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From: joes <noreply@example.org>
Newsgroups: comp.theory,sci.logic
Subject: Re: A state transition diagram proves ... GOOD PROGRESS
Date: Fri, 18 Oct 2024 17:00:38 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 18 Oct 2024 11:39:52 -0500 schrieb olcott:
> 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.
What? That is part of HHH, not 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.

>      If emulating termination analyzer HHH emulates its input DDD
>      until HHH determines that
>      its emulated DDD would never stop running unless aborted ...
But it would.

>>> *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.
Same as the outer HHH returning that the inner ones wouldn't.

>> 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.
-- 
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.