Practicing Validity: Using Rules of Inference to Prove Arguments

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The discussion revolves around using rules of inference to prove various logical arguments for an introductory logic class. The original poster presents ten arguments along with their conclusions, seeking assistance in validating them. Participants express concerns about the validity of the first argument, suggesting it may be invalid based on counterexamples. There is a focus on understanding how to group negations and apply logical rules correctly. Overall, the thread emphasizes the importance of effort and comprehension in mastering logical proofs.
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I have about 10 questions, I hope someone can take the time to help me with. The directions are: "Use the 18 rules of inference, supply a proof for each of the following arguments." So, here goes:

1.) 1. (R v X) > (A > B)
2. ~ Q > ~ C
3. ~ C > Z
4. R .Y
5. Q v A /Z v B

2.) 1. E . (P . B)
2. (E . B) >~ (P.~M) /E.M

3.) 1. ~(S v C)
2. ~(S . R) > (C v D) /D

4.) 1. D > P /(I . D) > P

5.) 1. P v (Y . H)
2. (P v Y) >~ (H v C)
3. (P .~ C) > (K . X) /X v T

6.) 1. A = J
2. A v J
3. A > (J > W) /W

7.) 1. ~Q> (C . B)
2. ~T> (B . H)
3. ~(Q . T) /B

8.) 1. (U . P) >Q
2. ~ O > U
3. ~ P > O
4. ~ O . T /Q

9.) 1. (J>K) . (~O>~P)
2. (L > J) . (~M>~O)
3. ~K> (L v~ M)
4. ~K . G /~P

10.) 1. (F . M) > (S v T)
2. (~S v A) > F
3. (~S v B) > M
4. ~S . G /T

Okay, that's all the questions. Now here is the legend key:
/ separates what the conclusion is supposed to be.
. conjunction
v disjunction
> implication
= biconditional
~ negation

This is for an introductory to logic class. I hope someone can help. Thank you.
 
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1.) 1. (R v X) > (A > B)
2. ~ Q > ~ C
3. ~ C > Z
4. R .Y
5. Q v A /Z v B
I read that as:
1. (R v X) > (A > B)
2. (~ Q) > (~ C)
3. (~ C) > Z
4. R .Y
5. Q v A
C. Z v B
By my calculations, this argument is invalid. Counterexample: (R, Y, C, Q) are true and (Z, B, A) are false (X is true or false). Did you copy it correctly? How are you grouping negations?
 
Here are some hints for the next three to get you started.
2) Can you derive ~(P . ~M)?
3) Can you derive (~S v ~R)?
4) Do you have the rule: ((P . Q) > R) = (P > (Q > R))?
 


honestrosewater said:
I read that as:
1. (R v X) > (A > B)
2. (~ Q) > (~ C)
3. (~ C) > Z
4. R .Y
5. Q v A
C. Z v B
By my calculations, this argument is invalid. Counterexample: (R, Y, C, Q) are true and (Z, B, A) are false (X is true or false). Did you copy it correctly? How are you grouping negations?
Yes I did copy it correctly. This is directly from my teacher, too. I don't think he feels it's invalid. And, the ones that you put parentheses around, he didn't. I don't know if that matters or not. I really don't get anything about this class at all. As far as how am I grouping negations, I have no idea on that either. Thanks.
 
completely lost said:
Yes I did copy it correctly. This is directly from my teacher, too. I don't think he feels it's invalid.
Well, regardless of what he may feel, I checked it again, and it is invalid. Do you know how to check an argument for validity?
And, the ones that you put parentheses around, he didn't. I don't know if that matters or not. I really don't get anything about this class at all. As far as how am I grouping negations, I have no idea on that either. Thanks.
I added the parentheses to make clear what was being negated.
If you have no idea what you're doing, there isn't much I can do for you today. If you have a problem understanding something specific, I'll try to help. But we don't do people's homework for them here, so you'll have to put in some effort.
 
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