On 13 Jun 2012, at 00:38, Russell Standish wrote:
On Tue, Jun 12, 2012 at 08:17:38PM +0200, Bruno Marchal wrote:
On 12 Jun 2012, at 00:47, Russell Standish wrote:
On Thu, Jun 07, 2012 at 01:33:48PM +0200, Bruno Marchal wrote:
In fact we have p/p for any p. If you were correct we would
On 6/13/2012 12:14 AM, Bruno Marchal wrote:
On 13 Jun 2012, at 00:38, Russell Standish wrote:
On Tue, Jun 12, 2012 at 08:17:38PM +0200, Bruno Marchal wrote:
On 12 Jun 2012, at 00:47, Russell Standish wrote:
On Thu, Jun 07, 2012 at 01:33:48PM +0200, Bruno Marchal wrote:
In fact we have
On 12 Jun 2012, at 00:47, Russell Standish wrote:
On Thu, Jun 07, 2012 at 01:33:48PM +0200, Bruno Marchal wrote:
In fact we have p/p for any p. If you were correct we would have []p
for any p.
This is what I thought you said the meta-axiom stated?
How else do we get p/[]p for Kripke
On Tue, Jun 12, 2012 at 08:17:38PM +0200, Bruno Marchal wrote:
On 12 Jun 2012, at 00:47, Russell Standish wrote:
On Thu, Jun 07, 2012 at 01:33:48PM +0200, Bruno Marchal wrote:
In fact we have p/p for any p. If you were correct we would have []p
for any p.
This is what I thought you
On Thu, Jun 07, 2012 at 01:33:48PM +0200, Bruno Marchal wrote:
In fact we have p/p for any p. If you were correct we would have []p
for any p.
This is what I thought you said the meta-axiom stated?
How else do we get p/[]p for Kripke semantics?
--
Bruno,
I've deleted the previous thread, so I've started a new one on modal
logic's metaaxiom.
IIUC, if I have p (true in a world), and by dint of whatever
convoluted steps, I have
p
--,
q
then it is also true that []q (q must be true in all worlds)?
Could this also be written
p
--- ?
[]q
On 07 Jun 2012, at 01:58, Russell Standish wrote:
Bruno,
I've deleted the previous thread,
?
It is possible to delete thread? Including my posts? That should not
be possible.
so I've started a new one on modal
logic's metaaxiom.
OK.
IIUC, if I have p (true in a world), and by
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