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Modal logic exercise
Modal logic exercise

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Modal logic exercise

frank2010
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Posted 10/13/09 - 11:07 PM:	#11
ClaudeHooper wrote: Is it allowed to rewrite P(A) as ~N(~A)? Yes, I guess, even if I'd write it as "Pp = ~N~p". In this simple case, you can bring the '~' inside the brackets (note that in the canonical form, if p is a propositional variable, you don't need brackets at all). Some examples to elucidate the rule at stake here. Of course, if you have, say, the wff 'P(p v q)', you can't rewrite it as '~P(~p v q)'. But, if you have the wff 'P(~p v ~q)' you can rewrite it as '~N(p & q)'.

ClaudeHooper
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Posted 10/14/09 - 03:21 AM:	#12
xzJoel wrote: What circumstance exists where P is possible and Q is possible where P & Q is not possible? I've only got a little bit of time before I leave for work, so I will revisit here this evening, but for now, let P be a contingent proposition, so that P(P) and P(~P) are both true. Clearly P(P&~P) is still false, though, since a contradiction is never possible.

xzJoel
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Posted 10/14/09 - 04:51 AM:	#13
ClaudeHooper wrote: I've only got a little bit of time before I leave for work, so I will revisit here this evening, but for now, let P be a contingent proposition, so that P(P) and P(~P) are both true. Clearly P(P&~P) is still false, though, since a contradiction is never possible. Fair enough. No need for the truth tables, your example is compelling. Thanks.
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ClaudeHooper
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Posted 10/14/09 - 04:00 PM:	#14
How is this for a proof? 1. (p => q) => ((~(q&r)) => (~(p&r))), propositional calculus 2. N((p => q) => ((~(q&r)) => (~(p&r)))), necessitation from 1 3. N(p => q) => N((~(q&r)) => (~(p&r))), axiom K, modus ponens from 2 4. N((~(q&r)) => (~(p&r))) => (N(~(q&r)) => N(~(p&r))), axiom K 5. ((A => B) & (B => C)) => (A => C), propositional calculus 6. N(p => q) => (N(~(q&r)) => N(~(p&r))), substitution {A := N(p=>q), B := N((~(q&r)) => (~(p&r))), C := (N(~(q&r)) => N(~(p&r)))} in 5, modus ponens from 3 & 4 7. (A => (B => C)) => ((A & ~C) => ~B), propositional calculus 8. (N(p => q) & ~N(~(p&r))) => ~N(~(q&r)), substitution {A := N(p=>q), B := N(~(q&r)), C := N(~(p&q))} in 7, modus ponens from 6 9. (N(p => q) & P(p&r)) => P(q&r), rewrite ~N(~x) as P(x) in 8

frank2010
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Posted 10/15/09 - 02:41 AM:	#15
Way to go! Thank you very very very much, Claude! How did you gain the crucial insight? Edited by frank2010 on 10/15/09 - 02:59 AM

ClaudeHooper
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Posted 10/15/09 - 03:19 AM:	#16
frank2010 wrote: Way to go! Thank you very very very much, Claude! How did you gain the crucial insight? At first, the "r" was making things complicated, so I started by trying to prove the simpler statement: N(p => q) & P(p) => P(q) And after I had solved that, I took a look at the proof to extend it to handle the original problem. The basic idea is that N(p => q) means that p implies q in all possible worlds that are accessible from the current world, so that in whatever world p is true, then q must be true in that world as well. There was only derived inference rule that mentioned P(), so I figured I would have better luck if I transformed everything into N(), at which point I had modal terms N(p => q), N(~p), and N(~q), and it struck me that using the contrapositive would be more helpful, N(~q => ~p).

frank2010
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