0 votes 0 votes L1 = {an bm | n ≤ m ≤ 2n} L2 = {an bm│m≠n & m≠2n} Which of the following is true? A. Only L1 is CFG B. Only L2 is CFG C. Both are CFG D. None of them is CFG Theory of Computation theory-of-computation context-free-grammar normal + – AmitPatil asked Jan 26, 2017 • retagged Jun 4, 2017 by Arjun AmitPatil 812 views answer comment Share Follow See all 12 Comments See all 12 12 Comments reply Show 9 previous comments Abbas2131 commented Jan 29, 2017 reply Follow Share Thank you , got it. Its a NCFL 0 votes 0 votes akhileshreddy commented Jul 13, 2017 reply Follow Share can anyone please provide cfg for the above languages as well. 0 votes 0 votes prayas commented Jan 12, 2018 reply Follow Share @ Sushant Gokhale This only proves that m!=2n.If we go by your algorithm we would never be able to say whether n!=m. 0 votes 0 votes Please log in or register to add a comment.