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|    sci.logic    |    Logic -- math, philosophy & computationa    |    262,912 messages    |
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|    Message 262,423 of 262,912    |
|    Mikko to olcott    |
|    =?UTF-8?Q?Re=3A_Boiling_G=C3=B6del=27s_1    |
|    08 Jan 26 12:21:15    |
      XPost: comp.theory, comp.ai.philosophy, comp.software-eng       XPost: sci.math       From: mikko.levanto@iki.fi              On 07/01/2026 15:06, olcott wrote:       > On 1/7/2026 6:10 AM, Mikko wrote:       >> On 06/01/2026 16:02, olcott wrote:       >>> On 1/6/2026 7:23 AM, Mikko wrote:       >>>> On 06/01/2026 02:24, Oleksiy Gapotchenko wrote:       >>>>> Just an external observation:       >>>>>       >>>>> A lot of tech innovations in software optimization area get       >>>>> discarded from the very beginning because people who work on them       >>>>> perceive the halting problem as a dogma.       >>>>       >>>> It is a dogma in the same sense as 2 * 3 = 6 is a dogma: a provably       >>>> true sentence of a certain theory.       >>>>       >>>       >>> ...We are therefore confronted with a proposition which       >>> asserts its own unprovability. 15 … (Gödel 1931:40-41)       >>>       >>> Gödel, Kurt 1931.       >>> On Formally Undecidable Propositions of       >>> Principia Mathematica And Related Systems       >>>       >>> F ⊢ G_F ↔ ¬Prov_F (⌜G_F⌝)       >>> "F proves that: G_F is equivalent to       >>> Gödel_Number(G_F) is not provable in F"       >>> https://plato.stanford.edu/entries/goedel-incompleteness/#FirIncTheCom       >>>       >>> Stripping away the inessential baggage using a formal       >>> language with its own self-reference operator and       >>> provability operator (thus outside of arithmetic)       >>>       >>> G := (F ⊬ G) // G asserts its own unprovability in F       >>>       >>> A proof of G in F would be a sequence of inference       >>> steps in F that prove that they themselves do not exist.       >>       >> From the way G is constructed it can be meta-proven that either       >       > Did you hear me stutter ?       > A proof of G in F would be a sequence of inference       > steps in F that prove that they themselves do not exist.              An F where such sequence really exists then in that F both G and       the negation of G are provable.              In an F where such sequnénce does not exist G is unprovable by       definition. However it is meta-provable frome the way it is       constructed and therefore true in every interpretation where       the natural numbers contained in F have their standard properties.              --       Mikko              --- SoupGate-Win32 v1.05        * Origin: you cannot sedate... all the things you hate (1:229/2)    |
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