27 27 votes State whether the following statements are TRUE or FALSE: The intersection of two CFL's is also a CFL. Theory of Computation gate1987 theory-of-computation context-free-language true-false + – Misbah Ghaya 6.5k views answer comment Share Follow Print See 1 comment 1 1 comment reply anon1 commented Dec 22, 2021 reply Follow flag Need not be. 2 2 replyShare Please log in or register to add a comment.
Best answer 43 43 votes No intersection of two CFLs may or may not be a CFL i.e. CFL is not closed under intersection operation. Example: $L_1: \{ a^nb^nc^m| n,m >=1 \} \cap L_2: \{ a^nb^mc^m | n,m >=1 \}$ $L_3= \{ a^mb^mc^m | m >=1 \}$, which is CSL. Prashant. answered Nov 9, 2016 • edited Apr 15, 2021 by Lakshman Bhaiya Prashant. comment Share Follow See all 10 Comments 10 10 Comments reply Show 7 previous comments Pranavpurkar commented Oct 26, 2022 reply Follow flag Abhrajyoti00so then , $L_{3} = {{a^nb^nc^n | n\geq 1}}$ is also correct. 1 1 replyShare Abhrajyoti00 commented Oct 26, 2022 reply Follow flag Yes, both are same of course. Just variable names 2 2 replyShare Pranavpurkar commented Oct 26, 2022 reply Follow flag Yup . Thanks again :) 0 0 replyShare Please log in or register to add a comment.
11 11 votes false anbnc* is cfl and anb*cn is also cfl but their intersection anbncn is not Anusha Motamarri answered Nov 9, 2016 Anusha Motamarri comment Share Follow 0 reply Please log in or register to add a comment.
2 2 votes False, context free language are not closed under intersection and complement. Tushar Garg answered Mar 15, 2018 Tushar Garg comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes FALSE, as CFL’s are not closed under intersection operation and hence intersection of two CFL’s may or may not be a CFL. It could be CSL also. manikantsharma answered Aug 27, 2022 manikantsharma comment Share Follow 0 reply Please log in or register to add a comment.