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I want to ask about reduction.

In the proof that Etm is undecideable in the definition of M1 is

1.if x!=w,reject

2.if x==w,run M on input w and accept if M does

In many proofs that I meet I see that bold line but i can not understand how can I do it because I do not know if it will stop.

I would be more than happy to know where am i wrong.

Thanks.

A.Shoob
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1 Answers1

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Accepting with a Turing Machine means to stop in an accepting configuration. So if you simulate M and it accepts, it will stop and you will be able to notice this.

If M does not stop, this means that it does not accept w. In this case you should not accept either. One way of not accepting is running forever. So if your simulation of M runs forever, this makes you run forever, which is just what you should do.

Thus there is no need for you to know whether M will stop. This does not work the other way. run M on input w and accept if M rejects would not be possible to compute, because you would need to detect infinite computations and accept their input.

Peter Leupold
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  • I mean how can i continue to give the answer for Atm if it is infinitie?how the process go after that? i can not imagine it continue. – A.Shoob Jul 04 '17 at 12:07
  • There is no "after that." The simulation of M runs forever. But it does not matter, "where" a computation runs forever, i.e. in which "part" of the programme. If it runs forever, this running forever is the result. And this result is not and does not need to be provided in an explicit manner. Running forever is rejecting. – Peter Leupold Jul 04 '17 at 13:11