[tempguy9999]

With respect, at least some of that's badly wrong.

From https://en.wikipedia.org/wiki/Prolog "Prolog has its roots in first-order logic"

I'm pretty sure quantification exists implicitly, but I'm too rusty. I think someone may better answer that than me (oh, "without some finite domain for X", ok, but that does not preclude it from being logic, and as for infinite domains, I doubt any any system can work with that without undecidability - and very quickly undecidable. But I'm no expert).

> TLDR: true logic languages let you talk about infinities; core Prolog does not.

I cannot accept the first part of that, therefore can't accept the implication that prolog isn't a true LL.

[colanderman]

> "Prolog has its roots in first-order logic"

And Erlang has its roots in Prolog. Yet I can't perform unification or backtracking in Erlang.

Quantification does exist implicitly, but (again, constraint solving extensions and dif/2 aside) it's not useful with respect to infinities. \+ (member(X, D), \+ P) works just fine for a finite ground D, but try with an unbound or partially-bound D and you get an instantiation error (at best) or infinite recursion (at worst).

SMT2 is perfectly happy with infinite domains. Yes, you can encounter undecidability (and obviously must in some cases), but it is not a given. You can absolutely prove theorems over infinite domains with it.

I'm basing my arguments on years of experience using Prolog, SMT2, and TLA+. You are free to disagree about the definition of a fuzzy English term, but terms are only valuable inasmuch as they are useful, and defining Prolog strictly as a logic language has not proved useful to me, given how much it differs in capability from true logic languages. The most effective programming style differs greatly between them when you realize that you must always be aware of terms and instantiation. Hence my assertion that it's much better to think of Prolog as a language for dealing with said terms and instantiation, than as a language for expressing logical statements.

To clarify: Prolog imbued with constraint-solving extensions I absolutely consider a logic language. You gain constraints and reasoning over infinities, and it allows you to stop thinking in terms of search and terms. See https://www.metalevel.at/prolog/purity for some examples of what this buys you (scroll down to "Applications to teaching Prolog"). It's enough of a game-changer to me as to warrant a different categorization.

[YeGoblynQueenne]

>> Quantification does exist implicitly, but (again, constraint solving extensions and dif/2 aside) it's not useful with respect to infinities. \+ (member(X, D), \+ P) works just fine for a finite ground D, but try with an unbound or partially-bound D and you get an instantiation error (at best) or infinite recursion (at worst).

The instantiation error would be raised by P being unbound, not D. You can walk over variables with member/2.

In Swi-Prolog:

  ?- member(X, D).
  D = [X|_3304] ;
  D = [_3302, X|_3310] ;
  D = [_3302, _3308, X|_3316] ;
  D = [_3302, _3308, _3314, X|_3322] .

In Sicstus Prolog:

  | ?- member(X,D).
  D = [X|_A] ? ;
  D = [_A,X|_B] ? ;
  D = [_A,_B,X|_C] ? ;
  D = [_A,_B,_C,X|_D] ? 
  yes

The instantiation error, e.g. in Swi:

  ?- (member(X,D), \+ P).
  ERROR: Arguments are not sufficiently instantiated

- is an implementation detail, not a feature of the language (although it's probably in some standard). You could write your own version of \+/2 that doesn't raise an error.

>> SMT2 is perfectly happy with infinite domains. Yes, you can encounter undecidability (and obviously must in some cases), but it is not a given. You can absolutely prove theorems over infinite domains with it.

You can prove theorems in infinite domains with Prolog. For instance, given the program P, below, the query Q terminates:

  P = 
  { s(0),
    s(s(N)):- s(N)
  }

  Q = { s(s(s(0))) }

Now that I think about it, member/2, append/2, etc list processing predicates also range over infinite domains.

  member(X,[X|T]).
  member(X,[H|T]):- 
    member(X,T).

etc. True for lists of arbitrary length, restricted only by your computer's memory.

