Challenging Natural Deduction Proof

I'll give you some hints. Take advantage of the middle term in each premise. Then manipulate the result.

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.