1 votes 1 votes Mathematical Logic mathematical-logic + – Samujjal Das asked Dec 15, 2016 • retagged May 12, 2021 by Shiva Sagar Rao Samujjal Das 1.1k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply santhoshdevulapally commented Dec 15, 2016 reply Follow Share i think all are correct? 0 votes 0 votes Samujjal Das commented Dec 15, 2016 reply Follow Share Ya D is the option. Can you explain why S1 is correct? S2 is correct and S3 is the negation of S1. 1 votes 1 votes thor commented Dec 15, 2016 reply Follow Share In above statement if everyone is not a politician then B converts to $false\; \rightarrow\; \exists_x Q(x)$ which is true but above statement yields false. How is it right?? 0 votes 0 votes santhoshdevulapally commented Dec 15, 2016 reply Follow Share Actual logic is $\forall xp(x)->\exists x Q(x)$ apply p->q=~pvq. $\sim \forall (x)p(x) V\exists (x) Q(x)$ $\exists x \sim p(x) V\exists (x) Q(x)$ $\exists x[ \sim p(x) V Q(x)]$ $\exists x[ p(x) -> Q(x)]$ 0 votes 0 votes Samujjal Das commented Dec 15, 2016 reply Follow Share Got it man!! Thanks a lot. 1 votes 1 votes thor commented Dec 15, 2016 reply Follow Share @santhoshdevulapally In above statement if everyone is not a politician then B converts to false→∃xQ(x) which is true but above statement yields false. How is it right?? 0 votes 0 votes santhoshdevulapally commented Dec 15, 2016 reply Follow Share instead of taking statement derive using predicates. otherwise take p(1)= F and p(2)=T. & Q(1)=T and Q(2)=F. $\forall xP(x)=p(1)\wedge p(2) \exists xP(x)=p(1)\vee p(2)$. apply for p->q and q->p then u get p=q. 1 votes 1 votes Shiva Sagar Rao commented May 12, 2021 reply Follow Share https://gateoverflow.in/103060/logic 0 votes 0 votes Please log in or register to add a comment.