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Logic proofs

Christim
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Posted 04/18/09 - 06:50 PM: Subject: Logic proofs	#1
Prove the following.. ~L ↔ [ J v (M & ~S)] A v (B v C) ∴ S → (S & S)

Ratheius Netheros
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Posted 04/18/09 - 07:58 PM:	#2
I'll try the second part as it's the only one I have any idea about. A -> A A V B A V B V C A V (B V C) Don't know which rule is which, but I assume you're allowed to start with A -> A for the second one.

Christim
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Posted 04/19/09 - 11:32 AM:	#3
\| = Assumption line designates scope of proof/subproof with open assumption. I'm not too sure if my proof is correct. I want to get S from ~L ↔ [ J v (M & ~S)], so I assume ~L to get J v(M&~S) through biconditional elimination rule. Since I can't do anything with Jv(M&~S), I make another assumption (~J) to get (M&~S) through Disjunction elimination rule. From there I can get ~S. After that I'm lost. I don't know how to get S → (S & S) using ~S. My guess to my problems would be that either the structure of my proof is wrong, or I'm missing another rule that I can use. Or both. I wonder if A v (B v C) is any use for solving the proof? I haven't used it at all. 1.~L ↔ [ J v (M & ~S)] 2. A v (B v C) want S → (S & S) \|3. ~L \|4. J v(M&~S) ↔E1,4 \|\|5. ~J \|\|6. M&~S vE3-4,5 \|\|7. M &E6 \|\|8. ~S &E6 I'm lost. I want something like this, but I don't know how to get to it. 1. S..................... Assume for →I 2. S & S.............. &I 3,3 3. S → (S & S)......→I 3-4 Here is my second guess.. Its probably way off. I Think this problem requires negation elimination rule (or for reductio), but I dont know how negation elimination works..Some ideas and example on negation elimination would be very helpful. 1.~L ↔ [ J v (M & ~S)] // The premises 2. A v (B v C) want S → (S & S) ------------------------------ 3. \|~L // First sub proof -------------- 4. \|J v(M&~S) ↔E1,4 5 \| \|~(M&S) // Sub proof within a Sub proof \| ------------ 6. \| \|~M v ~~S DeM5 (De Morgans Rule) 7. \| \| ~M v S DN6 (Double negation rule) 8. \| \|J 9. \| \| \|~~M // Sub proof within a Sub prof \| \| ------------- 10 \| \| \| M DN9 11.\| \| \| S vE7,9 12.\| \| \| S&S &I11 13. S → (S & S) I3-12 Edited by Christim on 04/19/09 - 07:25 PM. Reason: missing apostrophes, overzealous ellipses.

Ratheius Netheros
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Posted 04/21/09 - 10:26 AM:	#4
Yeah, my logic skills are lacking, honestly. I'm not even sure we're using the same logical system. I'm using a combination of first order logic and derivation. I haven't done first order logic in awhile so I'm probably using too much derivation. A v (B V C) seems useful to me because V allows you to add things for the sake if it.

xzJoel
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Posted 04/22/09 - 09:35 AM:	#5
Just to make sure I understand you, Premises are: ~L ↔ [ J v (M & ~S)] 1. A v (B v C) 2. Conclusion is S → (S & S) Right? Seems to me that the conclusion, if S then S and S, is a tautology. The premises are just red hearings. Proof would go like this in my world, \|S XXXXXXXXXXXXXXXXXXXX 3. Assumption \|________ \| \|S XXXXXXXXXXXXXXXXXXXX 4. Repetition of 3. \|S XXXXXXXXXXXXXXXXXXXX 5. Repetition of 4. \|S & S XXXXXXXXXXXXXXXXX 6. Conjunction of 4, 5. S -> (S&S) XXXXXXXXXXXXXX 7. Restatement of 3-6. You just need to determine what rules in your system allow you to say the things I just did. Does that make any sense?
Make a joyous noise onto the lord... Not a good one, just a joyous one.

xzJoel
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Posted 04/22/09 - 10:10 AM:	#6
For your reference, see "Tautology (2)" in wiki's article. http://en.wikipedia.org/wiki/Propo...ple_1._Simple_axiom_system
Make a joyous noise onto the lord... Not a good one, just a joyous one.

hhilliard
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Posted 11/03/09 - 05:31 PM:	#7
can anyone help me with indirect proofs?

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