Decidability

Question

L(M)={0}

We can have Tyes for {0} and Tno for Σ∗ ({0}⊂Σ∗{0}⊂Σ∗). Hence, L={M  ∣  L(M)={0}} is not Turing recognizable (not recursively enumerable)

I don’t understand why this is not decidable. We can easily create a turing that accepts this language

Comments

yes, it is like writing a program that will take another program (say Y) and print “yes” if “Y” prints “yes” for its input “0” and does not print “yes” for any other input.

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You are right @parth023 I completely misread the problem. If you find a solution then do post it here👍

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@rexritz you can watch this playlist, deepak sir has explained all the concepts beautifully. if i directly give you an answer you won’t learn a thing : )