Re: Modal Logic

From: Eric Cavalcanti <eric.cavalcanti.domain.name.hidden>
Date: Sun, 14 Aug 2005 10:38:14 +1000

Hi Bruno,

> > 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), or t.
(...)
> So once you have prove t, all the proposition of the shape
>
> <anyproposition> -> t
>
> is easily deducible, by applying modus ponens and the a fortiori axiom.
>
> 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 then any proposition of the form
B(q->p)->Bp
would be true as well. Why is that not an axiom?

Further, by your explanation above it seems that the
axiom should be more like Bp->B(q->p). Since as I
understood it, if you can justify p, then you can justify
any sentence of the form q->p.

On the other direction, suppose it is justifiable that
~p. That is, B(~p) and I suppose ~Bp with the excluded
middle. So it is true that Bp->q for any q, and in particular
for q=p. So B(Bp->p). But Bp is false by assumption so
B(Bp->p)->Bp is false. What's wrong?

Eric.
Received on Sat Aug 13 2005 - 20:40:00 PDT