Lecture15_Part5

Lecture15_Part5 - Demonstration 3 We assume that B is...

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Unformatted text preview: Demonstration 3. We assume that B is decidable and use this assumption to prove that A is decidable. In the following slides we assume that ATM is decidable and use this assumption to prove that HALTTM is decidable. 4. We conclude that B is undecidable. Demonstration 1. We know that A is undecidable. Now we pick HALTTM to play the role of A. 2. We want to prove B is undecidable. We pick ATM to play the role of B, that is: We want to prove that ATM is undecidable. 3. We assume that B is decidable and use this assumption to prove that A is decidable. 18 17 Discussion Discussion If however R rejects on input M , w , a decider for HALTTM cannot safely reject: M may be halting on w to reject it. So if M rejects w, a decider for HALTTM must accept M , w . Let R be a decider for ATM . Given an input for M , w , R can be run with this input : If R accepts, it means that M , w ATM . This means that M accepts on input w. In particular, M stops on input w. Therefore, a decider for HALTTM must accept M , w too. 19 20 ...
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This note was uploaded on 04/30/2009 for the course CSE 105 taught by Professor Paturi during the Winter '99 term at UCSD.

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