Challenging Natural Deduction Proof

Reply Mon 5 Dec, 2016 08:08 pm
I need help solving this proof with all the rules of inference and replacement rules, but I can't use indirect proof and conditional proof. Any help would be appreciated as I've been working on it for hours!
1. (A>A)>(B>B)
2. (B>C)>(A>A) / (B>B)
14. (B>B)
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Ding an Sich
Reply Mon 5 Dec, 2016 11:01 pm
I'll give you some hints. Take advantage of the middle term in each premise. Then manipulate the result.
Reply Mon 5 Dec, 2016 11:55 pm
@Ding an Sich,
3. ~(A>A)v(B>B) Implication 1
4. ~[(A>A) & ~(B>B)] De Morgan's 3

That's all I can figure out. I can't get the next line
Ding an Sich
Reply Tue 6 Dec, 2016 07:38 am
MT2070 wrote:

3. ~(A>A)v(B>B) Implication 1
4. ~[(A>A) & ~(B>B)] De Morgan's 3

That's all I can figure out. I can't get the next line

Here's what I did to start.

3. ~(B>B)>~(A>A) 1 contraposition
4. ~(A>A)>~(B>C) 2 contraposition
5. ~(B>B)>~(B>C) 3,4 hypothetical syllogism

Try and manipulate the last result to get a conditional by itself. I ended up getting (B>C), then used Modus Tollens on 5 to get ~~(B>B). Then use double negation to get (B>B).

I agree it's tricky since you need to pull apart the conditionals without negation, which is never fun. If it makes you feel any better, all proofs in first order logic can be solved with reductio ad absrudum. Makes life about 1000x easier.
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