0 votes 0 votes iita asked Jan 31, 2017 edited Jan 31, 2017 by sudsho iita 2.4k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply sudsho commented Jan 31, 2017 reply Follow Share B is wrong..compliment of e cant be f as join of e and f != upperbound i.e h and also meet of e and f != lower bound i.e a 0 votes 0 votes iita commented Jan 31, 2017 reply Follow Share they have also given as C is it...?? 0 votes 0 votes iita commented Jan 31, 2017 reply Follow Share how join of e and f is not h..?? they are connected to h 0 votes 0 votes Rahul Jain25 commented Jan 31, 2017 reply Follow Share Yes C is also false. 0 votes 0 votes sudsho commented Jan 31, 2017 reply Follow Share join of e and f is f and meet is e 0 votes 0 votes sudsho commented Jan 31, 2017 reply Follow Share yea c is also false 0 votes 0 votes iita commented Jan 31, 2017 reply Follow Share for complement of an element to other there sould not any direct edge is it..true.. 0 votes 0 votes sudsho commented Jan 31, 2017 reply Follow Share the statement i have written in the first line should satisfy fr compliments... 2 elements will be compliment of each other if their join = upper bound and their meet is = lower bound are u trying to learn concepts from questions and that too by solving ace? 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes option B AND C is right RAJESHWAR YADAV answered Jan 31, 2017 RAJESHWAR YADAV comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes GLB of e anf f is not a and LUB of e anf f is not h.hence complement of e is not f. GLB of d anf g is not a and LUB of d anf g is not h.hence complement of d is not g. Thus answer is B and C. Arnabi answered Jan 31, 2017 Arnabi comment Share Follow See all 0 reply Please log in or register to add a comment.