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Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: Mikko <mikko.levanto@iki.fi> Newsgroups: comp.theory Subject: Re: ChatGPT agrees that I have refuted the conventional Halting Problem proof technique --- Full 38 page analysis Date: Sun, 29 Jun 2025 12:27:52 +0300 Organization: - Lines: 78 Message-ID: <103r0uo$1evf8$1@dont-email.me> References: <103acoo$vp7v$1@dont-email.me> <728b9512cbf8dbf79931bfd3d5dbed265447d765@i2pn2.org> <103cvjc$1k41c$1@dont-email.me> <be0bff3b8d006e02858b9791d8508499992cbfda@i2pn2.org> <103edbp$22250$5@dont-email.me> <103g91n$2kugi$1@dont-email.me> <103h5dc$2rinm$4@dont-email.me> <103j6li$3dbba$1@dont-email.me> <103l1d7$3tktb$1@dont-email.me> <103lf9c$j25$1@dont-email.me> <103m99g$6dce$3@dont-email.me> <103olot$rfba$1@dont-email.me> <103os6c$rq7e$10@dont-email.me> MIME-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Sun, 29 Jun 2025 11:27:53 +0200 (CEST) Injection-Info: dont-email.me; posting-host="f2a24e9f48d9ee61bd9933efa533e484"; logging-data="1539560"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX19fIr2VlHAUmg4TF5rdXGt2" User-Agent: Unison/2.2 Cancel-Lock: sha1:wztVD7bO5bihBtwaHnlgS7uRnOU= On 2025-06-28 13:54:19 +0000, olcott said: > On 6/28/2025 7:04 AM, Mikko wrote: >> On 2025-06-27 14:19:28 +0000, olcott said: >> >>> On 6/27/2025 1:55 AM, Mikko wrote: >>>> On 2025-06-27 02:58:47 +0000, olcott said: >>>> >>>>> On 6/26/2025 5:16 AM, Mikko wrote: >>>>>> On 2025-06-25 15:42:36 +0000, olcott said: >>>>>> >>>>>>> On 6/25/2025 2:38 AM, Mikko wrote: >>>>>>>> On 2025-06-24 14:39:52 +0000, olcott said: >>>>>>>> >>>>>>>>> *ChatGPT and I agree that* >>>>>>>>> The directly executed DDD() is merely the first step of >>>>>>>>> otherwise infinitely recursive emulation that is terminated >>>>>>>>> at its second step. >>>>>>>> >>>>>>>> No matter who agrees, the directly executed DDD is mote than >>>>>>>> merely the first step of otherwise infinitely recursive >>>>>>>> emulation that is terminated at its second step. Not much >>>>>>>> more but anyway. After the return of HHH(DDD) there is the >>>>>>>> return from DDD which is the last thing DDD does before its >>>>>>>> termination. >>>>>>> >>>>>>> *HHH(DDD) the input to HHH specifies non-terminating behavior* >>>>>>> The fact that DDD() itself halts does not contradict that >>>>>>> because the directly executing DDD() cannot possibly be an >>>>>>> input to HHH in the Turing machine model of computation, >>>>>>> thus is outside of the domain of HHH. >>>>>> >>>>>> The input in HHH(DDD) is the same DDD that is executed in DDD() >>>>>> so the behaviour specified by the input is the behavour of >>>>>> directly executed DDD, a part of which is the behaour of the >>>>>> HHH that DDD calls. >>>>>> >>>>>> If HHH does not report about DDD but instead reports about itself >>>>>> or its own actions it is not a partial halt decideer nor a partial >>>>>> termination analyzer, as those are not allowed to report on their >>>>>> own behavour more than "cannot determine". >>>>> >>>>> Functions computed by Turing Machines are required to compute >>>>> the mapping from their inputs and not allowed to take other >>>>> executing Turing machines as inputs. >>>> >>>> There is no restriction on the functions. >>> >>> counter factual. >> >> That is not a magic spell to create a restriction on functions. >> >>>> A Turing machine is required >>>> to compute the function identified in its specification and no other >>>> function. For the halting problem the specification is that a halting >>>> decider must compute the mapping that maps to "yes" if the computation >>>> described by the input halts when directly executed. >>> >>> No one ever bothered to notice that because directly >>> executed Turing machines cannot possibly be inputs to >>> other Turing machines that these directly executed >>> Turing machines have never been in the domain of any >>> Turing machine. >> >> Irrelevant. They are the domain of the halting problem. > > That they are in the domain of the halting problem > and not in the domain of any Turing machine proves > that the requirement of the halting problem is incorrect. The halting problem can be partially solved with partial halt deciders. A computation that cannot be determined to halt or not to halt with some partical halt decier can be determined with another partial halt decider. -- Mikko