XPost: comp.theory, sci.math, comp.lang.prolog   
   From: mikko.levanto@iki.fi   
      
   olcott kirjoitti 10.12.2025 klo 16.10:   
   > On 12/10/2025 4:04 AM, Mikko wrote:   
   >> 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.   
   >   
   > ?- unify_with_occurs_check(G, not(provable(F, G))).   
   > false.   
   >   
   > Proves that   
   > G = not(provable(F, G)).   
   > would remain stuck in infinite recursion.   
   >   
   > unify_with_occurs_check() examines the directed   
   > graph of the evaluation sequence of an expression.   
   > When it detects a cycle that indicates that an   
   > expression would remain stuck in recursive   
   > evaluation never to be resolved to a truth value.   
   >   
   > BEGIN:(Clocksin & Mellish 2003:254)   
   > Finally, a note about how Prolog matching sometimes differs   
   > from the unification used in Resolution. Most Prolog systems   
   > will allow you to satisfy goals like:   
   >   
   > equal(X, X).   
   > ?- equal(foo(Y), Y).   
   >   
   > that is, they will allow you to match a term against an   
   > uninstantiated subterm of itself. In this example, foo(Y)   
   > is matched against Y, which appears within it. As a result,   
   > Y will stand for foo(Y), which is foo(foo(Y)) (because of   
   > what Y stands for), which is foo(foo(foo(Y))), and so on.   
   > So Y ends up standing for some kind of infinite structure.   
   >   
   > Note that, whereas they may allow you to construct something   
   > like this, most Prolog systems will not be able to write it   
   > out at the end. According to the formal definition of   
   > Unification, this kind of “infinite term” should never come   
   > to exist. Thus Prolog systems that allow a term to match an   
   > uninstantiated subterm of itself do not act correctly as   
   > Resolution theorem provers. In order to make them do so, we   
   > would have to add a check that a variable cannot be   
   > instantiated to something containing itself. Such a check,   
   > an occurs check, would be straightforward to implement, but   
   > would slow down the execution of Prolog programs considerably.   
   > Since it would only affect very few programs, most implementors   
   > have simply left it out 1.   
   >   
   > 1 The Prolog standard states that the result is undefined if   
   > a Prolog system attempts to match a term against an uninstantiated   
   > subterm of itself, which means that programs which cause this to   
   > happen will not be portable. A portable program should ensure that   
   > wherever an occurs check might be applicable the built-in predicate   
   > unify_with_occurs_check/2 is used explicitly instead of the normal   
   > unification operation of the Prolog implementation. As its name   
   > suggests, this predicate acts like =/2 except that it fails if an   
   > occurs check detects an illegal attempt to instantiate a variable.   
   > END:(Clocksin & Mellish 2003:254)   
   >   
   > Clocksin, W.F. and Mellish, C.S. 2003. Programming in Prolog   
   > Using the ISO Standard Fifth Edition, 254. Berlin Heidelberg:   
   > Springer-Verlag.   
      
   Thank you for the confirmation of my explanation of your error.   
      
   --   
   Mikko   
      
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