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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: A state transition diagram proves ... GOOD PROGRESS
Date: Fri, 18 Oct 2024 19:19:09 -0400
Organization: i2pn2 (i2pn.org)
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On 10/18/24 12:39 PM, olcott wrote:
> On 10/18/2024 9:41 AM, joes wrote:
>> Am Fri, 18 Oct 2024 09:10:04 -0500 schrieb olcott:
>>> On 10/18/2024 6:17 AM, Richard Damon wrote:
>>>> On 10/17/24 11:47 PM, olcott wrote:
>>>>> On 10/17/2024 10:27 PM, Richard Damon wrote:
>>>>>> On 10/17/24 9:47 PM, olcott wrote:
>>>>>>> On 10/17/2024 8:13 PM, Richard Damon wrote:
>>>>>>>> On 10/17/24 7:31 PM, olcott wrote:
>>
>>>>>>>>> When DDD is correctly emulated by HHH according to the semantics
>>>>>>>>> of the x86 language DDD cannot possibly reach its own machine
>>>>>>>>> address [00002183] no matter what HHH does.
>>>>>>>>> +-->[00002172]-->[00002173]-->[00002175]-->[0000217a]--+
>>
>>>>>>>> Except that 0000217a doesn't go to 00002172, but to 000015d2
>>
>>>> The Emulating HHH sees those addresses at its begining and then never
>>>> again.
>>>> Then the HHH that it is emulating will see those addresses, but not the
>>>> outer one that is doing that emulation of HHH.
>>>> And so on.
>>>> Which HHH do you think EVER gets back to 00002172?
>>>> What instruction do you think that it emulates that would tell it to do
>>>> so?
>>
>>>> At best the trace is:
>>>> 00002172 00002173 00002175 0000217a conditional emulation of 00002172
>>>> conditional emulation of 00002173 conditional emulation of 00002175
>>>> conditional emulation of 0000217a CE of CE of 00002172 ...
>>> OK great this is finally good progress.
>> The more interesting part is HHH simulating itself, specifically the
>> if(Root) check on line 502.
>>
> 
> That has nothing to do with any aspect of the emulation
> until HHH has correctly emulated itself emulating DDD.
> 
>>>> and if HHH decides to abort its emulation, it also should know that
>>>> every level of condition emulation it say will also do the same thing,
>>> If I understand his words correctly Mike has already disagreed with
>>> this.
>> He hasn't.
>>
>>> Message-ID: <rLmcnQQ3-N_tvH_4nZ2dnZfqnPGdnZ2d@brightview.co.uk>
>>> On 3/1/2024 12:41 PM, Mike Terry wrote:
>>>   > Obviously a simulator has access to the internal state (tape 
>>> contents
>>>   > etc.) of the simulated machine. No problem there.
>>> This seems to indicate that the Turing machine UTM version of HHH can
>>> somehow see each of the state transitions of the DDD resulting from
>>> emulating its own Turing machine description emulating DDD.
> 
>> Of course. It needs to, in order to simulate it. Strictly speaking
>> it has no idea of its simulation of a simulation two levels down,
>> only of the immediate simulation; the rest is just part of whatever
>> program the simulated simulator is simulating, which happens to be
>> itself.
>>
> 
>  From the concrete execution trace of DDD emulated by HHH
> according to the semantics of the x86 language people with
> sufficient technical competence can see that the halt status
> criteria that professor Sipser agreed to has been met.

Nope.

Proven previously and you accepted by default by not pointing out an error.

Your HHH neither "correctly simulated" per his definitions or correctly 
predicts the behavior of such a simulation, and thus never acheived the 
required criteria.

All you have done is proved you lie.

> 
> <MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
>      If simulating halt decider H correctly simulates its input D
>      until H correctly determines that its simulated D would never
>      stop running unless aborted then
> 
>      H can abort its simulation of D and correctly report that D
>      specifies a non-halting sequence of configurations.
> </MIT Professor Sipser agreed to ONLY these verbatim words 10/13/2022>
> 
> I will paraphrase this to use clearer language that directly applies
> to HHH and DDD.
> 
>      If emulating termination analyzer HHH emulates its input DDD
>      according to the semantics of the x86 language (including HHH
>      emulating itself emulating DDD) until HHH correctly determines
>      that its emulated DDD would never stop running unless aborted
>      then ...
> 
>      HHH can abort its emulation of DDD and correctly report that DDD
>      specifies a non-terminating sequence of x86 instructions.
> 
>>> *Joes can't seem to understand this*
>>> Only the outer-most HHH meets its abort criteria first, thus unless it
>>> aborts as soon as it meets this criteria none of them will ever abort.
> 
>> This is very simple to understand. Almost as simple as: even if only
>> the outermost HHH didn't abort, it would still halt, 
> 
> Yet that is based on the factually incorrect assumption
> that every instance of HHH does not use the exact same
> machine code.
> 
> Since you should know that this assumption is factually
> incorrect I could it as flat out dishonestly on your part.
> 
>> since it is
>> simulating a halting program: the nested version will abort.
>>
>>>> and thus the call HHH at 0000217a will be returned from, > and HHH has
>>>> no idea what will happen after that, so it KNOWS it is ignorant of the
>>>> answer.
> 
>