Help on Proving a sequent: Philosophy Forums

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Help on Proving a sequent

lii8080
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Posted 09/26/09 - 02:18 PM: Subject: Help on Proving a sequent	#1
Im stuck trying to prove a sequent: ( P -> R ) & ( Q -> R) \|- ( P v Q ) -> R All I know is that since Im trying to prove a conditional, I'll assume (P v Q ) for CP at line 2 1 (1) ( P -> R ) & ( Q -> R ) A 2 (2) ( P v Q ) A (for CP) 1 (3) ( P -> R ) 1, &E 1 (4) ( Q -> R ) 1, &E But I have two conditionals (3) and (4) having the same consequent R 5 (5) P A 1,5 (6) R 3,5,MPP 7 (7) Q A 1,7 (8) R 4,7,MPP it does not satisfy the proof if I went: 1,5 (9) ( P v Q ) -> R 2,6,CP 1,7 (10) ( P v Q ) -> R 2,8,CP What to do? Please help me!!

MathematicalPhysics Wizard
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Posted 09/27/09 - 01:03 AM:	#2
What rules are you allowed to use besides CP? Anyway, here's another CP proof, but using dobule negation, modus tollens and hypothetical syllogism. 1. ~R Assumption. 2. ~P 1+hypothsis MP. 3. ~Q 1+hypothsis MP. 4. ~p&~Q 2+3, conjunction. 5. ~(PvQ) 4, DeMorgan. 6. ~R->~(PvQ) 1-5, CP. 7. ~~(PvQ)->~~R 6, Modus Tollens. 8. PvQ->~~(PvQ) double negation. 9. ~~R->R double negation. 10. PvQ->R 7,8,9 Hypothetical Syllogism.
"What kind of a bomb is this? The exploding kind." Peter Sellers

Jean Francoise
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Posted 09/27/09 - 02:40 AM:	#3
Here's another approach ( P -> R ) & ( Q -> R) \|- ( P v Q ) -> R 1. ( P -> R ) & ( Q -> R) Hyp 2. ( P v Q ) Hyp 3. (~P v R) & (~Q v R) 1, by definition of material conditional 4. (~P&~Q) v R 3, by distributive laws 5. ~(P v Q) v R 4, by De Morgan's law 6. ~~ (P v Q) 2, double negation 7. R 5, 6, disjunctive syllogism 8. ( P -> R ) & ( Q -> R), ( P v Q )\|- R 1-7 9.( P -> R ) & ( Q -> R) \|- ( P v Q ) -> R 8, Deduction Theorem
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ClaudeHooper
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Posted 09/27/09 - 10:00 AM:	#4
Here is yet another proof, in yet another system: >> C1. (((P ==> R) and (Q ==> R)) ==> ((P or Q) ==> R)) by intro C1 \| +- H1. ((P ==> R) and (Q ==> R)) \| >> C2. ((P or Q) ==> R) \| by elim H1 \| \| \| +- H2. (P ==> R) \| \| H3. (Q ==> R) \| \| >> C2. ((P or Q) ==> R) \| \| by intro C2 \| \| \| \| \| +- H4. (P or Q) \| \| \| >> C3. R \| \| \| by elim H4 \| \| \| \| \| \| \| +- H5. P \| \| \| \| >> C3. R \| \| \| \| by elim H2 \| \| \| \| \| \| \| \| \| +- >> C3. R \| \| \| \| \| >> C4. P \| \| \| \| \| by hyp H5, C4 \| \| \| \| \| \| \| \| \| +- H6. P \| \| \| \| \| H7. R \| \| \| \| \| >> C3. R \| \| \| \| \| by hyp H7, C3 \| \| \| \| \| \| \| +- H5. Q \| \| \| \| >> C3. R \| \| \| \| by elim H3 \| \| \| \| \| \| \| \| \| +- >> C3. R \| \| \| \| \| >> C4. Q \| \| \| \| \| by hyp H5, C4 \| \| \| \| \| \| \| \| \| +- H6. Q \| \| \| \| \| H7. R \| \| \| \| \| >> C3. R \| \| \| \| \| by hyp H7, C3

ClaudeHooper
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Posted 09/27/09 - 10:04 AM:	#5
lii8080 wrote: What to do? Please help me!! You need to use vE (or elimination) against line (2).

lii8080
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Posted 09/27/09 - 11:07 AM:	#6
Interesting. How would I approach this problem if I was only to use the nine primitive rules of the system-L?

ClaudeHooper
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Posted 09/27/09 - 11:42 AM:	#7
Do you have an vE "or elimination" rule?

lii8080
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ClaudeHooper
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Posted 09/27/09 - 11:59 AM:	#9
What do you get when you apply the vE rule to your line (2) P v Q?

lii8080
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Posted 09/27/09 - 12:14 PM:	#10
Perhaps I do not exactly understand the vE rule. I am not sure what I would use vE for because the conclusion is a conditional ( P v Q ) -> R, and in order to use the CP rule the antecedent has to be an assumption (if I'm not mistaken). I am confused, please bear with me. Say line 5 and 7 is an assumption for vE, how would I proceed?

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