0 0 votes If L1 is context free and L2 is not context free, then L1 ∩ L2 is context free. Is this true or not? Theory of Computation + – atul_21 1.3k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes No..in worst case it will not be CFL.. L1 intersection L2= higher order language among two.. joshi_nitish answered Jun 29, 2017 joshi_nitish comment Share Follow See all 9 Comments 9 9 Comments reply Show 6 previous comments joshi_nitish commented Jun 29, 2017 reply Follow flag that's right way.. 0 0 replyShare popo040 commented Jun 29, 2017 reply Follow flag suppose if L2 was CSL?? then what would be intersection of the two?? CSL?? 0 0 replyShare student2018 commented Jun 30, 2017 reply Follow flag As per my understanding for union highest of two languages is the resultant language For intersection lowest of two languages (but not always true) I'm I wrong 0 0 replyShare Please log in or register to add a comment.
0 0 votes take an example L1= (a^n b^n | where n >=1 ) //CFL L2=(a^n b^n c^n| where n >=1 ) // NCFL there intersection are not CFL BUT if L1= (a^n b^n | where n >=0 ) //CFL L2=(a^n b^n c^n| where n >=0 ) // NCFL then there intersection will be empty language which is regular which is CFL hence we may contradict here BUT we always need do focus on worst case so this statement is FALSE akankshadewangan24 answered Jun 30, 2017 akankshadewangan24 comment Share Follow 0 reply Please log in or register to add a comment.