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|    comp.ai.philosophy    |    Perhaps we should ask SkyNet about this    |    59,235 messages    |
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|    Message 58,820 of 59,235    |
|    olcott to Richard Damon    |
|    Re: Proof that the halting problem is in    |
|    26 Dec 25 10:56:45    |
      XPost: comp.theory, sci.logic, sci.math       From: polcott333@gmail.com              On 12/26/2025 10:24 AM, Richard Damon wrote:       > On 12/26/25 10:20 AM, olcott wrote:       >> On 12/26/2025 9:05 AM, Richard Damon wrote:       >>> On 12/26/25 8:54 AM, olcott wrote:       >>>> On 12/26/2025 6:59 AM, Richard Damon wrote:       >>>>> On 12/25/25 11:51 PM, olcott wrote:       >>>>>> On 12/25/2025 10:32 PM, Richard Damon wrote:       >>>>>>> On 12/25/25 10:37 PM, olcott wrote:       >>>>>>>> On 12/25/2025 9:17 PM, Richard Damon wrote:       >>>>>>>>> On 12/25/25 10:12 PM, olcott wrote:       >>>>>>>>>> Three different LLMs have been totally convinced       >>>>>>>>>> a total of 50 times, you just don't understand.       >>>>>>>>>       >>>>>>>>> LLM LIE, so are not reliable sources.       >>>>>>>>>       >>>>>>>>       >>>>>>>> *Anyone that disagrees with this is not telling the truth*       >>>>>>>> "Any result that cannot be derived as a pure function       >>>>>>>> of finite strings is uncomputable."       >>>>>>>>       >>>>>>>       >>>>>>> But Halting *IS* a "pure function of finite strings"       >>>>>>>       >>>>>>> And it is uncomputable       >>>>>>>       >>>>>>       >>>>>> Not exactly. Usually ⟨M⟩ simulated by H == UTM(⟨M⟩)       >>>>>> Sometimes ⟨M⟩ simulated by H != UTM(⟨M⟩)       >>>>>       >>>>> Only if H doesn't CORRECTLY simulate (M).       >>>>>       >>>>       >>>> Correctly simulated is defined by the semantics       >>>> of C applied to the finite string input for       >>>> the N steps until H sees the repeating pattern.       >>>       >>> So, how does that differ from what the program actually does?       >>>       >>       >> Ah great this is the first time that you didn't       >> just dodge that out of hundreds of times.       >>       >> When-so-ever an input finite string ⟨M⟩ does not       >> cheat and call its own decider the input finite       >> string to H(⟨M⟩) is a valid proxy for UTM(⟨M⟩).       >>       >>       >       > So, you didn't answer the question.       >       > How does H CORRECTLY simulate the input and get a different result from       > what the program does?       >              The finite string P |
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