# Show that the equivalence problem for deterministic Turing machines and its complement is not semi-decidable

I would like to show that $$EQ_{DTM} = \{ (\langle M_1\rangle,\langle M_2\rangle) \mid M_1\text{ and } M_2 \text{ are DTMs and } L(M_1)=L(M_2)\}$$ and $$\overline{EQ_{DTM}}$$ are not semi-decidable. I think there must be a reduction from the complement of the halting problem.

I think an easier way to approach this is by reducing from $$ALL_{TM}=\{\langle M \rangle: M \text{ is a DTM and } L(M)=\Sigma^*\}$$
Showing that $ALL_{TM}$ is not in in RE nor coRE a standard exercise, and you can find the solution e.g. here
Then, we can reduce from $ALL_{TM}$ to $EQ_{TM}$ as follows: given input $\langle{M\rangle}$ for $ALL_{TM}$, we output $\langle{M,T\rangle}$ where $T$ is a fixed TM that accepts every input immediately. Thus, $L(T)=\Sigma^*$, and we have that $L(M)=L(T)$ iff $L(M)=\Sigma^*$, so the reduction is correct.