Source: cirosantilli/recursively-enumerable-language

= Recursively enumerable language
{tag=Chomsky hierarchy}
{wiki}

There is a <Turing machine> that halts for every member of the language with the answer yes, but does not necessarily halt for non-members.

Non-examples: https://cs.stackexchange.com/questions/52503/non-recursively-enumerable-languages

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