0 votes 0 votes Cfl is closed under intersection with regular language. Then resultant languages will be regular or cfl ? Let X is cfl Y is regular language L=X intersection Y Then L is what? Theory of Computation theory-of-computation closure-property + – Dhoomketu asked May 10, 2018 • edited Mar 14, 2019 by adeebafatima1 Dhoomketu 593 views answer comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Mk Utkarsh commented May 10, 2018 reply Follow Share CFL 1 votes 1 votes Dhoomketu commented May 10, 2018 reply Follow Share Thank you so much to clear my doubt. 0 votes 0 votes Kumar Iyer commented May 10, 2018 reply Follow Share "Cfl is closed under intersection with regular language." means it will always be the case any CFL with any regular language will always be regular. But we have a statement: "CFL is not closed under complementation" does not mean it will never be a CFL, It means it MAYor MAY NOT be a CFL. 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes CFL intersection REG ----> CFL is closed intersection with regular language . Means intersection may be regular hence it is also cfl. Intersection may be cfl . So it always gives CFL .but not above the CFL . abhishekmehta4u answered May 10, 2018 • selected May 10, 2018 by Dhoomketu abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Let's take an exaple X=$a^n$$b^n$|n>=0 -CFL Y=$a^m$$b^n$|m,n>=0 -regular $a^nb^n\bigcap a^mb^n=a^nb^n$ ,which is CFL Prateek Raghuvanshi answered May 10, 2018 Prateek Raghuvanshi comment Share Follow See all 0 reply Please log in or register to add a comment.