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From: olcott <polcott333@gmail.com>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sat, 29 Mar 2025 16:34:57 -0500
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On 3/29/2025 3:01 PM, joes wrote:
> Am Sat, 29 Mar 2025 10:28:15 -0500 schrieb olcott:
>> On 3/29/2025 4:56 AM, Mikko wrote:
>>> On 2025-03-28 22:41:12 +0000, olcott said:
>>>> On 3/28/2025 4:58 PM, Richard Damon wrote:
>>>>> On 3/28/25 2:13 PM, olcott wrote:
>>>>>> On 3/28/2025 8:50 AM, Richard Damon wrote:
>>>>>>> On 3/27/25 10:11 PM, olcott wrote:
>>>>>>>> On 3/27/2025 9:02 PM, Richard Damon wrote:
>>>>>>>>> On 3/27/25 9:10 PM, olcott wrote:
>>>>>>>>>> On 3/27/2025 7:47 PM, Richard Damon wrote:
>>>>>>>>>>> On 3/27/25 8:11 PM, olcott wrote:
>>>>>>>>>>>> On 3/27/2025 4:56 PM, joes wrote:
> 
>>>>>>>>>>>>> Yes, HHH is not a correct simulator.
>>>>>>>>>>>>>
>>>>>>>>>>>> You say that it is not a correct simulator on the basis of
>>>>>>>>>>>> your ignorance of the x86 language that conclusively proves
>>>>>>>>>>>> that HHH does correctly simulate the first four instructions
>>>>>>>>>>>> of DDD and correctly simulates itself simulating the first
>>>>>>>>>>>> four instructions of DDD.
> The x86 language or my supposed ignorance thereof doesn't prove shit.
> HHH does not simulate the infinite stack of recursive simulations,
> for obvious reasons.
> 
>>>>>>>>>>> It isn't a correct simulator,
>>>>>>>>>>
>>>>>>>>>> You know that you are lying about this or you would show how DDD
>>>>>>>>>> emulated by HHH would reach its final state ACCORDING TO THE
>>>>>>>>>> SEMANTICS OF THE X86 LANGUAGE.
>>>>>>>>>>
>>>>>>>>> It can't be, because your HHH doesn't meet your requirement.
>>>>>>>>>
>>>>>>>> You cannot show that because you know you are lying about that.
> One cannot show something impossible.
> 
>>>>>>> Sure we can, make a main that directly calls HHH and then DDD, then
>>>>>>> call HHH1(DDD)
>>>>>>> That HHH will return 0, saying that DDD is non-halting, but the DDD
>>>>>>> wll return, showing that DDD is halting.
>>>>>>> Look at the trace that HHH generates, and that HHH1 generates,
>>>>>>> HHH's will be a subset of the trace that HHH1 generates, showing
>>>>>>> that it is NOT proof that this program is non-halting as that exact
>>>>>>> same initial segment halts.
>>>>>>> Your argument about changing HHH shows that it doesn't halt is just
>>>>>>> invalid, as then you either changed the input, or demonstrated that
>>>>>>> you input was a class error as it didn't contain the COMPLETE
>>>>>>> representation of the code of DDD.
>>>>>>
>>>>>> I can't understand how that confused mess addresses the point of
>>>>>> this thread:
>>>>>> It is a verified fact that the finite string of machine code of DDD
>>>>>> emulated by HHH according to the semantics of the x86 language has
>>>>>> different behavior than DDD emulated by HHH1 according to the
>>>>>> semantics of the x86 language.
> Non sequitur.
> 
>>>>> Where did you "verify" that LIE. You claim fails the simple test:
>>>>> What is the first instruction actually correctly emulated by the
>>>>> rules of the x86 language by HHH and HHH1 that had a different
>>>>> result.
>>>>>
>>>> When DDD emulated by HHH calls HHH(DDD) this call NEVER returns.
>>>> When DDD emulated by HHH1 calls HHH(DDD) this call returns.
>>>
>>> When DDD is correctly emulated the call HHH(DDD) returns.
>>>
>> When are you going to understand that disagreeing with the semantics of
>> the x86 language IS NOT ALLOWED?
> Disagree with what semantics exactly? The call to HHH *must* return.
> 

Sure and any code placed inside of an infinite loop
must magically break out of this loop.

-- 
Copyright 2025 Olcott "Talent hits a target no one else can hit; Genius
hits a target no one else can see." Arthur Schopenhauer