# prove every language got a language that is harder

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.

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 $L$.
• I cant understand, if the language is not in $RE$ than how can I use the halting problem? – misha312 Feb 17 '18 at 13:24
• you mean some-thing like $Ld$ ? – misha312 Feb 17 '18 at 13:29