1 1 vote Is L={ ambncndm, m$\neq$ n} CFL na ? Theory of Computation theory-of-computation + – Pawan Kumar 2 647 views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply Pawan Kumar 2 commented Jan 3, 2018 reply Follow flag L={ ambncndm, m>n or m<n} thus we can push a , push b then pop b for every c and again pop a for d? Kindly correct me ... thanks 0 0 replyShare Ajay Jadhav commented Jan 3, 2018 reply Follow flag But in adding to m=m n=n you have to compare whether m!=n also, I don't think this is CFL 0 0 replyShare Please log in or register to add a comment.