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I need to prove whether L is decidable or not:

L={<M> | M is a TM and the union of L(M) and H_TM is in RE}

( H_TM={<M,w> | M is a TM that halts on w} )

ChikChak
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I suppose that <...> is the number of the TM in a Gödelization. L(M) is a set of words while H_TM is a set of pairs. Thus their union is disjoint, no element will appear in both. Concequently, the union is enumerable iff its two parts are. H_TM is enumerable, and thus the enumerability depends only on L(M). But being the language of a TM means being decidable thus clearly enumerable. Therefore the condition on M in the defintion of L is always true and thus L is the set of all TM descriptions, which is regular and clearly decidable.

Peter Leupold
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    I guess that my notations were not clear: is basically an encoidng for a turing machine,so when another machine get such as input it can simulate it.Moreover,H_TM is the halting problem,which you said is decidable,and I'm not sure why.. – ChikChak May 31 '16 at 15:04
  • Sorry! I wrote decidable when I was really referring to RE in the defintion of L. Of course, the halting problem is not decidable, but it is enumerable, I have corrected my answer. – Peter Leupold May 31 '16 at 15:30
  • I'm a bit confused-does the notation L(M) mean that M decide L?or maybe it only recognize L? – ChikChak May 31 '16 at 15:59
  • That is your notation ;-) Deciding and recognizing mean the same. Recognizing means that for any input the TM can decide whether it belongs to the language or not. – Peter Leupold May 31 '16 at 17:10