6 votes 6 votes Since Recursive languages are closed under intersection, therefore it decidable. Am I wrong? Theory of Computation recursive-and-recursively-enumerable-languages decidability + – nikhil_cs asked Jan 18, 2018 nikhil_cs 3.0k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply hs_yadav commented Jan 18, 2018 reply Follow Share nikhi... CFL are not closed under intrsection then it means it is undecidabel........?? actully decidability is seprate concept.... 0 votes 0 votes nikhil_cs commented Jan 18, 2018 reply Follow Share Here it is known that the intersection of two recursive languages is a recursive language, then can't we say that its decidable that intersection will be recursive one? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes See you already know the answer even before finding the actual intersection of the languages (closure property). That's when you call a problem trivially Decidable. Lakshay Kakkar answered Feb 19, 2018 Lakshay Kakkar comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes intersection of 2 recursive languages is always UNDECIDABLE suryaprakash answered Jun 1, 2018 suryaprakash comment Share Follow See 1 comment See all 1 1 comment reply Vamsi krishna satya commented Nov 24, 2018 reply Follow Share Can you explain it or give link to the article proving it? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Yes, It is decidable Refer this https://www.geeksforgeeks.org/decidability-table-in-theory-of-computation/ shashankrustagi answered Dec 1, 2020 shashankrustagi comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Intersection of two recursive gives a recursive language. This is decidable. rish1602 answered Aug 19, 2021 rish1602 comment Share Follow See all 0 reply Please log in or register to add a comment.
