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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: Tue, 13 May 2025 13:01:15 +0200
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Op 12.mei.2025 om 18:32 schreef olcott:
> On 5/12/2025 6:46 AM, dbush wrote:
>> On 5/12/2025 12:11 AM, olcott wrote:
>>> On 5/11/2025 10:54 PM, dbush wrote:
>>>> On 5/11/2025 11:50 PM, olcott wrote:
>>>>> On 5/11/2025 10:39 PM, dbush wrote:
>>>>>> On 5/11/2025 11:36 PM, olcott wrote:
>>>>>>>
>>>>>>> You try to get away with changing the subject
>>>>>>> because you know that you are lying about one
>>>>>>> or more steps of DDD emulated by HHH according
>>>>>>> to the rules of the x86 language 
>>>>>>
>>>>>> It is you who are changing the subject away from DDD emulated by 
>>>>>> HHH (not one or more steps of DDD emulated by HHH) which you have 
>>>>>> admitted for the record is not correct:
>>>>>>
>>>>>
>>>>> I am only referring to the hypothetical infinite
>>>>> set of pure x86 emulators specified below:
>>>>>
>>>>
>>>> Each of which is processing a distinct, different input.
>>>>
>>>> Changing the input is not allowed.
>>>
>>> I am examining all the elements of a infinite set
>>> such that each HHH/DDD pair has a specific fixed form.
>>
>> And each one has a distinctly different input, and as such the halt 
>> status of one is unrelated to the halt status of another.
>>
>> Changing the input is not allowed.
> 
> In other words you are clueless about mathematical induction.
> 
> It is true that when one step of DDD is emulated by HHH that
> DDD does not halt.
> 
> It is true that when N steps of DDD are emulated by HHH that
> DDD does not halt.
> 

But when enough steps of the *unchanged* DDD are emulated, then DDD does 
halt, as proven by HHH1. Only when you also change the input, then more 
steps are needed, but changing the input is not allowed.