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Unformatted text preview: Proof
S="On input M , w where M is a TM: 1. Run R on input M , w until it halts. 2. If R rejects, (i.e. M loops on w ) - reject. (At this stage we know that R accepts, and we conclude that M halts on input w.) 3. Simulate M on w until it halts. 4. If M accepts - accept, otherwise - reject. "
14 Proof (cont.) Assume by way of contradiction that HALTTM is decidable and let R be a TM deciding it. In the next slide we present TM S that uses R as a subroutine and decides ATM . Since ATM is undecidable this constitutes a contradiction, so R does not exist. 13 Another Example Another Example
Note: Since we already know that both ATM and HALTTM are undecidable, this new proof does not add any new information. We bring it here only for the the sake of demonstration. In the discussion, you saw how Diagonalization can be used to prove that HALTTM is not decidable. We can use this result to prove by reduction that ATM is not decidable. 15 16 ...
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- Winter '99