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These Proofs Giving Me Trouble
Can only use the following rules of inference...

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These Proofs Giving Me Trouble
Questioner
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Posted 08/20/08 - 12:20 AM:
Subject: These Proofs Giving Me Trouble 		quote post
#1
I am working on two problems that are giving me trouble.

I can only use the following rules of inference: Premise Introduction, &I, &E, MP, MT, CP, Rule of Assumptions (assumption introduction), Definition. Biconditional, RAA. I am doing this in Propositional Logic.

Please let me know if these are not clear. I'm a logic novice, so please explain.


1. ~(P -> Q) : P & ~Q
(Supposed to be able to do this in 12 lines)

2. ~P : P -> Q
(supposed to be able to do this in 9 lines)

In the second, I am not sure how to get Q from the premises (at least in usable form, vI doesn't help much, as far as I can tell).

Thanks for any help. (Although I am posting this on a logic homework forum, I am not doing this for any class. Just working through a logic book.)

Edited by Questioner on 08/20/08 - 12:31 AM
muxol
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0 of 2 people found this post helpful
Posted 08/20/08 - 08:53 AM:
quote post
#2
You say "at least in usuable form, vI doesn't help much...", but I thought you weren't allowed to use any disjunction rules. So can you or not? And what is the rule "Biconditional"? That could be a number of things. What is "Definition"? This is all very unclear.

Edited by Caldwell on 08/21/08 - 01:35 AM. Reason: Edited for content.
Questioner
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Posted 08/20/08 - 04:49 PM:
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#3
That was very rude, muxol... I am working through various books, without a teacher, entirely on my own, as a layman. I stated I was a novice. I wonder if that is how you learned logic, too. I came here for help, not to be insulted.

I left out vE and vI on accident. However, I have encountered two books which offer the rule of 'Definition. <->'. I stated that if things weren't clear, to ask for clarification. I assumed (inaccurately) that people on this board would be familiar with the rule 'Definition. <->', which allows one to go from (P -> Q) & (Q -> P) to (P <-> Q).

moonlight
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Posted 08/20/08 - 06:19 PM:
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#4
Questioner wrote:
That was very rude, muxol... I am working through various books, without a teacher, entirely on my own, as a layman. I stated I was a novice.


Pay no attention to such comments. Everyone was once a beginner. Even Kurt Godel.

Cordially,
moonlight.
All are lunatics, but he who can analyze his delusion is called a philosopher.
- Ambrose Bierce -
Questioner
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Posted 08/20/08 - 10:57 PM:
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#5
Thanks. I still need help with these, if anyone is able and willing to give it.
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Posted 08/21/08 - 12:29 AM:
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#6
Hi Questioner,

Well, I'm having some trouble with the method requested by your problem. The proof is pretty obvious, and one solution could be:

1) ~(P -> Q) premise
2) ~(~P v Q) reformulating in terms of 'v'
3) P & ~Q De Morgan's (conclusion)

or

2) ~~(P & ~Q) reformulating in terms of '&'
3) P & ~Q double negation elimination (conclusion)

But apparently you can't use De Morgan's, or such reformulations, or even double negation eliminations, which complicates the problem. I'll try to think about it some more later.

Cordially,
moonlight.
All are lunatics, but he who can analyze his delusion is called a philosopher.
- Ambrose Bierce -
muxol
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Posted 08/21/08 - 12:49 AM:
quote post
#7
Questioner wrote:
That was very rude, muxol... I am working through various books, without a teacher, entirely on my own, as a layman. I stated I was a novice. I wonder if that is how you learned logic, too. I came here for help, not to be insulted.

I left out vE and vI on accident. However, I have encountered two books which offer the rule of 'Definition. <->'. I stated that if things weren't clear, to ask for clarification. I assumed (inaccurately) that people on this board would be familiar with the rule 'Definition. <->', which allows one to go from (P -> Q) & (Q -> P) to (P <-> Q).


You should ask well-formulated questions then. For instance, what you call 'definition' might also be your rule in bidirectional form, or it might allow one or both directions of movement between (A & B) v (~A & ~B) and (A <-> B). Indeed it seems you mean 'definition biconditional' in which case it is likely the biconditional version of the rule you have, though you state it incorrectly as the unidirectional version.

Even with disjunction intelim (intro-elimination) rules, the rule system you present doesn't look promising for solving your problems. It is absolutely essential to a question like this that if you want help, you make it exactly clear what rules you're allowed to use.

Edited by Caldwell on 08/21/08 - 01:39 AM. Reason: Edited for content.
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Posted 08/21/08 - 01:01 AM:
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#8
Thanks for the attempt. I can't use De Morgans laws, unfortunately. I am familiar with Double Negation though (Double Negation Introduction and Double Negation Elimination - DNI and DNE, respectively.) If you could use that, would that make it easier? I'd like to see a proof with it in it. It might shed some light on what I am supposed to do here.
Questioner
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Posted 08/21/08 - 01:23 AM:
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#9
Questioner wrote:

I came here for help, not to be insulted.


Muxol wrote:

You should ask well-formulated questions then.


Asking a question that doesn't meet your criteria is not a license to insult someone. I asked people to ask for clarification if they needed it. From my view, I was being clear, although I did accidentally leave out vE and vI. Everyone is capable of asking for clarification without flinging insults.

Please don't post in this thread again.
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Posted 08/21/08 - 03:30 AM:
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#10
Here is an update on (2). here is my proof so far. I used DNE. I am unsure if am allowed to use this, but it was on the next page, so I started attempting to do this one with it and was able to narrow my proof down to 12 lines - apparently, it should be only 9.

Here it is. Let me know if you can find a better way of doing this with the rule set provided, and if you see a way of doing this without DNE.

~P: P -> Q
{1} 1. ~P Premise
{2} 2. P A
{3} 3. ~Q A
{2,3} 4. P & ~Q 2, 3, &I
{5} 5. P & ~Q A
{5} 6. P 5, &E
{1,5} 7. P & ~P 1, 6, &I
{1} 8. ~(P & ~Q) 5, 7, RAA
{1,2,3} 9. ~(P & ~Q) & (P & ~Q) 4, 8, &I
{1,2} 10. ~~Q 3, 9, RAA
{1,2} 11. Q 10, DNE
{1} 12. P → Q 2, 11, CP


Any help or suggestions would be greatly appreciated.
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