As far as I understand, a semi-decidable (recursively enumerable) problem could be: decidable (recursive) or 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?
Can a semi-decidable problem be also decidable?
terminology;computability;undecidability;semi decidability
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.