Question 1113762: I need help on constructing the proof to make the argument valid. The Commas are the breaks. Also, I don't know if I am doing this right
9.~~A, B-> ~A, A :. ~B
My work:
1.~~A
2. B->~A
3. A
4. :. ~B
ADD Step 1, 2
MT Steps 3, 4
Am I correct? I don't know if I am doing this right?
Answer by math_helper(2461)   (Show Source):
You can put this solution on YOUR website! I need help on constructing the proof to make the argument valid. The Commas are the breaks. Also, I don't know if I am doing this right
9.~~A, B-> ~A, A :. ~B
My work:
1.~~A
2. B->~A
3. A
4. :. ~B
ADD Step 1, 2
MT Steps 3, 4
Am I correct? I don't know if I am doing this right?
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I do not think you can use ADD 1,2. Disjunction Introduction (aka ADD) works like this:
1. Q
2. Q v P 1 ADD (essentially, if "Q" is true then "Q or P" is true)
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Here's how I'd do it, being careful to use Double Negation introduction because many formal logic texts
say "it is not the case that it is not raining" is weaker than "it is raining" and therefore treat ~~A as slightly weaker than A. The two are not directly interchangeable without DN elimination or DN introduction (sigh):
1. ~~A Premise
2. B-->~A Premise
3. A Premise
4. ~~A 3 Double Negation (DN)
5. ~B 2,4 Modus Tollens (MT) (Conclusion)
I used to be very good at formal logic, but I'm rusty. Another tutor may look at my solution and come up with a better one. I'm pretty certain about not using 1,2 ADD though.
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EDIT: As I was reviewing this, I noticed 3. A is probably one of your proof lines (not a premise). If so, then the proof can be shortened:
1. ~~A Premise
2. B-->~A Premise
3. ~B 1,2 Modus Tollens (MT) (Conclusion)
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