On 10/9/2013 12:26 AM, Bruno Marchal wrote:
On 08 Oct 2013, at 20:35, meekerdb wrote:
On 10/8/2013 2:51 AM, Russell Standish wrote:
On Mon, Oct 07, 2013 at 10:20:14AM +0200, Bruno Marchal wrote:
On 07 Oct 2013, at 07:36, Russell Standish wrote:
and Bp&p as "he knows p", so the person order of
the pronoun is also not relevant.
Yes, you can read that in that way, but you get only the 3-view of
Let us define [o]p by Bp & p
I am just pointing on the difference between B([o]p) and [o]([o]p).
Isn't B(Bp)=Bp so:
Bp -> B(Bp)
but B(Bp) does not necessarly imply Bp.
?? That seems like strange logic. How, in classical logic, can you prove that p is
provable and yet not conclude that p is provable. I understand that the set of true
propositions is bigger than the provable propositions, but I don't see that the set of
provably provable propositions is smaller than the provable propositions?
B(Bp & p) =? B(Bp & p) & (Bp & P)
Why would that be? [o](Bp & p) = B(Bp & p) & (Bp & p), but not B(Bp & p), because B(Bp &
p) does not imply Bp & p.
Not that I wrote =? meaning "is it equal?", not asserting it was equal, and I concluded
below they were not equal.
Bp =? Bp & p -> false
And so, this does not follow. (Keep in mind that Bp does not imply p, from the machine's
point of view). Think about Bf, if it implies f, we would have that the machine would
know that ~Bf, and knows that she is consistent. She can't, if she is correct.
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