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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: Who here understands that the last paragraph is Necessarily true?
Date: Fri, 19 Jul 2024 16:03:51 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Fri, 19 Jul 2024 09:54:07 -0500 schrieb olcott:
> On 7/19/2024 1:35 AM, Fred. Zwarts wrote:
>> Op 18.jul.2024 om 17:37 schreef olcott:
>>> On 7/18/2024 10:27 AM, joes wrote:
>>>> Am Thu, 18 Jul 2024 09:14:32 -0500 schrieb olcott:
>>>>> On 7/18/2024 3:25 AM, joes wrote:
>>>>>> Am Wed, 17 Jul 2024 15:36:24 -0500 schrieb olcott:
>>>>>>> On 7/17/2024 3:30 PM, joes wrote:
>>>>>>>> Am Wed, 17 Jul 2024 12:20:43 -0500 schrieb olcott:
>>>>>>>>> On 7/17/2024 12:16 PM, joes wrote:
>>>>>>>>>> Am Wed, 17 Jul 2024 08:27:08 -0500 schrieb olcott:
>>>>>>>>>>> On 7/17/2024 2:43 AM, Mikko wrote:
>>>>>>>>>>>> On 2024-07-16 18:24:49 +0000, olcott said:
>>>>>>>>>>>>> On 7/16/2024 3:12 AM, Mikko wrote:
>>>>>>>>>>>>>> On 2024-07-15 02:33:28 +0000, olcott said:
>>>>>>>>>>>>>>> On 7/14/2024 9:04 PM, Richard Damon wrote:
>>>>>>>>>>>>>>>> On 7/14/24 9:27 PM, olcott wrote:

>>>>>>>>>>>> The behaviour of HHH is specified outside of the input.
>>>>>>>>>>>> Therefore your "decider" decides about a non-input, which you
>>>>>>>>>>>> said is not allowed.
>>>>>>>>>>> HHH is not allowed to report on the behavior of it actual self
>>>>>>>>>>> in its own directly executed process. HHH is allowed to report
>>>>>>>>>>> on the effect of the behavior of the simulation of itself
>>>>>>>>>>> simulating DDD.
>>>>>>>>>> HHH must report on itself if its input calls it.
>>>>>>>>>> HHH does not directly simulate itself, it just executes.
>>>>>>>>>> It reports on DDD by simulating it.
>>>>>>>>> Its input cannot call its actual self that exists in an entirely
>>>>>>>>> different process.
>>>>>>>> Of course it doesn't make sense to return to a higher stack
>>>>>>>> frame. And of course a function can recursively call itself.
>>>>>>> A separate process is like a different program on a different
>>>>>>> computer.
>>>>>> It makes no sense to call a running program. DDD creates a new
>>>>>> instance of the same code with its own memory and code pointer.
>>>>> It is not that it makes no sense it is that it is impossible.
>>>> I mean, why are you talking about that?
>>> All of the halting problem proofs are incorrectly anchored in the
>>> behavior of the direct execution of the input thus not the behavior
>>> that this input specifies to a decider that this input invokes.
>> Exactly the same input is presented to the direct execution and the
>> simulation, namely the x86 code of the program.
>> The semantics of the x86 language does not change in these two cases,
>> so a correct simulator should interpret the x86 in the same way as the
>> direct execution.
>     Before HHH(DDD) aborts its emulation the directly executed DDD()
>     cannot possibly halt.
>     After HHH(DDD) aborts its emulation the directly executed DDD()
>     halts.
What do you mean "after"? The outer DDD called by main? It will halt
even before HHH has aborted, because it is deterministic and actually
does halt. It makes no sense to say that something that will, couldn't.

-- 
Am Fri, 28 Jun 2024 16:52:17 -0500 schrieb olcott:
Objectively I am a genius.