Premise 1: For all x in D [ (A(x) or B(x)) -> (M(x) or N(x)) ] Premise 2: There exists an x in D [ A(x) and ~N(x) ] Solution: There exists an x in D [ B(x) or M(x) ] Any help is appreciated. Thanks!
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Now you supply the reasons.
Thanks for your help. I understand all the reasons except the last step how did you get to B(t) or M(t)?
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