On 09 Oct 2013, at 22:02, meekerdb wrote:
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
the 1-view.
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.
Certainly. It came as a shock. It is the shock of Gödel and Löb
incompleteness results. Those are truly revolutionary.
If you find this strange and shocking, it means you begin to grasp!
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?
See below.
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.
You think like that because you know that "B" is correct, but "B" does
not know it.
In particular, like Bf -> f is true about "B", but not provable by
"B", B(Bf)-> Bf is also true about "B", but not provable by "B".
Here "B" designates the machine/person having "B" as provability
predicate.
Bruno
Brent
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.
Bruno
http://iridia.ulb.ac.be/~marchal/
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