0

# Challenging Natural Deduction Proof

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)
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14. (B>B)
• Topic Stats
• Top Replies
Type: Question • Score: 0 • Views: 2,012 • Replies: 3

Ding an Sich

3
Mon 5 Dec, 2016 11:01 pm
@MT2070,
I'll give you some hints. Take advantage of the middle term in each premise. Then manipulate the result.
MT2070

1
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

2
Tue 6 Dec, 2016 07:38 am
@MT2070,
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.
0 Replies

### Related Topics

How can we be sure? - Discussion by Raishu-tensho
Proof of nonexistence of free will - Discussion by litewave
Destroy My Belief System, Please! - Discussion by Thomas
Star Wars in Philosophy. - Discussion by Logicus
Existence of Everything. - Discussion by Logicus
Is it better to be feared or loved? - Discussion by Black King
Paradigm shifts - Question by Cyracuz

1. Forums
2. » Challenging Natural Deduction Proof