$L=\left \{ \left \langle M,D \right \rangle : M=TM\, ,\, D=DFA\, ,\, L(D)\neq \emptyset\, ,\, L(M)\subseteq L(D)\circ L(D) \right \}\notin RE$

Question

$L=\left \{ \left \langle M,D \right \rangle : M\, is\, a\, TM\, ,\, D\, is\, a\, DFA\, ,\, L(D)\neq \emptyset\, ,\, L(M)\subseteq L(D)\circ L(D) \right \}$

$L\notin R$ which can be shown for example with Turing reduction.

I thought that I build a recognizer to $L$, i.e showed that $L\in RE$, but there is some mistake that I don't see, because $L\in coRE$ (and there for $L\notin RE$).

Here is what I wrote:

Input = <M,D> s.t M = TM , D = DFA
1. run BFS scan on the states of D
2. if the BFS in step 1 didn't find accepting sates
3.    reject <M,D>
4. else
5.     build D', which will be a DFA for L(D)∘L(D)
6.     for i = 0 to ∞
7.         for j = 0 to i
8.             for every x∈Σ^j in lexicographical order:
9.                 run M on w for i steps
10.                 if M accepts w
11.                    run D' on w
12.                   if D' accepts w
13.                      accept

What is wrong with that ?

(This is a question from an exam on computability)

score 3 · Accepted Answer · answered Aug 01 '24 at 15:52

You are accepting as soon as you find a word that is in both $L(M)$ and $L(D) \circ L(D)$; that is, you are semideciding

$$\exists w. w \in L(M) \land w \in L(D) \circ L(D)$$

rather than

$$\forall w. w \in L(M) \to w \in L(D) \circ L(D)$$

These are of course not equivalent.

$L=\left \{ \left \langle M,D \right \rangle : M=TM\, ,\, D=DFA\, ,\, L(D)\neq \emptyset\, ,\, L(M)\subseteq L(D)\circ L(D) \right \}\notin RE$

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