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Let A be a Turing-recognizable language consisting of descriptions of Turing machines, {M1,M2,...}, where everyMiis a decider. Prove that some decidable languageDis not decided by any deciderMiwhose description appears in A. (Hint: You may find it helpful to consider an enumerator for A.)

Short Answer

Expert verified

The decidable language D is given that is not decided by any decider Mi

Step by step solution

01

Explain Turing machine

A Turing machine uses mathematics to simulate a machine that runs on tape mechanically. Such a machine is able to read and write symbols on a tape one at a time using a tape head. A limited number of simple instructions decide the operation completely.

02

Prove that D is not decidable

There is an enumerator E that counts A since it is Turing-recognizable. Allow Mi to be the ith output of E. Let all the possible strings be s1,s2,s3,…and so on. Construct a Turing Machine D as described below.

Reject if w does not belong to 0,1*.Otherwise, for some particular i, w is equal to si. Then, use E to numerate till Mi. After that, run Mi on w. Reject if Mi accepts. Otherwise, accept.

Thus, D is different fromany deciderMi and not in A.

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