Subproofs within subproofs: Philosophy Forums

Subproofs within subproofs

Parches
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Posted 10/14/09 - 06:03 PM: Subject: Subproofs within subproofs	#1
I am stuck on this question. I have the premises A v B and A v C and need to derive A v (B n C). The 'n' stands for AND. I need to use subproofs within subproofs and I can only use rules of elimination and introduction for negation, conjunction, disjunction and contradiction. I cannot cite DeMorgan's law. Your help will be greatly appreciated, as I have spent a long time trying out different combinations of rules. Thanks!

ClaudeHooper
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Posted 10/14/09 - 06:38 PM:	#2
What does your or-elimination rule look like?

frank2010
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Posted 10/15/09 - 01:33 AM:	#3
You said that you can use vE, but how can you use it without introducing conditionals? Please, double check your desiderata. Meanwhile, here's a proof that uses the Disjunctive Syllogism derived rule. (1) A v B Assumption (2) A v C Assumption (3) \| ~ (A v (B & C)) Hypothesis for ~I (4) \|\| A Hypothesis for ~I (5) \|\| A v (B & C) from 4 by vI (6) \|\| A v (B & C) & ~ (A v (B & C)) 3,5 by &I (7) \| ~A 4-6, ~I (8) \| B 1,7 Disjunctive Syllogism (9) \| C 2,7 Disjunctive Syllogism (10) \| B & C 8,9 &I (11) \| A v (B & C) 10 vI (12) \| (A v (B & C)) & ~(A v (B & C)) 3,11 &I (13) A v (B & C) 3-12 by ~I and Double negation

ClaudeHooper
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Posted 10/16/09 - 09:17 AM:	#4
Here is another proof, with a 5-way or-elimination rule: () 1. A v B, premise () 2. A v C, premise (3) 3. A, assumption (3) 4. A v (B & c), v-Intro, 3 (5) 5. B, assumption (5,6) 6. A, assumption (5,6) 7. A v (B & C), v-Intro, 6 (5,8) 8. C, assumption (5,8) 9. B & C, &-Intro, 5, 8 (5,8) 10. A v (B & C), v-Intro, 9 (5) 11. A v (B & C), v-Elim, 2, 6, 7, 8, 10 () 12. A v (B & C), v-Elim, 1, 3, 4, 5, 11

frank2010
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Posted 10/17/09 - 05:05 AM:	#5
Hi Claude, I've tried to rewrite your proof in the logical notation I use, to little effect. Could you double check my transcription? (1) A v B Premise (2) A v C Premise (3) \| A Hypothesis (4) \| A v (B & C) 3, vI (5) A -> (A v (B & C) 3-4, ->I (6) \| B Hypothesis (7) \|\| C Hypothesis (8) \|\| B & C 6,7 &I (9) \|\| (A v (B & C) 8, vI (10) \| C -> (A v (B & C)) 7-9 ->I (11) B -> (C -> (A v (B & C))) 8-10 ->I ...? I can't go on...

ClaudeHooper
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Posted 10/17/09 - 08:04 AM:	#6
frank2010 wrote: Hi Claude, I've tried to rewrite your proof in the logical notation I use, to little effect. Could you double check my transcription? Can you show me how you would write an or-elimination step?

frank2010
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Posted 10/17/09 - 08:08 AM:	#7
ClaudeHooper wrote: Can you show me how you would write an or-elimination step? A v B A-> C B-> C so C

ClaudeHooper
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Posted 10/17/09 - 09:09 AM:	#8
Here is my transcription of my proof, with new lines denoted by an "a" after the line number: 1. A v B, premise 2. A v C, premise 3. \| A, assumption 4. \| A v (B & C), v-Intro, 3 4a. A => (A v (B & C)), =>-Intro, 3, 4 5. \| B, assumption 6. \|\| A, assumption 7. \|\| A v (B & C), v-Intro, 6 7a. \| A => (A v (B & C)), =>-Intro, 6, 7 8. \|\| C, assumption 9. \|\| B & C, &-Intro, 5, 8 10. \|\| A v (B & C), v-Intro, 9 10a. \| C => (A v (B & C)), =>-Intro, 8, 10 11. \| A v (B & C), v-Elim, 2, 7a, 10a 11a. B => (A v (B & C)), =>-Intro, 5, 11 12. A v (B & C), v-Elim, 1, 4a, 11a Edited by ClaudeHooper on 10/17/09 - 09:44 AM

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