retagged by
261 views

1 Answer

0 votes
0 votes

By Rice Theorem 

For L2, we can have Tyes = Σand Tno = Ф but here we cannot have Tyes and Tno such that L(Tyes)⊂L(Tno). So L2 is not recursive.

For L1, we can have Tno = Σand Tyes = Ф and here L(Tyes)⊂L(Tno) So L1 is not RE.

Related questions