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   sci.logic      Logic -- math, philosophy & computationa      262,912 messages   

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   Message 261,962 of 262,912   
   Mikko to olcott   
   Re: A new foundation for correct reasoni   
   16 Dec 25 11:48:45   
   
   XPost: comp.lang.prolog, comp.theory, sci.math   
   From: mikko.levanto@iki.fi   
      
   On 15/12/2025 01:14, olcott wrote:   
   > On 12/14/2025 4:05 AM, Mikko wrote:   
   >> On 13/12/2025 16:43, olcott wrote:   
   >>> On 12/13/2025 4:19 AM, Mikko wrote:   
   >>>> olcott kirjoitti 12.12.2025 klo 16.19:   
   >>>>> On 12/12/2025 2:50 AM, Mikko wrote:   
   >>>>>> 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   
      
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