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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH is Correctly rejected as
 non-halting V2
Date: Wed, 24 Jul 2024 10:52:47 +0200
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Op 23.jul.2024 om 15:31 schreef olcott:
> On 7/23/2024 1:32 AM, 0 wrote:
>> On 2024-07-22 13:46:21 +0000, olcott said:
>>
>>> On 7/22/2024 2:57 AM, Mikko wrote:
>>>> On 2024-07-21 13:34:40 +0000, olcott said:
>>>>
>>>>> On 7/21/2024 4:34 AM, Mikko wrote:
>>>>>> On 2024-07-20 13:11:03 +0000, olcott said:
>>>>>>
>>>>>>> On 7/20/2024 3:21 AM, Mikko wrote:
>>>>>>>> On 2024-07-19 14:08:24 +0000, olcott said:
>>>>>>>>
>>>>>>>>> When we use your incorrect reasoning we would conclude
>>>>>>>>> that Infinite_Loop() is not an infinite loop because it
>>>>>>>>> only repeats until aborted and is aborted.
>>>>>>>>
>>>>>>>> You and your HHH can reason or at least conclude correctly about
>>>>>>>> Infinite_Loop but not about DDD. Possibly because it prefers to
>>>>>>>> say "no", which is correct about Infinte_loop but not about DDD.
>>>>>>>>
>>>>>>>
>>>>>>> *Because this is true I don't understand how you are not simply 
>>>>>>> lying*
>>>>>>> int main
>>>>>>> {
>>>>>>>    DDD();
>>>>>>> }
>>>>>>>
>>>>>>> Calls HHH(DDD) that must abort the emulation of its input
>>>>>>> or {HHH, emulated DDD and executed DDD} never stop running.
>>>>>>
>>>>>> You are the lying one.
>>>>>>
>>>>>> If HHH(DDD) abrots its simulation and returns true it is correct as a
>>>>>> halt decider for DDD really halts.
>>>>>>
>>>>>
>>>>> (b) We know that a decider is not allowed to report on the behavior
>>>>> computation that itself is contained within.
>>>>
>>>> No, we don't. There is no such prohibition.
>>>>
>>>
>>> Turing machines never take actual Turing machines as inputs.
>>> They only take finite strings as inputs and an actual executing
>>> Turing machine is not itself a finite string.
>>
>> The definition of a Turing machine does not say that a Turing machine
>> is not a finite string. It is an abstract mathematical object without
>> a specification of its exact nature. It could be a set or a finite
>> string. Its exact nature is not relevant to the theory of computation,
>> which only cares about certain properties of Turing machines.
>>
>>> Therefore It is not allowed to report on its own behavior.
>>
>> Anyway, that does not follow. The theory of Turing machines does not
>> prohibit anything.
>>
>>> Another different TM can take the TM description of this
>>> machine and thus accurately report on its actual behavior.
>>
>> If a Turing machine can take a description of a TM as its input
>> or as a part of its input it can also take its own description.
>> Every Turing machine can be given its own description as input
>> but a Turing machine may interprete it as something else.
>>
> In this case we have two x86utm machines that are identical
> except that DDD calls HHH and DDD does not call HHH1.
> 
> It is empirically proven that this changes their behavior
> and the behavior of DDD.
> 

The x86 code is exactly the same, therefore, the semantics of the x86 
does not change, which proves that your claim that HHH works according 
to the semantics of the x86 language is not true.
The semantics of the x86 language specify exactly the same behaviour.
That does not change when it is given to another simulator.
Your empirical proof is invalid, because it uses the wrong method, when 
it ignores the semantics of the x86 language.