# Whether language of all turing machines is decidable or undecidable or semi-decidable?

I recently came across this language:

$$L=\{| \text{TM accepts recursively enumerable languages}\}$$

It was asked in the question to find out whether language L is decidable or undecidable.

As far as I could reason out, language L is the set of all recursive enumerable languages or set of all turing machines.

But I can't figure out whether this language is decidable or not.

• Well, if, as you say, $L$ is simply the language of all TM encodings, then what do you need to check in order to decide it? – dkaeae Dec 4 '18 at 9:10
• cs.stackexchange.com/questions/37818/… – Sagnik Jan 22 '19 at 6:26

Let us express $$L$$ slightly more precisely. $$L=\{\; \mid \text{The language accepted by } TM \\ \text{ is a recursively enumerable language}\}$$
Since for any Turing machine $$TM$$, the language accepted by it is a recursively enumerable language by definition, that restriction clause for $$TM$$ does not have any effect at all. We have, as indicated by OP, $$L \text{ is the set of encodings of Turing machines.}$$
So the simplified answer is "$$L$$ is decidable". This is the generally accepted answer. It is especially suitable when the encoding of Turing machines is implied to be decidable such as in the book Introduction to the theory of computation by Michael Sipser. Or when a decidable encoding of Turing machines has been described explicitly and has been assumed to be the default such as in the book Introduction to automata theory languages and computation by John E. Hopcraft et. al.
(The complete answer would be "$$L$$ is decidable unless an unusual encoding of Turing machines is used", which sounds, however, pedantic and which is usually not a welcomed answer.)