XPost: comp.theory, sci.math, comp.lang.prolog   
   From: mikko.levanto@iki.fi   
      
   olcott kirjoitti 11.12.2025 klo 16.17:   
   > On 12/11/2025 2:42 AM, Mikko wrote:   
   >> 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.   
   >   
   >  >> 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.   
   > As I say non-terminating, thus never resolves to a truth value.   
      
   As according to Prolog rules foo(Y) isn't a truth value for any Y   
   the above is obviously just an attempt to deive with a distraction.   
      
   --   
   Mikko   
      
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