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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: Incorrect requirements --- Computing the mapping from the input
 to HHH(DD)
Date: Sat, 10 May 2025 10:30:38 +0200
Organization: A noiseless patient Spider
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Op 10.mei.2025 om 07:41 schreef olcott:
> On 5/10/2025 12:27 AM, wij wrote:
>> On Sat, 2025-05-10 at 00:19 -0500, olcott wrote:
>>> On 5/10/2025 12:13 AM, wij wrote:
>>>> On Sat, 2025-05-10 at 00:06 -0500, olcott wrote:>>
>>>>> When mathematical mapping is properly understood
>>>>> it will be known that functions computed by models
>>>>> of computation must transform their input into
>>>>> outputs according to the specific steps of an
>>>>> algorithm.
>>>>>
>>>>> _DDD()
>>>>> [00002172] 55         push ebp      ; housekeeping
>>>>> [00002173] 8bec       mov ebp,esp   ; housekeeping
>>>>> [00002175] 6872210000 push 00002172 ; push DDD
>>>>> [0000217a] e853f4ffff call 000015d2 ; call HHH(DDD)
>>>>> [0000217f] 83c404     add esp,+04
>>>>> [00002182] 5d         pop ebp
>>>>> [00002183] c3         ret
>>>>> Size in bytes:(0018) [00002183]
>>>>>
>>>>> For example HHH(DDD) only correctly map to the
>>>>> behavior that its input actually specifies by correctly
>>>>> emulating DDD according to the rules of the x86 language.
>>>>>
>>>>> This causes the first four instructions of DDD
>>>>> to be emulated followed by HHH emulating itself
>>>>> emulating the first three instructions of DDD.
>>>>>
>>>>> It is right at this recursive simulation just
>>>>> before HHH(DDD) is called again that HHH recognizes
>>>>> the repeating pattern and rejects DDD.
>>>>
>>>> Yes, but you still did not answer the question: Is POOH exactly 
>>>> about HP?
>>>>
>>>
>>>   >>>>> H(D)=1 if D() halt.
>>>   >>>>> H(D)=0 if D() not halt.
>>>
>>> Right now it is mostly about proving the
>>> above requirements are is mistaken.
>>>
>>
>> Why is the requirement invalid?
>>
>> H(D)=1 if D() halt.
>> H(D)=0 if D() not halt.
>>
> 
> The notion that the behavior specified by the finite
> string input to a simulating termination analyzer
> does sometimes differ from the behavior of its direct
> execution. It is a provably different sequence of steps.
> 


Only when the simulation ignores the most relevant part of the input and 
halts before it can see it. Even then the only difference is that all 
steps that are simulated are exactly the same as the direct execution 
and there is no reason to think that the behaviour specified in the 
input would be different in the following steps.
It seems difficult for you to accept verifiable facts when they disturb 
your dreams.