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From: "Fred. Zwarts" <F.Zwarts@HetNet.nl>
Newsgroups: comp.theory
Subject: Re: DDD correctly emulated by HHH cannot possibly halt
Date: Fri, 12 Jul 2024 15:02:35 +0200
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Op 12.jul.2024 om 14:08 schreef olcott:
> On 7/12/2024 3:05 AM, joes wrote:
>> Am Thu, 11 Jul 2024 22:30:00 -0500 schrieb olcott:
>>> On 7/11/2024 10:18 PM, Richard Damon wrote:
>>>> On 7/11/24 10:28 PM, olcott wrote:
>>>>> On 7/11/2024 9:08 PM, Richard Damon wrote:
>>>>>> On 7/11/24 10:51 AM, olcott wrote:
>>>>>>> On 7/10/2024 8:50 PM, Richard Damon wrote:
>>>>>>>> On 7/10/24 9:21 PM, olcott wrote:
>>>>>>>>> On 7/10/2024 8:11 PM, Richard Damon wrote:
>>>>>>>>>> On 7/10/24 9:01 PM, olcott wrote:
>>>>>>>>>>> On 7/10/2024 7:37 PM, Richard Damon wrote:
>>>>>>>>>>>> On 7/10/24 8:24 PM, olcott wrote:
>>>>>>>>>>>>> On 7/10/2024 7:01 PM, Richard Damon wrote:
>>>>>>>>>>>>>> On 7/10/24 9:41 AM, olcott wrote:
>>>>>>>>>>>>>>> On 7/10/2024 8:27 AM, joes wrote:
>>>>>>>>>>>>>>>> Am Tue, 09 Jul 2024 23:19:25 -0500 schrieb olcott:
>>>>>>>>>>>>>>>>> On 7/9/2024 11:01 PM, joes wrote:
>>
>>>>>>>>>>>>>>>>> *DDD NEVER HALTS*
>>>>>>>>>>>>>>>> DDD ONLY calls HHH...
>>>>>>>>>>>>>>> DDD correctly emulated by any pure function HHH that
>>>>>>>>>>>>>>> correctly emulates 1 to ∞ lines of DDD can't make it to the
>>>>>>>>>>>>>>> second line of DDD no matter what.
>>>>>>>>>>>>>> Nope, DDD does if HHH(DDD) returns.
>>>>>>>>>>>>> DDD correctly emulated by any pure function HHH that correctly
>>>>>>>>>>>>> emulates 1 to ∞ lines of DDD can't make it to the second line
>>>>>>>>>>>>> of DDD no matter what.
>>>>>>>>>>>> WRONG, you don't seem to understand the difference between DDD
>>>>>>>>>>>> and HHH's emualtion of it.
>>>>>>>>> We stipulate that the only measure of a correct emulation is the
>>>>>>>>> semantics of the x86 programming language. By this measure when 1
>>>>>>>>> to ∞ steps of DDD are correctly emulated by each pure function x86
>>>>>>>>> emulator HHH (of the infinite set of every HHH that can possibly
>>>>>>>>> exist) then DDD cannot possibly reach its own machine address of
>>>>>>>>> 00002174 and halt.
>>>>>>>> By the semantic of the x86 programming language, the only correct
>>>>>>>> simulation is a FULL simulation
>>>>>>> In other words you are trying to get away with the lie that when 1
>>>>>>> step of DDD is correctly emulated that 0 steps of DDD are correctly
>>>>>>> emulated.
>>>>> When 1,2,3... ∞ steps of DDD are correctly emulated by HHH it is a lie
>>>>> to say that this many instructions were not correctly emulated and you
>>>>> know it.
>>>> But only N instructions "correctly emulated" is NOT a CORRECT
>>>> emulaition of the instructions of DDD/HHH
>>> I didn't limit it to N. Is this your ADD? I say 1 to infinity steps !!!
>> Please don't insult ADD people.
> 
> This does not say from 1 to 8 steps
> when 1 to ∞ steps of DDD are correctly emulated
> 
> When I say the same words 150 and Richard does not
> see these words I have to know why this is.
> My aim is effective communication. I can't fix
> the issue unless I know what the issue it.
> 
> The two possibilities Richard's ADD, and Richard
> is a Liar. If is is Richards's ADD then repeating
> the same sentence a dozen times seems to help.
> 
> If Richard is being a liar then calling him a Liar
> and telling him where this leads seems to help.
> 
>> You did talk of an HHH that only simulated a fixed number of steps.
>> They do not provide a correct (full) simulation.
> 
> when 1 to ∞ steps of DDD are correctly emulated in
> the infinite set of every HHH/DDD pair and no DDD
> halts then we can say that DDD DOES NOT HALT.

If 1 to ∞ steps are simulated by HHH and none of these simulations is 
correct, then none of the simulations is correct.
If N cycles are simulated by HHH when N+1 cycles must be simulated to 
make the simulation correct, then the simulation is incorrect for N = 1 
to ∞ (meaning in the limit of arbitrarily large N).
So, you definitely proved that none of the HHH can possible simulate 
itself correctly.
Every DDD in this set halts, but none of the simulations of HHH by 
itself in this set is able to show the halting, because the simulation 
is always aborted one cycle too soon and therefore, the simulation 
misses the last cycle, the abort and the return. So, the simulation does 
not show the full behaviour of the input.