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Modal logic exercise
Modal logic exercise

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Modal logic exercise

frank2010
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Posted 10/11/09 - 06:51 AM: Subject: Modal logic exercise	#1
Hi all, could you guys help me with this exercise? Prove in K the following wff ('L' stands for 'necessarily' and 'M' stands for 'possibly'): (L(p->q) & M(p & q)) -> N(q & r) Thank you for your time

frank2010
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Posted 10/11/09 - 07:31 AM:	#2
Does this work? PC [L(p->q)/p, M(p&q)/q, M(q&r)/r] (1) (L(p->q) & M(p&q)) -> M(q&r) Is it that simple??

frank2010
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Posted 10/13/09 - 08:41 AM:	#3
frank2010 wrote: Does this work? PC [L(p->q)/p, M(p&q)/q, M(q&r)/r] (1) (L(p->q) & M(p&q)) -> M(q&r) Is it that simple?? No, this solution is completely wrong. Need some help here...

xzJoel
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Posted 10/13/09 - 11:42 AM:	#4
frank2010 wrote: No, this solution is completely wrong. Need some help here... Could you please re-write the problem? (L(p->q) & M(p & q)) -> N(q & r) Just want to make sure that you intend the premise to be: [ Necessarily( P then Q) & Possibly ( P & Q ) ] The conclusion is? "N" is undefined and r is not in the premise. The problem is much tougher if you really have "r" in the conclusion.
Make a joyous noise onto the lord... Not a good one, just a joyous one.

frank2010
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Posted 10/13/09 - 11:54 AM:	#5
xzJoel wrote: Could you please re-write the problem? (L(p->q) & M(p & q)) -> N(q & r) Just want to make sure that you intend the premise to be: [ Necessarily( P then Q) & Possibly ( P & Q ) ] The conclusion is? "N" is undefined and r is not in the premise. The problem is much tougher if you really have "r" in the conclusion. Thank you for your post. Sorry for the typo. The problem is the following: prove that the relevant wff follows from the axioms and the theorems of the modal system K. The wff is: ( Necessarily (if P then Q) & Possibly ( P & R )) -> Possibly (Q & R)

xzJoel
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Posted 10/13/09 - 12:42 PM:	#6
Sorry mate, wish I could help but I don't have a logic book in front of me. If you can remind me what the rules are for modal introduction, elimination, etc., I can see what I can do. Also need the rules for the distribution of possiblity. (I know that the defining criteria of K is that you can distribute necessity.) The proof itself looks pretty straight forward to me. If P is possible and R is possible, then it stands to reason that if P then Q is necessary, that q is also possible. Combine the possible P and possible R together and then take the possible out to a general term, and you've got Possibly (Q & R). 1. \| N(P > Q) & P(P & R) Ass. 2. \| N(P > Q) &E 1 3. \| (P > Q) N Elim 2 4. \| P(P & R) &E 1 5. \| P(P) & P(R) P Dist 4 6. \| P(P) &E 5 7. \| P(Q) ----- This is the one line that I need the rules for. getting a possible antecedent to combine with a necessary conditional so that you can wind up with a possible consequent via modes ponens. 8. \| P(R) &E 5 9. \| P(Q) & P(R) &I 7,8 10.\| P(Q & R) P Dist 9 _______ [N(P > Q) & P(P & R)] > P(Q & R)] That is a sketch with some of the rules. I am pretty sure that there are rules that you do each of the steps, but I can't remember their names and the exact way to symbolize them. Can't you do a Possibility introduction in a way very similar to conditional introduction?
Make a joyous noise onto the lord... Not a good one, just a joyous one.

frank2010
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Posted 10/13/09 - 01:27 PM:	#7
xzJoel wrote: Sorry mate, wish I could help but I don't have a logic book in front of me. If you can remind me what the rules are for modal introduction, elimination, etc., I can see what I can do. Also need the rules for the distribution of possiblity. (I know that the defining criteria of K is that you can distribute necessity.) The proof itself looks pretty straight forward to me. If P is possible and R is possible, then it stands to reason that if P then Q is necessary, that q is also possible. Combine the possible P and possible R together and then take the possible out to a general term, and you've got Possibly (Q & R). 1. \| N(P > Q) & P(P & R) Ass. 2. \| N(P > Q) &E 1 3. \| (P > Q) N Elim 2 4. \| P(P & R) &E 1 5. \| P(P) & P(R) P Dist 4 6. \| P(P) &E 5 7. \| P(Q) ----- This is the one line that I need the rules for. getting a possible antecedent to combine with a necessary conditional so that you can wind up with a possible consequent via modes ponens. 8. \| P(R) &E 5 9. \| P(Q) & P(R) &I 7,8 10.\| P(Q & R) P Dist 9 _______ [N(P > Q) & P(P & R)] > P(Q & R)] That is a sketch with some of the rules. I am pretty sure that there are rules that you do each of the steps, but I can't remember their names and the exact way to symbolize them. Can't you do a Possibility introduction in a way very similar to conditional introduction? Thanks, man. Actually it is not allowed to assume the line (1). Indeed I have to deduce "(N(P > Q) & P(P & R))>P(Q & R)" from the following axioms PC: if alpha is a propositional calculi theorem, then it is a theorem also in K K: if N(P > Q) then NP > NQ MP (Modus ponens) US (Uniform Substitution) N (Necessitation rule): if alpha is a theorem, then N(alpha) is a theorem DR1 (Derived Rule 1): if alpha > beta is a theorem, then N(alpha) > N(beta) is so DR2: if alpha iff beta is a theorem, then N(alpha) iff N(beta) is so DR3: if alpha > beta is a theorem, then P(alpha) > P(beta) is so These are the main axioms and rules in K.

ClaudeHooper
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Posted 10/13/09 - 03:23 PM:	#8
xzJoel wrote: 9. \| P(Q) & P(R) &I 7,8 10.\| P(Q & R) P Dist 9 Step 10 isn't sound, because it could be the case that Q is true in some alternative worlds and R is true in different alternative worlds, but that there is no alternative world in which Q and R are both true at the same time. Edited by ClaudeHooper on 10/13/09 - 03:33 PM

ClaudeHooper
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Posted 10/13/09 - 04:09 PM:	#9
frank2010 wrote: These are the main axioms and rules in K. Is it allowed to rewrite P(A) as ~N(~A)?

xzJoel
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Posted 10/13/09 - 06:49 PM:	#10
ClaudeHooper wrote: Step 10 isn't sound, because it could be the case that Q is true in some alternative worlds and R is true in different alternative worlds, but that there is no alternative world in which Q and R are both true at the same time. You know much more about this than I do, C, so please don't take this as anything other than a request for indulgence. Is it possible for you to do a truth tree for P(P), P(Q), and P(P & Q) ? In my head, I can't see why it doesn't follow that if both P is possible and Q is possible, that it doesn't follow that both P and Q are possible. I know that one can actually be while the other is not, but why can't you say that both are possible together? What circumstance exists where P is possible and Q is possible where P & Q is not possible? I don't have access to my books and I don't remember offhandedly what truth trees look like in modal logic. Thank you.
Make a joyous noise onto the lord... Not a good one, just a joyous one.

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