Rice theorem - $EQ_{TM}$

Question

Could use some help understanding if Rice’s theorem applies to the following language: $ = \{\langle \rangle \lvert M \text{ } () \subseteq _{TM} \}$ (where $EQ_{TM} =\{\langle M,N\rangle| M,N \text{ are TMs and } L(M)= L(N)\} $ EQ is in none to my understanding)

My first intuition is no, because C (the property) is not exactly a property of a language. If this is indeed false then I should be able to construct two TMs that have the same language but only one of them belongs to L (struggling with that)

Answer 1

You seem to have a hard time understanding what it means for a language to be semantic. Formally

A language $L$ is semantic, if for TMs $M_1, M_2$ with $L(M_1) = L(M_2)$ holds $\langle M_1 \rangle \in L \iff \langle M_2 \rangle \in L$.

But intuitively, this only means something like this

A language $L$ is semantic, if whether $\langle M \rangle \in L$ depends only on $L(M)$, not of $M$ itself.

Your language for example is semantic, because whether $\langle M \rangle \in L$ is decided only by $L(M) \subseteq EQ_{TM}$, it doesn't really matter what $M$ looks like.

You can also show this using the definition, assume $M_1, M_2$ are TMs with $L(M_1) = L(M_2)$. Then either $L(M_1) = L(M_2) \subseteq EQ_{TM}$ or $L(M_1) = L(M_2) \nsubseteq EQ_{TM}$. Now, is it possible for only one of $\langle M_1 \rangle, \langle M_2 \rangle$ to be in $L$?

Answer 2

Yes, you can apply Rice' Theorem. You need that $L$ is of the form

$$ L = \{\langle M \rangle \mid L(M)\in U \}$$

Where $U$ is a proper, nonempty subset of the r.e. languages. (In this case we say the $U$ is not trivial.) In your case we have that $$U = \{ L \mid \text{$L$ is r.e. und $L \subseteq EQ_{TM}$ }\}$$

No Matter how you did the encoding in $EQ_{TM}$ this set is not $\Sigma^*$. It follows that $\emptyset$ is in $U$, but $\Sigma^*$ is not in $U$. Thus, $U$ has the tow required properties (proper subset because $\Sigma^*$ is missing, not empty because it contains $\emptyset$).