Thank you for your previous answer it was very helpful.
In relation to part b of the question, I seek clarification on a specific aspect of the reduction process. we know that L1 is not recursively enumerable due to Rice's theorem. Subsequently, I utilized this information to demonstrate that L2 is also not REC in part b.
My specific question pertains to the handling of the relationship between L1 and L2 in the reduction. In the (yes-yes part), can I substitute L(M) from L1 = L(M) and L(M prime) into ∅? Similarly, in the (no-no part), is it acceptable to introduce LR = ∅?
Yes-Yes Part:
if enc(M) ∈ L1 then L(M)= Σ* and since L(M)=L(M) and we have that L(M) = Σ* so we will have that L(M) U L(M PRIME) U LR = Σ* so enc(M,M prime) ∈ L2
No-No Part:
if enc(M) ∉ L1 then L(M)≠ Σ* then since L(M)=L(M) and we have that L(Mprime) and LR = ∅ so that L(M) U L(M prime) U LR ≠ Σ* so then enc(M,M prime) ∉ L2