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?
Take any proposition you can prove, for example the tautology (p -> p),
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
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
In particular Bt -> t is justified.
And thus B(Bt -> t) is true. (or, by necessitation B(Bt -> t)
So an example of such a p making B(Bp -> p) true, is p = t.
But any p you actually prove will do.
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.