Path: news.eternal-september.org!eternal-september.org!.POSTED!not-for-mail From: Mikko Newsgroups: comp.theory Subject: Re: ChatGPT totally understands exactly how I refuted the conventional halting problem proof technique Date: Thu, 26 Jun 2025 13:00:36 +0300 Organization: - Lines: 116 Message-ID: <103j5o4$3d3gm$1@dont-email.me> References: <1037cr1$1aja4$1@dont-email.me> <1038iil$enlc$1@dont-email.me> <10394o5$j159$2@dont-email.me> <103av83$140ie$1@dont-email.me> <103bq8n$1a3c8$4@dont-email.me> <103brmh$1bfio$1@dont-email.me> <103bvt3$1cjeg$1@dont-email.me> <103do8b$1ti9d$1@dont-email.me> <103easr$22250$1@dont-email.me> <103ekj4$22qb$1@news.muc.de> <103elhi$24lrk$1@dont-email.me> <103enru$22qb$2@news.muc.de> <103eo8q$25hsi$1@dont-email.me> <995bace8fe29b576c0d9410f991981143fd20046@i2pn2.org> <103epev$25ucn$1@dont-email.me> <9103e4719abf89a6964453318d3f52878a718788@i2pn2.org> <103f62i$292tp$1@dont-email.me> <103g811$2knml$1@dont-email.me> <103h4f4$2rinm$1@dont-email.me> MIME-Version: 1.0 Content-Type: text/plain; charset=utf-8; format=flowed Content-Transfer-Encoding: 8bit Injection-Date: Thu, 26 Jun 2025 12:00:36 +0200 (CEST) Injection-Info: dont-email.me; posting-host="9532ec9cca79e56225651e6eb36dbe45"; logging-data="3575318"; mail-complaints-to="abuse@eternal-september.org"; posting-account="U2FsdGVkX1/7f4SQof2m4PvF5kUEQtfx" User-Agent: Unison/2.2 Cancel-Lock: sha1:hiwyRlDuYUS+p/jrgeFgLQx+OuA= On 2025-06-25 15:26:28 +0000, olcott said: > On 6/25/2025 2:21 AM, Mikko wrote: >> On 2025-06-24 21:41:37 +0000, olcott said: >> >>> On 6/24/2025 4:07 PM, joes wrote: >>>> Am Tue, 24 Jun 2025 13:06:22 -0500 schrieb olcott: >>>>> On 6/24/2025 12:57 PM, joes wrote: >>>>>> Am Tue, 24 Jun 2025 12:46:01 -0500 schrieb olcott: >>>>>> >>>>>>> It is an easily verified fact that no *input* to any partial halt >>>>>>> decider (PHD) can possibly do the opposite of what its corresponding >>>>>>> PHD decides. In all of the years of all of these proofs no such >>>>>>> *input* was ever presented. >>>>>> >>>>>> You should clarify that you don't even think programs can be passed as >>>>>> input. >>>>>> >>>>> It is common knowledge the no Turing Machine can take another directly >>>>> executed Turing Machine as an input. >>>> So common that nobody would suggest such. You are the king of strawmen. >>> >>> *From the bottom of page 319 has been adapted to this* >>> https://www.liarparadox.org/Peter_Linz_HP_317-320.pdf >>> >>> When Ĥ is applied to ⟨Ĥ⟩ >>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.∞ >>>    if Ĥ applied to ⟨Ĥ⟩ halts >>> Ĥ.q0 ⟨Ĥ⟩ ⊢* Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ ⊢* Ĥ.qn >>>    if Ĥ applied to ⟨Ĥ⟩ does not halt >>> >>> Ĥ applied to ⟨Ĥ⟩ does not have embedded_H reporting on >>> the behavior specified by its input ⟨Ĥ⟩ ⟨Ĥ⟩ it has embedded_H >>> reporting on its own behavior. >> >> As made clear in the source text, embedded_H does the same as >> H when given the same input. The only difference is that if >> that same behaviour reaches its qy state then H halts there >> but Ĥ runs forever in a tight loop. > > *You are not getting the main point* > The fact that Ĥ.embedded_H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to ⟨Ĥ.qn⟩ is > not contradicted by the fact that Ĥ.embedded_H itself halts. That is not the main point. The main point is that Ĥ ⟨Ĥ⟩ halts if iH ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qn but not if H ⟨Ĥ⟩ ⟨Ĥ⟩ transitions to Ĥ.qy. Anything said about embedded_H is merely an intermediate step in the proof of the main point if not totally irrelevant. > Because Ĥ.embedded_H cannot possibly take any directly > executing TM as its input that makes the behavior of > Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H. Irrelevant. It can and does take the same input as H and from that computes the same as H. That is all that is needed for the proof. >>> Since Turing Machines cannot take directly executing >>> Turing Machines as inputs this means that the directly >>> executed Ĥ applied to ⟨Ĥ⟩ is not in the domain of >>> Ĥ.embedded_H, *thus no contradiction is ever formed* >> >> False. That Turing Machines cannot take directly executing >> Turing Machnes as inputs is irrelevant. > > Directly executing TM's are not in the domain of any > halt decider. The definition of a halt decider requires that a halt decider correctly predicts whether a direct execution halts if the computation specified by the input will be directly executed. >> The definition of >> "halting decider" requires that the decider thakes a >> description of a Turing machine and a an input to it. > > Yes. > >> From the construction of Ĥ follows that the domain of Ĥ is >> the same as the required domain of a halt decider. As the > > Maybe, IDK. What I do know is that > Ĥ applied to ⟨Ĥ⟩ outside of the domain of Ĥ.embedded_H. The proof does not specify a domain fro embedded_H. Instead it specifies that for every input, including ⟨Ĥ⟩ ⟨Ĥ⟩, and including every input outside of the domain of H, the result embedded_H gives is the same as the result H gives. Otherwise embedded_H is not correctly constructed. >> proof proves H does not do what a halting decider is >> required to do > > It is required to take a directly executing TM as input. > and > It is not allowed to take a directly executing TM as input. > >> when the input is <Ĥ> <Ĥ>, contradicting >> the claim that H is a halting decider. > > *It never has been doing any such thing* You have not shown that a premise is not true. You have not shown that an inference is not truth preserving. Therefore you have not shown that the conclusion is not known to be true. > When Ĥ.embedded_H is a simulating partial halt decider > then Ĥ.embedded_H applied to ⟨Ĥ⟩ ⟨Ĥ⟩ specifies recursive > simulation that cannot possibly reach its own simulated > final halt state of ⟨Ĥ.qn⟩. Which is perfectly compatible with the hypothesis that H is not a halting decider. -- Mikko