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I have a Turing-Machine M and i have proven that M is not a decider. I have then proven A=L(M) or that the language A that M recognizes. I have now been asked "Is the language (A) Turing-decidable".

my question is, if i have already proven that M is not a decider could i not use that to imply that the language (A) is not Turing-Decidable? The way i see it, the language for machine M would consist of, not only accepted languages, but also infinity long strings that never halt. That would make the language also not Turing-decidable?

Thanks for any advice.

skerr4311
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    The existence of a non-deciding TM for a language A does not imply that the language is not Turing-decidable. A language is decidable if there exists a machine that decides it. A language is not decidable if there exists no machine that decides it. For example, take the regular language `a*`. You can make a TM that halts when it reads any number of `a`, but loops forever when it reads a `b`. This is not a decider, but the language `a*` is decidable. What you need to show is that **no** machine can decide A, or that **a** machine can decide A. – Welbog May 10 '19 at 13:04
  • Thank you for that, so all I would have to do is say that there exists a TM that decides A and as a simple example, I could use the TM I already have and remove the state that loops endlessly on itself and redirecte it to the reject state. Thank you for the clarification – skerr4311 May 11 '19 at 20:49

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I think there are some ambiguities in the question that should be clarified. To be in the same page, let me clarify some definitions:

Definitions

  1. A formal language L is called "Turing-decidable" if there exists a "decider" for it.
  2. A "decider" is a TM that halts on all strings of Sigma-star (Sigma is the alphabet of this TM.). To be specific, when we input strings from L (that should be accepted), or from L-bar (that should be rejected), the machine halts. Note that all of these strings are from Sigma-star. We are not allowed to input strings outside of Sigma-star.
  3. “String” is a finite sequence of symbols from alphabet. Therefore, in your question: “... infinite long strings ...” is an invalid statement in formal languages because of strings must be finite.

So, when you say, you’ve proven M is not a decider, it means, you’ve proven that M falls in an infinite loop for “at least one string of Sigma-star”. This string can be in the set A or A-bar.

My point here is that you cannot prove a TM is, per se, decider or non-decider without any language.

Now, based on this clarification, if you got your answer, that’s excellent, but if not, can you please rephrase your question more precisely.

Henry Yik
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Ahmad
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