5 votes 5 votes If $L$ and $P$ are two recursively enumerable languages then they are not closed under Kleene star $L^*$ of $L$ Intersection $L \cap P$ Union $L \cup P$ Set difference Theory of Computation isro2017 set-theory theory-of-computation recursive-and-recursively-enumerable-languages closure-property + – sh!va asked May 7, 2017 edited Feb 25, 2019 by Devwritt sh!va 6.8k views answer comment Share Follow See 1 comment See all 1 1 comment reply sourav. commented Oct 14, 2017 reply Follow Share what is wrong in this question? 2 votes 2 votes Please log in or register to add a comment.
Best answer 10 votes 10 votes Set difference=$L-P =L\cap P^{C}$. Since, recursively enumerable languages are closed under intersection but not under complement, Set difference of these two language is not closed. Refer : This also REGGIE S answered May 7, 2017 edited May 8, 2017 by Prashant. REGGIE S comment Share Follow See 1 comment See all 1 1 comment reply shyamalenduk commented Apr 24, 2020 reply Follow Share Intersection of Recursive and Recursively enumerable language is recursively enumerable (https://gateoverflow.in/237873/recursive-and-recursively-enumerable). If L and P are recursively enumerable then L−P=L∩P' (complement). P' must be recursive. Thus L−P=L∩P' is recursively enumerable (). Thus recursively enumerable is closed under set difference. But you are saying L-P is not closed. Please clarify. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes D) SET DIFFERNCE pawan kumarln answered May 8, 2017 pawan kumarln comment Share Follow See 1 comment See all 1 1 comment reply Ishan Jawa commented May 8, 2017 reply Follow Share recursively enumerabke languages are closed under union i tersection and closure..the operation L-P i.e difference can be written as L and ~P since recursively enumerable languages are not closed under complement.so option d is correct 1 votes 1 votes Please log in or register to add a comment.