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NNTP-Posting-Date: Wed, 14 Aug 2024 20:56:37 +0000
Subject: Re: key error in all the proofs --- Mike's correction
Newsgroups: comp.theory
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From: Mike Terry <news.dead.person.stones@darjeeling.plus.com>
Date: Wed, 14 Aug 2024 21:56:36 +0100
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On 14/08/2024 18:45, olcott wrote:
> On 8/14/2024 11:31 AM, joes wrote:
>> Am Wed, 14 Aug 2024 08:42:33 -0500 schrieb olcott:
>>> 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:
>>>>>>>
>>>>>>> *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.
>> What do we care about a complete simulation? HHH isn't doing one.
>>
> 
> Please go read how Mike corrected you.
> 

Lol, dude...  I mentioned nothing about complete/incomplete simulations.

But while we're here - a complete simulation of input D() would clearly halt.  You have seen that 
yourself, e.g. with main() calling DDD(), or UTM(DDD), or HHH1(DDD).  [All of those simulate DDD to 
completion and see DDD return.  What I said earlier was that HHH(DDD) does not simulate DDD to 
completion, which I think everyone recognises - it aborts before DDD() halts.


Mike.