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Consider the following problem:

  • Input: A Turing Machine M and a DFA D.
  • Question: Is $L(D) \subseteq L(M)$?

Of course, this problem is not decidable. Because it is known that judging whether a word belong to a Turing Machine is not decidable. If this problem is decidable, then we can write the word as regular expression and construct DFA from this regular expression, and then check the inclusion relationship between DFA and Turing Machine. So this is not decidable.

However, I feel difficult to prove or disprove whether it is recursively enumerable.

Any ideas are welcome.

John L.
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Audra Jacot
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2 Answers2

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No, $\{\langle D,M\rangle\mid D\text{ is a DFA }\land M\text{ is a Turing Machine }\land L(D)\subseteq L(M)\}$ is not recursively enumerable.

Let $Z=\{0^{n}\mid n\ge0\}$. Call a Turing Machine (TM) $M$ a $Z$-TM if $Z\subseteq L(M)$.

Claim. $\{\langle M\rangle\mid M\text{ is a } Z\text{-TM}\}$ is not recursively enumerable.
Proof (by diagonalization): Towards a contradiction suppose $\{\langle M\rangle\mid M\text{ is a } Z\text{-TM}\}$ is recursively enumerable. That implies there is Turing machine $A$ that enumerates all $Z$-TMs, i.e., given a natural number $n$ as input, $A$ will output (the encoding of ) a $Z$-TM, which is denoted by $C_n$ such that the set of all $C_n$ is the set of all $Z$-TMs.

Let us construct TM $B$ so that on input $w$, $B$ will check first whether $w$ contains only $0$s.

  • If not, $B$ accepts. (The behavior of $B$ here does not matter.)
  • If yes, $B$ computes $n$, the number of $0$s in $w$. Then $B$ will use $A$ to obtain $C_n$. Then $B$ will simulate $C_n$ on input $0^n$, until $C_n$ accepts. Then $B$ will run one more step and accepts.

$B$ is different from each $C_n$ for all $n$, but $Z\subseteq L(B)$. This means $A$ misses $B$ in its enumeration of all $Z$-TMs. $\quad\checkmark$

If we can enumerate all pairs of $D$ and $M$ such that $L(D)\subseteq L(M)$, we can further select the pairs such that $L(D)=Z$, since it is decidable whether $L(D)=Z$. Then we can enumerate $Z$-TMs. This contradicts the claim above.

John L.
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  • Almost understand but a few points need to be clarified.

    1. "Z-TMs are not recursively enumerable", is it means that "given a Turing Machine description, if it belong to Z-TMs"?

    2. definition about Z-TMs, is it formally ${ M: \exists w \in Z, w \in L(M) }$?

     – Audra Jacot May 16 '22 at 06:59
  • Yes. It is common for "a Turing machine" to mean " the encoding of a Turing machine". ""Z-TMs are not recursively enumerable" is shorthand for "the set of encodings of $Z$-TMs is recursively enumerable." 2. ${ M: \forall w \in Z, w \in L(M)}$
    • – John L. May 16 '22 at 12:21
  • @AudraJacot If a proof is highly formalized, the idea may become less "intuitive". If not formalized enough, the proof could be ambiguous. It is not easy to find the right balance. I just made the answer more formal. Is this version good enough? – John L. May 16 '22 at 13:33
  • @AudraJacot Next time when you post a question, you may want to tell something like which section of which textbook you are reading (also for consistent notations and terminologies) or you are preparing the final exam. – John L. May 16 '22 at 13:34
  • Yes, I totally agree with you. Usually, when I write a solution, I will write the formal definition following by casual comments. – Audra Jacot May 16 '22 at 23:20
  • So, in summarize, your central idea is "If ZTMs is recursively enumerable, then ZTMs will be uncountable. All the set of Turing Machine is countable, so ZTMs is not recursively enumerable.", is that right? – Audra Jacot May 16 '22 at 23:47
  • @AudraJacot Your summary sounds reasonable. – John L. May 17 '22 at 01:12