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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: Defining a correct simulating halt decider
Date: Thu, 5 Sep 2024 22:35:10 -0400
Organization: i2pn2 (i2pn.org)
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On 9/5/24 3:48 PM, olcott wrote:
> On 9/5/2024 2:41 PM, joes wrote:
>> Am Thu, 05 Sep 2024 13:10:13 -0500 schrieb olcott:
>>> On 9/5/2024 12:22 PM, joes wrote:
>>>> Am Thu, 05 Sep 2024 12:17:01 -0500 schrieb olcott:
>>>>> On 9/5/2024 11:56 AM, joes wrote:
>>>>>> Am Thu, 05 Sep 2024 11:52:04 -0500 schrieb olcott:
>>>>>>> On 9/5/2024 11:34 AM, joes wrote:
>>>>>>>> Am Thu, 05 Sep 2024 11:10:40 -0500 schrieb olcott:
>>>>>>>>> On 9/5/2024 10:57 AM, joes wrote:
>>
>>>>>>> The directly executed HHH correctly determines that its emulated DDD
>>>>>>> must be aborted because DDD keeps *THE EMULATED HHH* stuck in
>>>>>>> recursive emulation.
>>>>>> Why doesn’t the simulated HHH abort?
>>>>> The first HHH cannot wait for its HHH to abort which is waiting for
>>>>> its HHH to abort on and on with no HHH ever aborting.
>>>> But why does HHH halt and return that itself doesn’t halt?
>>> First agree that you understand the first part so that we don't
>>> endlessly digress away from the point.
> 
>> I smell evasion but fine, I understand that HHH cannot wait.
> 
> Do you really understand this?
> 
> It took far too long to get to this point we cannot simply
> drop it without complete closure before moving on.
> 
> <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
> 
> Thus this criteria has been met.

No it hasn't, because your decider doesn't do a "correct" (which means 
complete) simulation, and doesn't correct determine the results of a 
correct (which means complete) simulation of *THIS* input (which calls 
the version of the decider that does what this one finally decides to do).

Since if you do choose to abort, you will return to your caller and thus 
this DDD will halt, you can't have correctly proven that it doesn't

Sorry, you are just proving that you are just a liar.

> 
>> Now that I think about it, HHH could recognise itself (wouldn’t it
>> need to be a quine?)… no, the copy would do the same.
>>
> 
>