To prove that the set $\mathrm{DIAG} = \{\langle M\rangle \mid \langle M\rangle \notin L(M)\}$ is not decidable, we can assume, to the contrary, that there is a Turing Machine $M$ such that $L(M) = \mathrm{DIAG}$. Consider the string $\langle M\rangle$. By definition of $\mathrm{DIAG}$, $\langle M\rangle \in \mathrm{DIAG} \Leftrightarrow \langle M\rangle \notin L(M)$. But since $L(M) = \mathrm{DIAG}, \langle M\rangle \notin L(M) \Leftrightarrow \langle M\rangle \notin \mathrm{DIAG}$. So $\langle M\rangle \in \mathrm{DIAG} \Leftrightarrow \langle M\rangle \notin \mathrm{DIAG}$, which is impossible.

I don't know how to extend this proof to prove that $DD = \{\langle M\rangle \mid \langle M\rangle\langle M\rangle \notin L(M)\}$ is not decidable. Can someone show how to do this?

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    $\begingroup$ Can you can apply rice's theorem? $\endgroup$ – Daniel Saad Jan 24 '18 at 12:06
  • $\begingroup$ @DanielSaad Why give a misleading hint? Rive's theorem doesn't apply here. $\endgroup$ – Raphael Jan 24 '18 at 18:29
  • $\begingroup$ @Raphael Why Rice's Theorem can't be applied here? The predicate isn't a non-trivial property? You should be more careful with your words, even if I'm wrong I was trying to help. It was neither an answer, nor a misleading tip, It was a question. Please be more polite. $\endgroup$ – Daniel Saad Jan 24 '18 at 18:50
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    $\begingroup$ @DanielSaad Neither DIAG nor DD are index sets. See e.g. here and here. (It's nice that you're trying to help, but you should make sure you know the material yourself beforehand. Even though your comment is grammatically a question, how could it be read but as a hint?) $\endgroup$ – Raphael Jan 24 '18 at 20:24
  • $\begingroup$ What I am thinking is construct some language which contains halt which can be reduced from Diag, and then try to reduce it to DD, but I can not reduce it to DD. $\endgroup$ – Zhou He Jan 24 '18 at 22:30

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