Semi-decidability of the language $\overline{L_{\epsilon}}$

Question

Firstly consider the problem: given $L_H = \{R(M)w : M \in TM_0, w\in L(M)\}$ where $R(M)$ are encoded transitions of $M \in TM_0$. Assume for contradiction $\overline{L_{H}}$ is semi-decidable, then there is $Q \in TM_0$ with $L(Q) = \overline{L_{H}}$ therefore for every $M \in TM_0$ we have the following $$Q \ accepts \ input \ R(M)w \iff M \ does \ not \ accept \ input \ w \ \ \ \ (1)$$ Then we construct machine $Z$ s.t. doubles the input and runs machine $Q$. Observe the following for arbitrary $M \in TM_0$: \begin{alignat*}{2} Z \ accepts \ input \ R(M) &\iff Q\ accepts \ input \ R(M)R(M)\\ &\iff M \ does \ not \ accept \ input \ R(M) \end{alignat*} Taking $M = Z$ will yield us a contradiction. Hence, $\overline{L_{H}}$ is not semi-decidable.

The same technique I am trying to apply for the case $L_{\epsilon} = \{R(M) : M \in TM_0 \ \text{s.t. $M$ accepts $\epsilon$}\}$ where $R(M)$ are encoded transitions of $M \in TM_0$. But I am facing some troubles. Assume for contradiction $\overline{L_{\epsilon}}$ is semi-decidable, then there is $Q \in TM_0$ with $L(Q) = \overline{L_{\epsilon}}$ therefore for every $M \in TM_0$ we have the following $$Q \ accepts \ input \ R(M) \iff M \ does \ not \ accept \ input \ \epsilon$$ Then I cannot really find appropriate $Z$ for this case, as doubling the input simply won't work. Any suggestions are appreciated.

Answer 1

Assume for contradiction $\overline{L_{\epsilon}}$ is semi-decidable. It's easily proven that $L_{\epsilon}$ is semi-decidable by reduction to the HALTING problem, i.e. $L_{H}$ is semi-decidable. If $L_{\epsilon}$ and $\overline{L_{\epsilon}}$ are semi-decidable, hence $L_{\epsilon}$ is decidable. This is a contradiction, as $L_{\epsilon}$ as well as $L_{H}$ are semi-decidable.

In this prove the following claim was used:

$\textbf{Claim:} \text{ If $L$ and $\overline{L}$ are semi-decidable, then $L$ is decidable}$