Algorithmic problem of regular, context-free, and recursively enumerable languages

Question

Consider a language $L_1$ that is recursively enumerate, $L_2$ that is regular, and $L_3$ that is context-free.

Are the following problems algorithmically decidable?

1. Is $L_1 \cap L_2 = L_3$?
2. Is $L_1 \cap L_3 = L_2$?

I think problem 1 is undecidable since $L_1 \cap L_2$ is a recursively enumerable language it can be expressed as a Turing machine and since $L_3$ is context free it can be expressed as a Grammar or Turing machine. Since we cannot determine the equivalence of two Turing machines, the problem is undecidable.

I think problem 2 is also undecidable for similar reasons, only the right hand side ($L_2$) is regular instead of context-free.

Am I correct?

Answer 1

Both questions are undecidable. Notice that the empty language is both regular and context-free. Thus we can state that the problem of determining if $L_1 \cap L_2$ is context-free is at least as hard as determining if $L_1$ is context-free (and similarly determining if $L_1 \cap L_3$ is a regular language is at least as hard as determining if $L_1$ is a regular language). By Rice's Theorem any non-trivial property about a Turing-Machine is undecidable and thus your two questions are undecidable.

For a detailed proof/answer to your second question look at this post. IT shows how you would apply Rice's Theorem in your cases.