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.