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I am prety stuck over here:

prove or disprove that every $L$ got $L'$ s.t $L'\geq L$ and for every $L''\geq L$

$L''\ngeq L'$

basically it means L' is the hardest...

my intuition tells me that this is correct, but I cant prove why. I tried to prove using counting but got stuck:

  1. we got $\aleph_0$ pairs of functions that we can use in the reduction for both ways

  2. we got $\aleph$ languages so there must be a language that got a function to compute the redaction in one way and doesnt got a function for the reduction to the other way

Its feels terribly incorrect any help will be appreciated

EDIT: I just realized that I didn't write the question correctly please read it again.

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The halting problem for a class of machines that can decide $L$ is always harder than $L$. You already know this when $L$ is computable and the proof of the more general result is essentially the same. For every $L$, you need to think of a class of machines that can decide $L$.

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  • $\begingroup$ I cant understand, if the language is not in $RE$ than how can I use the halting problem? $\endgroup$ – misha312 Feb 17 '18 at 13:24
  • $\begingroup$ Don't use the halting problem for ordinary Turing machines. You need to think of a class of machines that can decide the langauge you're working with. $\endgroup$ – David Richerby Feb 17 '18 at 13:26
  • $\begingroup$ you mean some-thing like $Ld$ ? $\endgroup$ – misha312 Feb 17 '18 at 13:29
  • $\begingroup$ I don't know what you mean by that. $\endgroup$ – David Richerby Feb 17 '18 at 13:44
  • $\begingroup$ all the TM machines that doesn't accept their encoding $\endgroup$ – misha312 Feb 17 '18 at 13:48

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