Prove that $L = \{a^i \;:\; (\exists x \in \mathrm{Lang}(M_i))\;[ xx \notin \mathrm{Lang}(M_i) ] \}$ not recursively enumerable

Question

Past year paper question:

Let $M_i$ denote the Turing machine with code $i$ using the alphabet $\Sigma=\{a,b\}$.

Show that the following language is not recursively enumerable:

$L = \{a^i \;:\; (\exists x \in \mathrm{Lang}(M_i))\;[ xx \notin \mathrm{Lang}(M_i) ] \}$

I tried to show that we can reduce $L_e$, the set of Turing machines that accept no strings, to $L$. To do this, I need to find a transformation $f$ such that is $\mathrm{Lang}(M)$ is empty, then the machine represented by $f(M)$ has some string $x$ but not $xx$, and if $\mathrm{Lang}(M)$ is not empty, then the machine represented by $f(M)$ contains $xx$ if it contains $x$.

But I am not sure how to find such a transformation. Can someone show a transformation to solve this problem?

Answer 1

First, recall that the problem "$L(M)\neq\emptyset$" is semidecidable.

Define $f(M)$ as $a^i$ where $M_i$ is a TM such that

• it accepts $aa$ iff $L(M)\neq\emptyset$,
• it accepts any other string (including $a$).

For the first step: when $M_i$ receives $aa$ as input, it semi-decides $L(M)\neq\emptyset$. If that holds, the semi-decider halts in finite time, so $M_i$ can accept. If that does not hold, the semi-decider diverges, making $M_i$ diverge. This is fine, since $aa$ is not accepted.

Note that we never reject $aa$, because we do not have a full decider for $L(M)\neq\emptyset$. This however does not matter: we do not have to reject $aa$, we only have to avoid accepting it, and diverging suffices.