3 votes 3 votes $L1$ is a Context free language (CFL), $L2$ is a Deterministic Context free language (DCFL) and , $L = L1 \cap\overline{L2}$ then $L$ isa) Need not be CFL b) not CFL c)DCFL Theory of Computation context-free-language closure-property + – srestha asked Dec 2, 2015 retagged Jul 4, 2017 by Arjun srestha 740 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 3 votes 3 votes L2 = DCFL , which is closed under complement so L2' = DCFL L1 = CFL , L = CFL ∩ (DCFL )' = CFL ∩ DCFL BUT ,DCFL AND CFL both are not closed in intersection ...so it may or may not be CFL. so answer is A) http://gatecse.in/wiki/Closure_Property_of_Language_Families minal answered Dec 2, 2015 selected Dec 2, 2015 by srestha minal comment Share Follow See all 3 Comments See all 3 3 Comments reply srestha commented Dec 2, 2015 reply Follow Share @Sonam I agree. and also ur ans is correct. But my confusion is "DCFL AND CFL both are not closed in intersection" right? So, it should be not CFL why u told that it may CFL? what is that "may " case I want to know 0 votes 0 votes minal commented Dec 2, 2015 reply Follow Share not closed means if we perform operation then they might be in or may not .... we cant conclude that it always not in cfl or dcfl whatever ..thats we here option A) is correct ... if i got any eg i will tell you .. 3 votes 3 votes srestha commented Dec 2, 2015 reply Follow Share ok 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes i guess language which is not in dcfl than it cannot be even in cfl so ans is (b) from expressive power theory correct me! abhishek14893 answered Oct 8, 2016 abhishek14893 comment Share Follow See all 0 reply Please log in or register to add a comment.