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How does one prove that some arbitrary language $L$ is not recursively enumerable?

I know I can prove that the language $L$ is recursively enumerable by constructing a Turing machine $M$ that accepts all words in the language (and the language would be even recursive if $M$ halts on all inputs).

But it is not clear to me how to prove that language in not RE. I was thinking about showing the fact, that such TM could not be constructed for a given language, but proving non-existence is always difficult.

Discrete lizard
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Smajl
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2 Answers2

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Here are two methods.

Consider the complement

Theorem. If a language $L$ and its complement are both RE, they are both recursive.

Proof. Decide whether $w\in L$ by enumerating $L$ and its complement in parallel and accept/reject as soon as $w$ appears in one of the enumerations. $\Box$

So, if you can prove that $L$ is not recursive but its complement is RE, then $L$ is not RE.

Halting problems

Theorem. Let $\mathcal{M}$ be the class of Turing machines equipped with an oracle for the ordinary Turing machine halting problem. The halting problem for $\mathcal{M}$ is not RE.

Proof. Essentially the same as the proof that the ordinary Turing machine halting problem is not recursive. $\Box$

So, if you can reduce the halting problem for $\mathcal{M}$ to your problem, your problem is not RE.

Community
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David Richerby
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Some common techniques include:

We start by picking any $L'$ which is known to be non RE, e.g. we let $L'$ to be the complement of the halting problem. Then we prove the m-reduction $L' \leq_m L$. If we can do that, we can conclude that $L$ is not RE, since otherwise $L'$ would be RE -- contradiction.

The Rice-Shapiro theorem is a very convenient and widely applicable method to establish "non-RE" properties. It is not a silver bullet which always applies, but many common languages are covered by it.

chi
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