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From: Mikko <mikko.levanto@iki.fi>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Sun, 30 Mar 2025 12:16:12 +0300
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On 2025-03-29 15:28:15 +0000, olcott said:

> 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:
>>>>>>>>>>>> Am Thu, 27 Mar 2025 13:10:46 -0500 schrieb olcott:
>>>>>>>>>>>>> On 3/27/2025 6:02 AM, Richard Damon wrote:
>>>>>>>>>>>>>> On 3/26/25 11:47 PM, olcott wrote:
>>>>>>>>>>>>>>> On 3/26/2025 10:28 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 3/26/25 11:09 PM, olcott wrote:
>>>>>>>>>>>>>>>>> On 3/26/2025 8:22 PM, Richard Damon wrote:
>>>>>>>>>>>> 
>>>>>>>>>>>>>>>>>> Non-Halting is that the machine won't reach its final staste even
>>>>>>>>>>>>>>>>>> if an unbounded number of steps are emulated. Since HHH doesn't do
>>>>>>>>>>>>>>>>>> that, it isn't showing non-halting.
>>>>>>>>>>>>>>>>> DDD emulated by any HHH will never reach its final state in an
>>>>>>>>>>>>>>>>> unbounded number of steps.
>>>>>>>>>>>>>>>> But DDD emulated by an actually correct emulator will,
>>>>>>>>>>>>>>> If you were not intentionally persisting in a lie you would
>>>>>>>>>>>>>>> acknowledge the dead obvious that DDD emulated by HHH according to the
>>>>>>>>>>>>>>> semantics of the x86 language cannot possibly correctly reach its
>>>>>>>>>>>>>>> final halt state.
>>>>>>>>>>> 
>>>>>>>>>>> _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]
>>>>>>>>>>> 
>>>>>>>>>>>> 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.
>>>>>>>>>>> 
>>>>>>>>>> 
>>>>>>>>>> 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.
>>>>>>> 
>>>>>> 
>>>>>> 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.
>>>>>> 
>>>>>> Sorry, This is what you have been told for years, but you refuse to 
>>>>>> look at the truth, because you have been brainwashed by your lies.
>>>>>> 
>>>>>> Look
>>>>> 
>>>>> 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.
>>>>> 
>>>> 
>>>> 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?

My opinion does not matter. It only matters that when an x86 processor
runs your DDD the last executed instruction of DDD will be ret.

-- 
Mikko