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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: Mon, 31 Mar 2025 13:59:52 -0500
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On 3/31/2025 6:19 AM, Richard Damon wrote:
> On 3/30/25 11:16 PM, olcott wrote:
>> On 3/30/2025 9:40 PM, Richard Damon wrote:
>>> On 3/30/25 10:17 PM, olcott wrote:
>>>> On 3/30/2025 7:35 PM, Richard Damon wrote:
>>>>> On 3/30/25 5:56 PM, olcott wrote:
>>>>>> On 3/30/2025 4:05 PM, Richard Damon wrote:
>>>>>>> On 3/30/25 4:32 PM, olcott wrote:
>>>>>>>> On 3/30/2025 1:52 PM, Richard Damon wrote:
>>>>>>>>> On 3/30/25 2:27 PM, olcott wrote:
>>>>>>>>>> On 3/30/2025 3:12 AM, joes wrote:
>>>>>>>>>>> Am Sat, 29 Mar 2025 16:46:26 -0500 schrieb olcott:
>>>>>>>>>>>> On 3/29/2025 3:14 PM, dbush wrote:
>>>>>>>>>>>>> On 3/29/2025 4:01 PM, olcott wrote:
>>>>>>>>>>>
>>>>>>>>>>>>>> We can know that when this adapted UTM simulates a finite 
>>>>>>>>>>>>>> number of
>>>>>>>>>>>>>> steps of its input that this finite number of steps were 
>>>>>>>>>>>>>> simulated
>>>>>>>>>>>>>> correctly.
>>>>>>>>>>>>> And therefore does not do a correct UTM simulation that 
>>>>>>>>>>>>> matches the
>>>>>>>>>>>>> behavior of the direct execution as it is incomplete.
>>>>>>>>>>>> It is dishonest to expect non-terminating inputs to complete.
>>>>>>>>>>> A complete simulation of a nonterminating input doesn't halt.
>>>>>>>>>>>
>>>>>>>>>>>>>>> 2) changing the input is not allowed
>>>>>>>>>>>>>> The input is unchanged. There never was any indication 
>>>>>>>>>>>>>> that the input
>>>>>>>>>>>>>> was in any way changed.
>>>>>>>>>>>>> False, if the starting function calls UTM and UTM changes, 
>>>>>>>>>>>>> you're
>>>>>>>>>>>>> changing the input.
>>>>>>>>>>>> When UTM1 is a UTM that has been adapted to only simulate a 
>>>>>>>>>>>> finite
>>>>>>>>>>>> number of steps
>>>>>>>>>>> So not an UTM.
>>>>>>>>>>>
>>>>>>>>>>>> and input D calls UTM1 then the behavior of D simulated
>>>>>>>>>>>> by UTM1 never reaches its final halt state.
>>>>>>>>>>>> When D is simulated by ordinary UTM2 that D does not call 
>>>>>>>>>>>> Then D reaches
>>>>>>>>>>>> its final halt state.
>>>>>>>>>>> Doesn't matter if it calls it, but if the UTM halts.
>>>>>>>>>>>
>>>>>>>>>>>>> Changing the input is not allowed.
>>>>>>>>>>>> I never changed the input. D always calls UTM1.
>>>>>>>>>>>> thus is the same input to UTM1 as it is to UTM2.
>>>>>>>>>>> You changed UTM1, which is part of the input D.
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> UTM1 simulates D that calls UTM1
>>>>>>>>>> simulated D NEVER reaches final halt state
>>>>>>>>>>
>>>>>>>>>> UTM2 simulates D that calls UTM1
>>>>>>>>>> simulated D ALWAYS reaches final halt state
>>>>>>>>>>
>>>>>>>>>
>>>>>>>>> Only because UTM1 isn't actually a UTM, but a LIE since it only 
>>>>>>>>> does a partial simulation, not a complete as REQURIED by the 
>>>>>>>>> definition of a UTM.
>>>>>>>>>
>>>>>>>>
>>>>>>>> _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]
>>>>>>>>
>>>>>>>> DDD EMULATED BY HHH DOES SPECIFY THAT IT
>>>>>>>> CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
>>>>>>>>
>>>>>>>> THAT IS WHAT IT SAYS AND ANYONE THAT DISAGREES
>>>>>>>> IS A DAMNED LIAR OR STUPID.
>>>>>>>>
>>>>>>>
>>>>>>> How is that DDD correctly emulated beyond the call HHH 
>>>>>>> instruction by a program that is a pure function, and thus only 
>>>>>>> looks at its input?
>>>>>>>
>>>>>>
>>>>>> *THE ENTIRE SCOPE IS*
>>>>>> DDD EMULATED BY HHH DOES SPECIFY THAT IT
>>>>>> CANNOT POSSIBLY REACH ITS OWN FINAL HALT STATE.
>>>>>
>>>>>  From where? Remember, the Halting problem is SPECIFICALLY 
>>>>
>>>> OFF F-CKING TOPIC. WE ABOUT ONE F-CKING STEP OF MY PROOF.
>>>> WE HAVE BEEN TALKING ABOUT ONE F-CKING STEP OF MY PROOF
>>>> FOR THREE F-CKING YEARS.
>>>>
>>>> DDD correctly emulated by HHH DOES NOT F-CKING HALT !!!
>>>>
>>>>
>>>
>>> Your proof is just off topic ranting.
>>>
>>> The problem is that DDD is NOT correctly emulated by HHH, 
>>
>> You are a damned liar when you try to get away
>> with implying that HHH does not emulate itself
>> emulating DDD in recursive emulation according
>> to the semantics of the x86 language.
>>
>>
> 
> Of course it doesn't CORRECTLY emulate itself emulating DDD (and 
> omitting that CORRECTLY is a key point to your fraud), as it stops part 
> way, and CORRECT emulation that determines behavior doesn't stop until 
> the end is reached.

It is ALWAYS CORRECT for any simulating termination
analyzer to stop simulating and reject any input
that would otherwise prevent its own termination.

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