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Conditional Proof and Reduction Ad Absurdum
Some problems I need help with

Conditional Proof and Reduction Ad Absurdum

Markh5000
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Posted 04/07/08 - 03:19 PM: Subject: Conditional Proof and Reduction Ad Absurdum	#1
These are homework problems that our professor said wouldn't be on any exam so admittedly I didn't pay as much attention in lecture but this homework will be graded so any help would be appreciated. 1. Using conditional proof (A & K) v L /: ~L → K 2. Using Reductio Ad Absurdum M v P ~D ⇔ ~P ~M /: D 3. Using conditional proof H → I J → K H v J /: I v K 4. Using Reductio Ad Absurdum (A & B) v (A & C) D → (C → ~B) E & ~C /: ~D v ~C 5. Using conditional proof ~L v ~H H v (M → K) ~N → ~K /: L → (M → N)

Brian Bosse
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Posted 04/09/08 - 10:01 AM:	#2
Hello Mark, I am going to attempt to point you in the right direction without actually doing the homework for you. 1. You want to show ~L → K. Th first step in conditional proof is to assume the antecedent (~L) and then derive the consequent (K). As such, you start with ~L. Now, right away the assumption of ~L allows you to use Modus Tollendo Ponens on your given (A & K) v L. From this an application of simplification is all you will need. 2. A Reductio Ad Absurdum assumes the negation of what it is trying to prove and from this derives a contradiction. You are trying to prove 'D'. As such, assume ~D. This will allow you to derive M via Modus Ponens and Modus Tollendo Ponens. Since ~M is a given, then you will have derived your contradiction. 3. I am not sure what your instructor means by using conditional proof here. You are going to derive (I v K). (~I → K) is logically equivalent to (I v K). This logical law is sometimes referred to as the rule of conditional disjuction. As such, if you assume ~I what happens? You can derive K. In other words, you set out to show (~I → K) - conditional proof - and then once this is proved you can use the rule of conditional dusjunction to end up with (I v K). I suspect this is what your teacher wants. 4. Just like (2) above, you assume the negation of the conclusion and derive a contradiction. In other words, assume ~(~D v ~C) and see where it leads. The first step is show that this leads to (D & C) . This will immediately lead to the contradiction of (E & ~C). You will have both C and ~C. 5. Like (1) above, assume L and see if you can derive (M → N). You will use the first given, and from this you will then use the second given and, from this you will use your third given to derive (M → N). Sincerely, Brian

Markh5000
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Posted 04/10/08 - 01:00 PM:	#3
I very much appreciate the help the last poster has given me but if anyone else has anything to add that might help me that would be great. I've tried to learn this stuff and its just not my forte...these are the last problems I'll need help with as we're moving on to non-sentential logic type stuff after this chapter.

Brian Bosse
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Posted 04/10/08 - 01:52 PM:	#4
Hello Mark, I can do the problems for you, but I am not sure that would help. I will work the first one for you in an explicit step-by-step manner. After this, if you still need more help, then I need you to post what you have tried and where you are getting stuck. Problem 1 Given: ((A ∧ K) v L) 1. Show ¬L → K Since we are going to use a conditional proof, then we assume that antecedent, '¬L', and derive the consequent, 'K'. Therefore, step 2 is the assumption '¬L'. ---------------------2. Assume ¬L Notice, that I indented step 2. The reason for this is that the proposition expressed in this step has not been established. It's use is only valid at this level. It kind of marks a "make believe" area. However, if we are able to derive 'K' from this "make believe" area, then we have shown step 1. If by assuming '¬L' I can derive 'K', then I can validly conclude '¬L → K'. At this point, we are going to bring our given into our make believe area. This accounts for step 3. ---------------------3. ((A ∧ K) v L) [Given] We are now able to apply the logical law of Modus Tollendo Ponens. This law states that if we have 'φ v ψ' and '¬ψ', where 'φ' and 'ψ' are formulas, then we can conclude 'φ'. In step 3, 'φ' stands for '(A ∧ K)' and 'ψ' stands for ' L'. Our '¬ψ' is found in step 2. As such, we can conclude 'φ', which is '(A ∧ K)'. This accounts for step 4. ---------------------4. (A ∧ K) [Modus Tollendo Ponens - 2, 3] At this point, it is easy to see how to derive 'K'. We use the law sometimes called simplification. It states that if you have 'φ ∧ ψ', then you can conclude 'ψ'. You can also conclude just 'φ' by the same law. As such, we can conclude 'K' in step 5. ---------------------5. K [Simplification - 4] At this point, we assumed '¬L' and derived 'K'. As such, by conditional proof, this constitues a derivation of '¬L → K' from our given. I can now strike through the 'Show' in step 1since it has been demonstrated. Q.E.D. Here is what the proof looks like without the comments... 1. ~~Show~~ ¬L → K ---------------------2. Assume ¬L ---------------------3. ((A ∧ K) v L) [Given] ---------------------4. (A ∧ K) [Modus Tollendo Ponens - 2, 3] ---------------------5. K [Simplification - 4]

Markh5000
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Posted 04/14/08 - 10:31 AM:	#5
OK, again, thanks. I guess what I'm having trouble with are the names of the arguments on which I'm basing my statements. Such as, with #4, I have D & C as being equal to ~(~D v ~C) but what is the name of the rule that implies that? I guess our instructor didn't go over that too well. Anyway, I will post what I have for the rest of the problems; I haven't gotten far...like I said, this isn't my forte. I'm not a logic major. #2 1. M v P 2. ~D ⇔ ~P 3. ~M /: D 4. ~D AFR <----our instructor told us to use this for RAA. Stands for Assumed for Reductio 5. M MP #3 1. H → I 2. J → K 3. H v J /: I v K 4. ~I ACP <----and this stands for Assumed for Conditional Proof #4 1. (A & B) v (A & C) 2. D → (C → ~B) 3. E & ~C /: ~D v ~C 4. ~(~D v ~C) AFR 5. D & C #5 1. ~L v ~H 2. H v (M → K) 3. ~N → ~K /:L → (M → N) 4. L ACP

Brian Bosse
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Posted 04/14/08 - 12:43 PM:	#6
Hello Mark, Thank you for putting down what you have so far. This post will deal with #2. What are you trying to derive? You are deriving 'D'. As such, I will make this the first step... 1. Show D You are correct that your next step in a Reductio Ad Absurdum proof is to assume the negation of what you are trying to show. This is my step 2. -----2. Assume ¬D Now what? Well, look at your givens. Is there anything in your givens that allows you to use a rule of inference with the assumption in step 2? Pay particular attention to the given ¬D ⇔ ¬P. Although directly you cannot use this given, this given will lead you to the being able to apply Modus Ponens using step 2. Can you see how? Step 3 should be to simply list the given ¬D ⇔ ¬P. Step 4 should be an immediate inference from step 3 whose conclusion will allow you to use Modus Ponens in step 5. Good luck! Brian

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