1| A⊃(~BVC) ASS 2| B ASS 3| A⊃~C ASS ------------------
So far I have:
1| A⊃(~BVC) ASS 2| B ASS 3| A⊃~C ASS 4| |A ASS / ⊃E 5 |~BVC 1,4 ⊃E
I know I need to derive a contradiction to get ~A, and I see it hiding in ~B and B, and ~C and C, but I can't seem to get it out of those last lines...Any help would be appreciated! (and i'll hit the thank you button ;))
Mar 21st 2012, 06:38 AM
Re: Derive ~A
Assume A. Then ~B \/ C from 1. Also, ~C from 3. Using disjunction elimination on ~B \/ C, assume ~B. This gives a contradiction with B. Similarly, assume C. This gives a contradiction with ~C. Thus, we get a contradiction from assuming A.