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I can't find how to prove the decibility with a reduction.

EDIT: I've tried the reduction from the halting problem and the aceptance problem. Stopping for at least one entry has infinite inputs (you have to check all possible inputs) but the halting problem only has one input for the TM.

I don't understand how can i formally define a machine that using a machine that checks all inputs solves all cases of the halting problem.

dipzza
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2 Answers2

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It is not decidable. You can prove that by reducing the Halting problem to it as follows: Given a tuple $(M,x)$ and the question whether $M$ halts on $x$, we can construct a Turing machine $M_x$ which simply simulates $M$ on $x$ irrespective of the actual input $y$ of $M_x$. It is clear that $M_x$ will halt on any input iff $M$ halts on $x$.

prime_hit
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I found the answer to my question in this similar problem Halting problem reducing to the blank tape halting problem You can just erase the 'for any input' problem using a Turing Machine that always erase the input and write w. This way you can make an equivalence between an instance (M, w) of the halting problem and an instance (M) of this problem.

dipzza
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