0 votes 0 votes Mathematical Logic mathematical-logic + – vaishali jhalani asked Dec 16, 2016 vaishali jhalani 655 views answer comment Share Follow See all 16 Comments See all 16 16 Comments reply Sushant Gokhale commented Dec 23, 2016 i edited by Sushant Gokhale Dec 24, 2016 reply Follow Share is it option C ? 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share i am not able to use any identity for statement 1. p V q--{1} (-p . q)-> r --(ii) from ii,i can only write p v -q v r which propertty to use now?? 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share Just convert A->B to ~AVB. Then used kmap method for 'pqrs' where above statements are in POS form. 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share by k-map mehtod,i am getting (p+r).(p+q) so,from simplification, we can get (p V r) which means ~r -> p is this correct? btw,i did nt know this kmap method for this thing. 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share If the term is pVq then put 0 for p=0, q=0. Now, Fill the k-map for all such terms. Now, see the meaning of A->B . If A is true, B must be necessarily true. You needn't check for B seperately. Use this definition for kmap. Now, for 1st example, fill the kmap using the premises. Now, test the places in kmap for the terms of the conclusion. If they have 0's filled in, then the statement is valid. 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share Should I show the kmap for 1st example? 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share please if you can..i will have much clear idea then..:-) 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share 1st example. 1st premise: Vx ( p(x) V q(x) ) (terms in kmap shown by blue color) 2nd premise: Vx ( (~p(x) ^ q(x) ) -> r(x) ) = Vx ( ( p(x) V ~q(x) ) V r(x) ) (terms in kmap shown by green color) Conclusion: Vx ( ~r(x) -> p(x) ) = Vx ( r(x) V p(x) ) (terms in kmap shown by red color) Now, when the premises are true (i.e the terms are 0 in kmap), is the conclusion always true(i.e. terms for conclusion also 0)? Yes, always, right? 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share yea...got that..very nice approach..thanks a lot ..:-) one more thing like,here there were only universal quantifiers,but in second option,there are existential also..caan we adopt same way for them also? 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share Yes, its just slightly different. Try out. 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share in second option ,we can write (i) p -> Q ^ R (universal instatiation) can we write (ii) as P ^ S? because then from (ii),we will get P (iii)& S (iv) through simplification. from (iii) and (i), P -> Q ^ R P we get Q^ R(modes ponens) (v) from (v),we can get Q (vi) & R (vii)(simplification) with (vii) and (iv) i.e R and S,we get R^ S(conjunction) correct me if i am wrong 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share For 2nd premise, its given existential quantifier. So, P^S is true only for some instances. 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share alright,so how will we know that for which intances we need to mark 0 in our k-map..?? so,surely this k-map is wrong cuz here,i have marked 0 for every P and evry S.caan you tell at which one to mark 0..? 0 votes 0 votes Sushant Gokhale commented Dec 24, 2016 reply Follow Share P^S is SOP term, right? But there is existential quantifier. So, when p=1 and S=1, the entries would be 1 only sometimes. Now, Lets ignore P^S. So, remaining entries ( excluding the 2 premises)would be 1 or dont care, right? Now, lets look at the conclusion. $\exists$x { r(x) ^ s(x) } which corresponds to third column from left. So, you want 3rd column to be all 1's atleast once or for some instances. Lets go back to $\exists$x { P(x) ^ S(x) } So, now mark all 1's for this function because there exists atleast one instance where both P(x) and S(x) is true. Now, for this instance, just combine the 1's with dont care. So, 2nd statement is also valid. I think I gave wrong answer initially :) 0 votes 0 votes Akriti sood commented Dec 24, 2016 reply Follow Share yes..i got that..thankyou so much..:-) 1 votes 1 votes Shiva Sagar Rao commented May 7, 2021 reply Follow Share https://gateoverflow.in/107257/mock1-ace 0 votes 0 votes Please log in or register to add a comment.