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|    Message 262,096 of 262,912    |
|    olcott to Richard Damon    |
|    Re: Carol's question + my Prolog are a c    |
|    22 Dec 25 11:59:52    |
      XPost: comp.theory, comp.lang.prolog, sci.math       From: polcott333@gmail.com              On 12/22/2025 11:41 AM, Richard Damon wrote:       > On 12/22/25 12:30 PM, olcott wrote:       >> On 12/22/2025 11:23 AM, Richard Damon wrote:       >>> On 12/22/25 12:19 PM, olcott wrote:       >>>> On 12/22/2025 11:11 AM, Mikko wrote:       >>>>> On 22/12/2025 18:39, olcott wrote:       >>>>>       >>>>>> % This sentence is not true.       >>>>>> ?- LP = not(true(LP)).       >>>>>> LP = not(true(LP)).       >>>>>       >>>>> The Prolog implementation's opinion is that it is true.       >>>>>       >>>>       >>>> % This sentence is not true.       >>>> ?- LP = not(true(LP)).       >>>> LP = not(true(LP)).       >>>> ?- unify_with_occurs_check(LP, not(true(LP))).       >>>> false.       >>>>       >>>> By erasing the last line you seem to be dishonest       >>>> was that your intention?       >>>>       >>>> Also you do not seem to understand exactly       >>>> what unify_with_occurs_check() means even       >>>> when I quoted Clocksin & Mellish on this.       >>>>       >>>       >>> It means that the input sentence didn't obey Prologs non-recursvie       >>> nature.       >>>       >>       >> No that is not what it means.       >> It means that the evaluation of LP is stuck       >> in infinite recursion. LLMs are smart enough       >> to immediately see this.       >       > BECAUSE Prolog, and the simplistic logic it uses, can't handle that       > statement.              Counter-factual.              Prolog (and Olcott's Minimal Type Theory) detects       cycles in the directed graph of the evaluation       sequence of an expression.                     > Prolog can't handle LP = not(not(true(LP)))       > which isn't a contradiction either.       >       > I suspect it can't handle       >       > LP := true(LP) or (X or not(x))       >       > either, unless your version parallel searches both sides.       >       >              The Truth-teller also fails to be semantically       grounded thus specifies infinite recursion.              --       Copyright 2025 Olcott |
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