Define languages L0L0 and L1L1 as follows :
L0={⟨M,w,0⟩∣M halts on w}L0={⟨M,w,0⟩∣M halts on w}
L1={⟨M,w,1⟩∣M does not halts on w}L1={⟨M,w,1⟩∣M does not halts on w}
Here ⟨M,w,i⟩⟨M,w,i⟩ is a triplet, whose first component MM is an encoding of a Turing
Machine, second component ww is a string, and third component ii is a bit.
Let L=L0∪L1L=L0∪L1. Which of the following is true?
1. L is recursively enumerable, but L′ is not
2.L′ is recursively enumerable, but L is not
3. Both L and L′ are recursive
4. Neither L nor L′ is recursively enumerable
@srestha @Anu007 @others
As L0 is recursive as it halt
And L1 is recursive enumerable as it does not halt
So recursive U recursive enumerable will recursive enumerable rt?