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From: Richard Damon <richard@damon-family.org>
Newsgroups: comp.theory
Subject: Re: HHH maps its input to the behavior specified by it --- key error
in all the proofs --- Mike --- basis
Date: Wed, 14 Aug 2024 19:40:52 -0400
Organization: i2pn2 (i2pn.org)
Message-ID: <2337974ddf2b276988bb81b0c30d8616a1d76079@i2pn2.org>
References: <v8jh7m$30k55$1@dont-email.me> <v99870$14mlk$1@dont-email.me>
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On 8/14/24 9:42 AM, olcott wrote:
> On 8/14/2024 2:30 AM, Mikko wrote:
>> On 2024-08-13 13:30:08 +0000, olcott said:
>>
>>> On 8/13/2024 6:23 AM, Richard Damon wrote:
>>>> On 8/12/24 11:45 PM, olcott wrote:
>>>>>
>>>>> void DDD()
>>>>> {
>>>>> HHH(DDD);
>>>>> return;
>>>>> }
>>>>>
>>>>> *DDD correctly emulated by HHH cannot possibly reach its*
>>>>> *own "return" instruction final halt state, thus never halts*
>>>>>
>>>>
>>>> Which is only correct if HHH actuallly does a complete and correct
>>>> emulation, or the behavior DDD (but not the emulation of DDD by HHH)
>>>> will reach that return.
>>>>
>>>
>>> A complete emulation of a non-terminating input has always
>>> been a contradiction in terms.
>>>
>>> HHH correctly predicts that a correct and unlimited emulation
>>> of DDD by HHH cannot possibly reach its own "return" instruction
>>> final halt state.
>>
>> That is not a meaningful prediction because a complete and unlimited
>> emulation of DDD by HHH never happens.
>>
>
> A complete emulation is not required to correctly
> predict that a complete emulation would never halt.
>
But since HHH is both a Halt Decider and an emulator, it must obey both
criteria, so it MUST actually predict what the complete emulation will do.
You don't seem to understand that both view of HHH must exist in the
same code, and it must satisfy both requrements at the same time, which
it can't.