XPost: comp.theory, sci.math, comp.lang.prolog   
   From: mikko.levanto@iki.fi   
      
   olcott kirjoitti 8.12.2025 klo 21.09:   
   > On 12/8/2025 3:13 AM, Mikko wrote:   
   >> olcott kirjoitti 5.12.2025 klo 19.43:   
   >>> On 12/5/2025 3:38 AM, Mikko wrote:   
   >>>> olcott kirjoitti 4.12.2025 klo 16.06:   
   >>>>> On 12/4/2025 2:58 AM, Mikko wrote:   
   >>>>>> Tristan Wibberley kirjoitti 4.12.2025 klo 4.32:   
   >>>>>>> On 30/11/2025 09:58, Mikko wrote:   
   >>>>>>>   
   >>>>>>>> Note that the meanings of   
   >>>>>>>>   ?- G = not(provable(F, G)).   
   >>>>>>>> and   
   >>>>>>>>   ?- unify_with_occurs_check(G, not(provable(F, G))).   
   >>>>>>>> are different. The former assigns a value to G, the latter does   
   >>>>>>>> not.   
   >>>>>>   
   >>>>>>> For sufficiently informal definitions of "value".   
   >>>>>>> And for sufficiently wrong ones too!   
   >>>>>>   
   >>>>>> It is sufficiently clear what "value" of a Prolog variable means.   
   >>>>   
   >>>>> % This sentence cannot be proven in F   
   >>>>> ?- G = not(provable(F, G)).   
   >>>>> G = not(provable(F, G)).   
   >>>>> ?- unify_with_occurs_check(G, not(provable(F, G))).   
   >>>>> false.   
   >>>>>   
   >>>>> I would say that the above Prolog is the 100%   
   >>>>> complete formal specification of:   
   >>>>>   
   >>>>> "This sentence cannot be proven in F"   
   >>>>   
   >>>> The first query can be regarded as a question whether "G =   
   >>>> not(provable(   
   >>>> F, G))" can be proven for some F and some G. The answer is that it can   
   >>>> for every F and for (at least) one G, which is not(provable(G)).   
   >>>>   
   >>>> The second query can be regarded as a question whether "G =   
   >>>> not(provable   
   >>>> (F, G))" can be proven for some F and some G that do not contain   
   >>>> cycles.   
   >>>> The answer is that in the proof system of Prolog it cannot be.   
   >>>   
   >>> No that it flatly incorrect. The second question is this:   
   >>> Is "G = not(provable(F, G))." semantically sound?   
   >>   
   >> Where is the definition of Prolog semantics is that said?   
   >   
   > Any expression of Prolog that cannot be evaluated to   
   > a truth value because it specifies non-terminating   
   > infinite recursion is "semantically unsound" by the   
   > definition of those terms even if Prolog only specifies   
   > that cannot be evaluated to a truth value because it   
   > specifies non-terminating infinite recursion.   
      
   Your Prolog implementation has evaluated G = not(provablel(F, G))   
   to a truth value true. When doing so it evaluated each side of =   
   to a value that is not a truth value.   
      
   --   
   Mikko   
      
   --- SoupGate-Win32 v1.05   
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