Can a semi-decidable problem be also decidable?

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Problem Detail: As far as I understand, a semi-decidable (recursively enumerable) problem could be:

  1. decidable (recursive) or
  2. undecidable (nonrecursively enumerable)

This post made me wonder if this is not conventionally followed. This is my answer to it and as far as I understand it is correct:

A semidecidable problem (or equivalently a recursively enumerable problem) could be: Decidable: If the problem and its complement are both semidecidable (or recursively enumerable), then the problem is decidable (recursive). Undecidable: If the problem is semidecidable and its complement is not semidecidable (that is, is not recursively enumerable). Important note: Remember that a decidable (recursive) problem is also semidecidable (recursively enumerable). Conversely, if a problem is not recursively enumerable (semidecidable), then is not recursive (decidable). What the Wikipedia entry says is that: Partially decidable problems THAT ARE NOT DECIDABLE are called undecidable. In general, a semidecidable problem (recursively enumerable) could be decidable (recursive) or undecidable (nonrecursively enumerable). Also note that a problem and its complement could both (or just one of them) be not even semi-decidable (nonrecursively enumerable). Also note that, if a problem is recursive, its complement is also recursive.

Is it conventionally (always) understood this way? Is there some literature that presents semi-decidability (partially decidable, recursively enumerable) problem as an equivalent of undecidability?

Asked By : PALEN

Answered By : David Richerby

Yes, a recursively enumerable language may be either decidable or undecidable. To see this, you ust look at the definitions of the terms. A language $L$ is recursive (aka decidable) if there is a Turing machine that halts for all inputs, accepting every word in $L$ and rejecting every word not in $L$. $L$ is recursively enumerable (aka semi-decidable) if there is a Turing machine that halts and accepts any input in $L$ and, for any input not in $L$, it either halts and rejects or it does not halt. Therefore, every recursive language is recursively enumerable. The machine that decides the recursive language is a special case of the machine required for a recursively enumerable language: specifically, it is allowed to either loop forever or reject for words not in $L$; in fact, it always rejects and never uses the option of looping forever.
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Question Source : http://cs.stackexchange.com/questions/19641

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