Le 13-août-05, à 08:38, Eric Cavalcanti a écrit :

Can you give a particular example of a sentence p such that B(Bp->p) is true?

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`Take any proposition you can prove, for example the tautology (p -> p),`

`or t.`

`Now we talk classical logic ok? So p -> (q -> p) is itself a tautology`

`(the a fortiori`

axiom, which BTW is criticized by the Relevance Logicians and by the Quantum logicians! (Note that it is accepted by intuitionist logicians) Note for Lee:

`The use of the "a fortiori axiom" has made me possible to simplify a`

`bit recently some of the "intractable" problems, I say this for Lee`

`Corbin who seems to infer I was stuck on necessarily intractable`

`problem, it could still be that only my incompetence has to be`

`blamed!).`

End note for Lee (apology to Eric). <I must go now> So once you have prove t, all the propositions of the shape <anyproposition> -> t

`are easily deducible, by applying modus ponens and the a fortiori`

`axioms.`

In particular Bt -> t is justified.

`And thus B(Bt -> t) is true. (or, by necessitation B(Bt -> t)`

`follows).`

So an example of such a p making B(Bp -> p) true, is p = t. But any p you actually prove will do. Note:

`I recall if necessary that Lob formula, B(Bp -> p) -> Bp, is *the*`

`main axiom of G, the basic logic of arithmetical self-reference.`

`You could search Lob or Santa Klaus in the archive. See the proof in`

`the SANE paper.`

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