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L = { <M> | M is a Turing machine over {0, 1}, and <M>||<M> (not in) L(M)}

How do I prove that L is not recognizable? Any ideas?

I've proven the L compliment is recognizable:

Set Turing machine to J 
1. Run J on input <M>||<M>
2. TM J accepts then accept, it reject the reject.

<M>||<M> is the concatenation of the encoding of the Turing machine.
Hackaholic
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johny59
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1 Answers1

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You can reduce the (diagonal) acceptance problem to this problem. I try to use your own notation

D = { <M> | M is a Turing machine over {0, 1}, and <M> (not in) L(M)}

Suppose to fix an encoding of machine M, and consider a new program that takes in input a string w and accepts it just in case <M> in L(M) (so it has a constant behaviour, independent from the input string, and only dependent on <M>). The previous program can be build parametrically and effectively in <M>, that is we have a total computable function h such that the previous program has code h(<M>). Formally, I am using the smn theorem here, but since I am not sure you are confident with it, I prefer not mentioning it.

Now the question is if h(<M>) is in L.

If <M> in D, then by construction the machine h(<M>) does not accept any string, and in particular does not accept h(<M>)||h(<M>), so h(<M>) in L.

Conversely, if <M> not in D, then by construction the machine h(<M>) does accept any string, and in particular it accepts h(<M>)||h(<M>), so h(<M>) not in L.

If we had a way to decide L, we would have a way to decide D, and we know that D is not decidable (in fact, it is productive, similarly to L).

Andrea Asperti
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