1 votes 1 votes How to prove this.. CFL is closed under intersection with regular languages. Theory of Computation theory-of-computation context-free-language closure-property + – vaishali jhalani asked Dec 14, 2016 retagged Jul 4, 2017 by Arjun vaishali jhalani 722 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply santhoshdevulapally commented Dec 14, 2016 reply Follow Share i)CFL=$a^{n}b^{n}$ and REG=$\phi$ CFL $\cap$ REG=$\phi$----This is Reg and Cfl. ii)CFL=$a^{n}b^{n}$ and REG=$(a+b)^{*}$ CFL $\cap REG=$a^{n}b^{n}$ ----This is CFL. so we can conclude. 0 votes 0 votes Rupendra Choudhary commented May 29, 2017 reply Follow Share It's wrong way. How can you prove some closure property because you find one example satisfying that property . In that way very soon you will prove CFL is close under intersection and complement. 0 votes 0 votes Please log in or register to add a comment.
–2 votes –2 votes Absolutely not. CFL is not closed under intersection with regular language, Suppose we have a L1=a^nb^n(CFL) L2=ϕ(regular languages) L1UL2=CFl But in the case of DCFL .It will be right. Paras Nath answered Dec 23, 2016 Paras Nath comment Share Follow See 1 comment See all 1 1 comment reply Rupendra Choudhary commented May 29, 2017 reply Follow Share brother CFL is close under regular intersection , regular union , regular difference. by close we mean the output of those operations would be CFL for sure. 0 votes 0 votes Please log in or register to add a comment.