late. Computer problems did make my mailbox quite`

`messy, and also I'm rather busy.`

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`The "B" is a unary connector, like the negation "~". ~p is intended for`

`"not p" or "p is false".`

`"Bp" is intended for "provable p", or more precisely "provable by some`

`(fixed) sound lobian machine" (and a lobian machine is just a machine`

`"sufficiently rich" to prove the main elementary arithmetical`

`propositions (including the induction axioms pages: search the Podnieks`

`web page for more on this:`

http://www.ltn.lv/~podnieks/

`Obviously "Bp -> p" is true (by definition) for a *sound* lobian`

`machine. The interesting and fundamental point is that "Bp -> p",`

`although true, cannot be *proved* by the lobian machine, making the`

`logic of "Bp & p" (p provable and p true) different from the logic of`

`Bp, and this is what I exploit to distinguish formally the first and`

`third person discourse in the translation of UDA in a subset of`

`arithmetic (as a lobian machine's natural language). And this is needed`

`for isolating the logic of physical (observable) propositions in the`

`comp frame.`

`Solovay has succeeded in formalizing completely the true`

`(propositional) logic of B by a modal logic named G* and the provable`

`part of by the (weaker) modal logic G. It is the difference between G`

`and G* which makes comp formally consistent (and then necessary bu the`

`Universal Dovetailer Argument (UDA)).`

`A variant of quantum logic appears exactly where the UDA shows it needs`

`to appear would the comp assumption be correct.`

`Hope this helps. Don't hesitate to ask questions even very basic one. I`

`know we can lose time for simple notation problem (but then they are`

`simple to answer, also!)`

Regards, Bruno Le 14-août-05, à 00:40, Aditya Varun Chadha a écrit :

Hi Bruno, Can you please say exactly the meanings of B and Bp? does Bp mean "B and p"? or B parametrized by p? without their meanings specified your derivation isn't obvious to me. On 8/13/05, Bruno Marchal <[EMAIL PROTECTED]> wrote:And thus B(Bt -> t) is true. (or, by necessitation B(Bt -> t) follows).-- Aditya Varun Chadha adichad AT gmail.com http://www.adichad.com

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