(F&~H)→G
~A→~H
~HvC
~(B→E)
C→D
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{~G→[F→(D&A)]}&B
Can someone solve this, I believe the answer is that it is valid. But I don't know how to negate a ~A on a separate branch. I have to work through it some more but would love some help thanks in advance =)
Search found 1 match
- Thu Oct 08, 2015 11:23 pm
- Forum: Logic and Philosophy of Mathematics
- Topic: Challenging Truth Tree problem
- Replies: 1
- Views: 1283