XPost: comp.theory
From: 643-408-1753@kylheku.com
On 2025-11-20, dart200 wrote:
> On 11/20/25 1:09 PM, Richard Heathfield wrote:
>> On 20/11/2025 20:53, Mike Terry wrote:
>>
>>
>>
>>> I think you're just saying that HP cannot be solved by having two
>>> deciders e.g. my H0/H1, such that given any input M, one of H0/H1
>>> decides M correctly. That's true, as the HP asks for /one/ TM
>>> decider, not two, and crucially we don't have a way to combine H0/H1
>>> into a single decider that satisfies HP.
>>
>>
>>
>> Let's say we did.
>>
>> It would necessarily have its own diagonal case, which it couldn't solve.
>>
>> The End.
>>
>
> yeah i think my point is trying to be that...
>
> you can build this "diagonal case"/semantic paradox for trying to solve
> the HP with any N deciders, because to solve HP you need to funnel the
> semantic meaning thru one single interface which is where the paradox
> will invariably target...
"N deciders" and "1 decider built out of combining the decision of N deciders"
are completely different beasts.
One produces a tuple of N bools, the other a bool.
There is a diagonal case for every one of the N deciders individually,
as well as for the combined decider built on N of them.
There isn't a diagonal case against N decisions combined into
a tuple for N > 1, except when all N produce the same value.
If H0(D) yields false, and so does H1(D) and so does H2(D) --- all
unanimously assert that D doesn't halt, then D can contradict
them all by halting. If any two disagree, then no.
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