2 votes 2 votes Which of the following is true for the predicate logic P ? ~ $\forall z[P(z) \rightarrow ($~$Q(z)\rightarrow P(z)) ]$ a.) P is satisfiable b.) P is Tautology c.) P is Contradiction d.) None of these G.K.T asked Jan 16, 2018 G.K.T 560 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply gauravkc commented Jan 16, 2018 reply Follow Share Contradiction? 0 votes 0 votes G.K.T commented Jan 16, 2018 reply Follow Share Yes given ans is contradiction but my doubt is if I am propagating the negation inside it would become $\exists z[$~$P(z)\rightarrow (Q(z) \rightarrow $~$P(z))]$ so further solving this it comes $\exists z$ (True) what is wrong with this approach ? 0 votes 0 votes gauravkc commented Jan 16, 2018 reply Follow Share ∀z [A(z)] -> For all z, A[z] is true ~ would be ∃z for which A[z] will be false. I thought in statements. Taking negation in terms of statement would be for all z A[z] is false. 0 votes 0 votes G.K.T commented Jan 16, 2018 reply Follow Share if everyone is true its negation would be atleast one is false like in a village everyone says truth only so for negating this statement atleast one person saying lie would suffice hope you get it 0 votes 0 votes Anu007 commented Jan 16, 2018 i edited by Anu007 Jan 16, 2018 reply Follow Share ~∀z[P(z)→(~Q(z)→P(z))] you are solving problem in your ways (by creating new thing). when you propagate negation then why doing partiality with inside expression. ~ ∀z[ some thing] what you are solving is ~ ∀z[P(z)]→(~Q(z)→P(z)) Hope you got it. 0 votes 0 votes G.K.T commented Jan 16, 2018 reply Follow Share sorry, I didn't get you. your expression seems different from what I have posted . 0 votes 0 votes Anu007 commented Jan 16, 2018 reply Follow Share Now check... 0 votes 0 votes G.K.T commented Jan 16, 2018 reply Follow Share But I have propagated the negation all the way till end in the expression I have negated Q an P as well 0 votes 0 votes gauravkc commented Jan 16, 2018 reply Follow Share The expression in square brackets is true. It's a tautology. But how to evaluate ~ ∀z [tautology] in logic terms? 0 votes 0 votes Please log in or register to add a comment.