Re: modal logic's meta axiom
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 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? Because if p is true in all worlds, then []p is true in all worlds OK? No. I didn't say that. p means p is true in a world. p true in all worlds would be written []p. But in logic, if p appears in a deduction, p is true in all worlds. Take as example a formalization of classical propositional calculus. The axioms have to be tautologies, and so are true in all worlds (valuations, interpretation). The modus ponens concerves tautologicalness, so all theorems (the formula appearing in the deduction) are true in all worlds. And p/[]p means that if p is true in all worlds (like if it is proved) then []p is true in all worlds. If you want to mean that p is true in a world, or the actual world, you can say it, but not in deduction. Usually you will name that world, by saying that p is true in alpha, at some meta level. Bruno -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/everything-list?hl=en . http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
Re: modal logic's meta axiom
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 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? Because if p is true in all worlds, then []p is true in all worlds OK? No. I didn't say that. p means p is true in a world. p true in all worlds would be written []p. But in logic, if p appears in a deduction, p is true in all worlds. You mean if p is a tautology. It may be a deduction from premises that are not true in all worlds. Is Russell thinking of p=Socrates is mortal while you're thinking of p=If all men are mortal and Socrates is a man then Socrates is mortal. Take as example a formalization of classical propositional calculus. The axioms have to be tautologies, An axiom is just a proposition taken to be true for purposes of inference. Why can't Socrates is a man be an axiom? Brent and so are true in all worlds (valuations, interpretation). The modus ponens concerves tautologicalness, so all theorems (the formula appearing in the deduction) are true in all worlds. And p/[]p means that if p is true in all worlds (like if it is proved) then []p is true in all worlds. If you want to mean that p is true in a world, or the actual world, you can say it, but not in deduction. Usually you will name that world, by saying that p is true in alpha, at some meta level. Bruno -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en. http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
Re: modal logic's meta axiom
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 semantics? Because if p is true in all worlds, then []p is true in all worlds OK? If p is true in all worlds (validity) then p is true in particular in all worlds accessible from alpha, and so []p is true in alpha, and this works for any alpha, so []p is true in all worlds. In a deduction, without proviso, p means that we have already proved p, (or in some case, that it is an assumption but you have to be careful). The meta-semantics of a rule is dependent on the theory, and the way we define what constitutes a proof. In modal logic it usually mean, in p/[]p that p occur at the end of a proof. Self-reference is confusing because deduction p/q get represented by [](p-q), or []p-[]q. []p is the machine's language for the machine asserts (believes, proves) p. All universal machine, when they asserts p, will soon or later asserts []p. So []p - [][]p is true about all universal (correct) machines. The löbian machine are those for which []p - [][]p is not only true, but they actually can justify why they know that []p - [][]p. It is the kind of truth they can communicate. So the fact p/q is modeled by []p-[]q, in the language of the machine. Löbian machines are close for the Löb rule. If they asserts []p - p, they will inevitably asserts soon or later p. And they know that, which means they can prove []([]p-p)-[]p, for each p arithmetical sentences, in the arithmetical interpretations. Bruno -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com . For more options, visit this group at http://groups.google.com/group/everything-list?hl=en . http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
Re: modal logic's meta axiom
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 said the meta-axiom stated? How else do we get p/[]p for Kripke semantics? Because if p is true in all worlds, then []p is true in all worlds OK? No. I didn't say that. p means p is true in a world. p true in all worlds would be written []p. -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
Re: modal logic's meta axiom
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? -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
modal logic's meta axiom
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 Cheers -- Prof Russell Standish Phone 0425 253119 (mobile) Principal, High Performance Coders Visiting Professor of Mathematics hpco...@hpcoders.com.au University of New South Wales http://www.hpcoders.com.au -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
Re: modal logic's meta axiom
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 dint of whatever convoluted steps, I have p --, q then it is also true that []q (q must be true in all worlds)? Not necessarily. It can depend on what axioms are used on p/q. Nor is it necessary that you get p - q, nor p- []q. Modal logic does not obey to the deduction theorem. In fact we have p/p for any p. If you were correct we would have []p for any p. Could this also be written p --- ? []q I don't see any reason for this. Bruno http://iridia.ulb.ac.be/~marchal/ -- You received this message because you are subscribed to the Google Groups Everything List group. To post to this group, send email to everything-list@googlegroups.com. To unsubscribe from this group, send email to everything-list+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/everything-list?hl=en.
