Re: DDD INcorrectly emulated by HHH is INCorrectly rejected as non-halting V2

On 7/16/24 2:10 PM, olcott wrote:

On 7/16/2024 2:47 AM, Mikko wrote:

On 2024-07-15 13:39:07 +0000, olcott said:
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On 7/15/2024 3:09 AM, Mikko wrote:

On 2024-07-14 14:00:55 +0000, olcott said:
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According to the theory of computation the DDD that calls
HHH(DDD) is not in the domain of HHH.

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The theory of computation does not say what the domain of HHH is.

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Sure it does. Where the Hell have you been?
It says that the halting problem is defined in terms
of finite strings that encode Turing machines.

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No, it does not. The halting problem is not a part of any theory of
computation. It is a question that one maight expect the theory of
computation to answer.
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Note that the halting problem does not specify how Turing machines
should be encoded to finite strings. It meresly requires that the
solution includes encoding rules so that every Turing machine can be
encoded.
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 The theory of computation only allows finite string inputs.
It does not allow the currently executing Turing Machine
to be its own input.
 

But it does allow the finite string represention of that machine to be given as the input, and that string represent that machine, and the behavior that machine does when it is run,
You are just stuck in your own world of lies and you words just show how stupid and ignorant you have made yourself.
A decider CAN be asked about itself (by giving the description of itself to it) or about a program that uses a copy of that decider (again, by given the descirption of that machine). You just need to try to pretend that it isn't allowed because that it what the proof uses.
Of cource. just like the Diagonalization proof you claimed to have and then called nonsense, you will not be able to show any actual rule like what you quote, mostly because you are just totally ignorant about the rules of computation, and all you think you know are made up "facts" based on your casual reading about it.

Unless the specificaiton of HHH says otherwise HHH should be able
to handle every input that can be given to it,

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No halt decider is allowed to report on the computation
that it is contained within for several different reasons
one of them is that computations are not finite strings.

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The halting problem requires that every Turing machine computation
can be given as input.
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A partial halt decider may fail to answer for some computations.
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