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From: joes <noreply@example.org>
Newsgroups: comp.theory
Subject: Re: No decider is accountable for the computation that itself is
 contained within
Date: Tue, 30 Jul 2024 19:52:34 -0000 (UTC)
Organization: i2pn2 (i2pn.org)
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Am Tue, 30 Jul 2024 11:24:35 -0500 schrieb olcott:
> On 7/30/2024 2:24 AM, joes wrote:
>> Am Mon, 29 Jul 2024 15:32:44 -0500 schrieb olcott:
>>> On 7/29/2024 3:17 PM, joes wrote:
>>>> Am Mon, 29 Jul 2024 11:32:00 -0500 schrieb olcott:
>>>>> On 7/28/2024 3:40 AM, Mikko wrote:
>>>>>> On 2024-07-27 14:21:50 +0000, olcott said:
>>>>>>> On 7/27/2024 2:46 AM, Mikko wrote:
>>>>>>>> On 2024-07-26 16:28:43 +0000, olcott said:
>> 
>>>>> Halt deciders are not allowed to report on the behavior of the
>>>>> actual computation that they themselves are contained within. They
>>>>> are only allowed to compute the mapping from input finite strings.
>>>> What if the input is the same as the containing computation?
>>> It always is except in the case where the decider is reporting on the
>>> TM description that itself is contained within.
> 
>> I don't understand. "The input is not the same as the containing
>> computation when deciding on the description of the containing
>> computation"?
I mean: is that an accurate paraphrase?

> An executing Turing machine is not allowed to report on its own
> behavior. Every decider is only allowed to report on the behavior that
> its finite string input specifies.
And what happens when those are the same?

-- 
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.