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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: Sat, 29 Mar 2025 09:31:33 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
Message-ID: <45b3405a167984b8649777fdc0804b124b21e19b@i2pn2.org>
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Am Fri, 28 Mar 2025 14:27:36 -0500 schrieb olcott:
> On 3/28/2025 2:17 PM, dbush wrote:
>> On 3/28/2025 3:02 PM, olcott wrote:
>>> On 3/28/2025 1:12 PM, dbush wrote:
>>>> On 3/28/2025 1:57 PM, olcott wrote:
>>>>> On 3/27/2025 9:33 PM, dbush wrote:
>>>>>> On 3/27/2025 10:10 PM, olcott wrote:
>>>>>>> On 3/27/2025 8:24 PM, dbush wrote:
>>>>>>>> On 3/27/2025 9:21 PM, olcott wrote:
>>>>>>>>> On 3/27/2025 8:09 PM, dbush wrote:
>>>>>>>>>> On 3/27/2025 9:07 PM, olcott wrote:
>>>>>>>>>>> On 3/27/2025 7:38 PM, dbush wrote:

>>>>>>>> Good, because that's all that's required for a solution to the
>>>>>>>> halting problem:
>>>>>>>>
>>>>>>> There are sometimes when the behavior of TM Description D
>>>>>>> correctly simulated by UTM1 does not match the behavior correctly
>>>>>>> simulated by UTM2.
>>>>>>
>>>>>> Irrelevant, because to satisfy the requirements, the behavior of
>>>>>> the described machine when executed directly must be reported.
>>>>>
>>>>> I HAVE PROVED THAT THE REQUIREMENT IS WRONG NITWIT.
According to what? WE require it. YOU are answering a different question.

>>>> Category error.
>>>> I want to know if any arbitrary algorithm X with input Y will halt
>>>> when executed directly.
>>>
>>> It is 100% impossible for any TM to take another executing TM as its
>>> input.
Quit that.

>> But it can take a complete description of a TM that
> 
> Is not always a perfect proxy for the behavior of the direct execution
> of the underlying machine.
Uh yes it is.

> I have proven this hundreds and hundreds of times over several years.
> PATHOLOGICAL SELF-REFERENCE CANNOT SIMPLY BE IGNORED. IT IS EITHER
> MORONIC OR DISHONEST TO DO SO.
Simulation by the called simulator is not 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.