Unformatted text preview: Proof
Given an instance for ATM , M , w , we may try to run R on this instance. If R accepts, we know that L(M ) = . In particular, M does not accept w so a decider for ATM must reject M,w . Proof Assume by way of contradiction that ETM is decidable and let R be a TM deciding it. In the next slides we devise TM S that uses R as a subroutine and decides ATM . 29 30 Proof
M1 M1 Description of___ We start with a TM satisfying L(M 1 ) = L(M ) .
nqaccept What happens if R rejects? The only conclusion we can draw is that L(M ) . What we need to know though is whether w L(M ) . M
qstart qaccept qreject nqreject In order to use our decider R for ETM , we once again modify the input machine M to obtain TM M 1 : 31 32 ...
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- Winter '99
- Subroutine, TM, Proof by contradiction, ETM