1 votes 1 votes HeadShot asked Sep 6, 2018 HeadShot 948 views answer comment Share Follow See all 17 Comments See all 17 17 Comments reply arvin commented Sep 6, 2018 reply Follow Share c : both 0 votes 0 votes sakharam commented Sep 6, 2018 reply Follow Share In the second statement do we consider conditional operator as right associative or left associative? 0 votes 0 votes HeadShot commented Sep 6, 2018 reply Follow Share @arvin Key - D (None) 0 votes 0 votes Tesla! commented Sep 6, 2018 reply Follow Share S2 is correct X is best friend of some y, and for all z except y, x and z are not best friend 0 votes 0 votes arvin commented Sep 6, 2018 reply Follow Share and statement 1 says that if x and y are best friends and x and z are not best friends then y is not equal to z i think this must be also true... 0 votes 0 votes srestha commented Sep 6, 2018 reply Follow Share I too think S2 is correct 0 votes 0 votes Bikash Singh commented Sep 6, 2018 reply Follow Share Statement S1 states that if x and y are the best friends and x and z are not best friend than y and z must not be the same person Statement S2 states that if x and y are the best friends then for all z, y and z are not the same people implies x and z are not the best friends. I think the answer will be c: both 0 votes 0 votes Soumya29 commented Sep 6, 2018 reply Follow Share Answer is D only. "Everyone has exactly one best friend" can be represented as - $\forall x \exists y(BF(x,y) \wedge \forall z(BF(x,z) \rightarrow (y=z)))\\OR\\\forall x \exists y(BF(x,y) \wedge \forall z( (y \neq z)\rightarrow \neg BF(x,z)))$ Counter Ex for S1- put x=a , y=b ,z=c and check S2 is not true because of that big braces. right most z is not bounded by $\forall$. 1 votes 1 votes Tesla! commented Sep 6, 2018 reply Follow Share @Bikash Singh @arvin @srestha I think this clears everything (chapter 1 subsection 1.5 Rosen 7th edition page number 62). Option D is correct 2 votes 2 votes srestha commented Sep 6, 2018 reply Follow Share @Tesla I understood that but check S1 once more it is telling if x and y are best friends and x and z are not best friends then y and z are not same person why is it incorrect? 0 votes 0 votes srestha commented Sep 6, 2018 reply Follow Share @Soumya can u elaborate , where is incorrect in S1? 0 votes 0 votes Soumya29 commented Sep 6, 2018 reply Follow Share @Srestha, Suppose- a) x and y are best friends. b)x and z are best friends. Now, in this case, LHS of implication becomes false so no matter what the RHS is, given statement will also be true. So because of this 2 cases arise- i)$y=z$, It makes RHS false and overall statement true. This condition is desirable too. ii)$y \neq z$, It makes RHS true and overall statement true. But This condition is $not$ desirable because in this case, x can have 2 different best friends. Also, note that S1 is inverse of the correct statement. A statement and its inverse can't be equivalent. 1 votes 1 votes Tesla! commented Sep 6, 2018 reply Follow Share S1 is incorrect because consider the following case X is a friend of Y now statement (BF(X, Y))->True and ~BF(X,Y)->False (it is possible for z=y as till now no restriction on Z) so compound statement (BF(X, Y)^~BF(X, Y) become false making the entire proposition true because till now we don't require to check the condition z$\neq$y 1 votes 1 votes Abhinavg commented Sep 6, 2018 reply Follow Share Only S2 is true , S1 becomes true for a case if everyone has exactly 2 friends 0 votes 0 votes Tesla! commented Sep 6, 2018 reply Follow Share S2 will be true even if x and y not best friend and x and z are also not best friend or to make it simple S2 will be true even if x has no best friend 0 votes 0 votes srestha commented Sep 6, 2018 reply Follow Share @Soumya ,@Tesla in S1 there is no bracket So, why statement will evaluate from left side and not from right side? 0 votes 0 votes Tesla! commented Sep 7, 2018 reply Follow Share Logical "and" has higher priority then 'implication" as per Rosen 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes https://gateoverflow.in/74933/logic Mohit Kumar 6 answered Sep 8, 2018 Mohit Kumar 6 comment Share Follow See all 0 reply Please log in or register to add a comment.