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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: DDD specifies recursive emulation to HHH and halting to HHH1
Date: Mon, 31 Mar 2025 20:30:46 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Mon, 31 Mar 2025 13:59:52 -0500 schrieb olcott:
> 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:


>>>>>>>>>>>>> It is dishonest to expect non-terminating inputs to
>>>>>>>>>>>>> complete.
>>>>>>>>>>>> A complete simulation of a nonterminating input doesn't halt.
You missed something here.

>>>>>>>>>>>>> When UTM1 is a UTM that has been adapted to only simulate a
>>>>>>>>>>>>> finite number of steps
>>>>>>>>>>>> So not an UTM.
And here.

>>>>>>>>>>>>> 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.

>>>>>>>>>>>>> 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 EMULATED BY HHH DOES SPECIFY THAT IT CANNOT POSSIBLY REACH
>>>>>>>>> ITS OWN FINAL HALT STATE.
Only if HHH is not a decider.

>>>>>>>> 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.
You have, only you. We asked your for other steps.

>>>>> 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.
Not correct is returning the wrong value. It should just say "I can't
simulate it, but it halts". I don't care about a supposed simulator
that does not say anything about the direct execution. 

-- 
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.