1 votes 1 votes abhishekmehta4u asked Dec 16, 2018 abhishekmehta4u 325 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply Shobhit Joshi commented Dec 16, 2018 reply Follow Share assuming $S_{2}$ means $(\exists_{x}H(x)\,\wedge\,\forall_{x}(H(x)\rightarrow M(x)))\,\Rightarrow\,\forall_{x}M(x)$ $LHS$ means "For some $x, H(x)$ is $true$ and $for\,all\,x\,if\,H(x)\,is\,true\,then\,M(x)\,is\,true$" LHS can be $true$ when M(x) is true only when H(x) is true. So, $for\,all\,x,\,M(x)\,may\,not\,be\,true$. So, there is a condition when LHS is $true$ and RHS is $false$. So, $S_{2}$ is not valid 0 votes 0 votes Navneet Kalra commented Dec 16, 2018 reply Follow Share in second one if you write first statement then that can be written as H(a) second can be written as H->M which can also be written as H(a)->M(a) now if u apply inference rule you will get M(a) which is (for some x)M(x) but in question it is given (For all x) M(x) so statement becomes invalid 0 votes 0 votes Navneet Kalra commented Dec 16, 2018 reply Follow Share For first statement take some of true false values for P(x) and Q(x) and verify it will be coming true 0 votes 0 votes Chaitrasj commented Jan 19, 2019 reply Follow Share I am getting S1 to be false, LHS saying "There exists some x, for which P(X) is true, then it implies for all x, Q(X) is also True". If we apply Existential Specification and Universal Specification, I am getting P(b)->Q(b) (b is some fixed variable). And after applying Universal Specification on RHS, it's coming as P(a)->Q(a). (As in for all a this is valid). But it;s not true, because LHS is only for some b and not for all. Can anyone please help where i am going wrong? Thanks in advance. 0 votes 0 votes Please log in or register to add a comment.