[colanderman]

I don't think you are interpreting my claims in good faith (moreso in your other post where you are nitpicking about the syntax I am using when talking about cross-language concepts to make a high-level point), so I am not going to address all your statements. But to clarify the specific example I had in mind:

    ?- \+ (member(X, D), \+ (X < 2)).

fails with an instantiation error on X, as it ought (since </2 requires ground arguments). That's not an "implementation detail", that's how Prolog works. Core Prolog can't handle constraints outside of term structure. There's no way to express "declare an infinite vector D, whose elements are all less than 2" which is trivial to express in FOL.

You can't "write your own version of \+" to cause the above program to give the answer one would expect from FOL, without changing the language, and I'm not making claims about "YeGoblynQueenne's Imagined Prolog".

Your example on naturals breaks down as soon as one tries to make a statement ranging over the domain of all naturals. Even something simple like:

    ?- forall(s(N), (N = 0; N = s(M), s(M))).

dives headlong into infinite recursion. My only claim is that, it's much easier to understand why this happens if one thinks of Prolog as a search language over structural terms, rather than as a logic language where such a statement could be expected to be useful (if not necessarily decidable). It's beyond me why anyone considers this contentious.

[YeGoblynQueenne]

>> It's beyond me why anyone considers this contentious.

Could you tone this sort of thing down please? Its needlessly confrontational tone only serves to add irrelevant noise to the converstation. Thank you in advance.

>> ?- L = [_,_,_], \+ (member(X, L), \+ (X < 2)).

Thank you for clarifying your intention with the original example. In that case, you could redefine </2 to avoid raising an error. Off the top of my head:

  1 < 2.
  2 < 3.
  3 < 4.

etc. Again it's a matter of implementation - in this case, of the predicate </2, rather than \+/2.

>> ?- forall(s(N), (N = 0; N = s(M), s(M))).

In your previous comment you said that "You can absolutlely prove theorems over infinite domains with [SMT2]", constrasting it with Prolog that if I understand correctly, you claim cannot. In my reply to your comment, I gave examples of theorems over infinite domains (integers and lists), that can be proven with Prolog.

You given an example of how one of those theories can "go infinite" if you make the right query. It can, but if you make a different query, it doesn't go infinite and Prolog proves it. So, Prolog can prove theories over infinite domains.

EDIT:

>> My only claim is that, it's much easier to understand why this happens if one thinks of Prolog as a search language over structural terms, rather than as a logic language where such a statement could be expected to be useful (if not necessarily decidable).

Your claim would be clearer if you explained what you mean by "logic language". Given what you've said so far I honestly have no idea what you mean by that.

In any case, what you say above sounds like the well-understood fact that Prolog programs have a declarative and a procedural reading and that one must be aware of the internal machinery of the interpreter to avoid infinite recursion.

[colanderman]

> Could you tone this sort of thing down please? […] Thank you in advance.

These sorts of comments (and your earlier notation nitpicking) come across as smug and condescending. Why start a conversation with someone you don't appear to respect?

> In any case, what you say above sounds like the well-understood fact that Prolog programs have a declarative and a procedural reading and that one must be aware of the internal machinery of the interpreter to avoid infinite recursion.

Yes, hence why I don't understand why my statement is so contentious. I wasn't being hyperbolic with that remark; I really don't understand the strong negative reaction to my off-the-cuff explanation of the above fact.

You're clearly an intelligent person, and I like to think I am too, and since intelligent people should not disagree, let me restate my thesis in clearer terms:

By "logic language" I mean, a language which allows one to express and reason about logical relations (roughly in the sense of first-order logic). Yes, my definition is intentionally vague, as are all language classifications.

Core Prolog (meaning, Horn clauses + terms + SLD + ISO definitions of predicates) allows expressing only the logical relations of equality (=/2) and inequality (dif/2; to contrast, </2 is nonlogical in core Prolog), and then, only over structured terms (i.e., atoms + compounds), and even then, is limited by its resolution strategy. This stands in contrast to languages like SMT2 and TLA+, which permit arbitrary relations over various, even abstract, domains (such as SMT2's capability to express relations on uninterpreted functions) and which do not specify, nor are associated with, a particular resolution strategy. (TLA+ for example, is strongly associated with TLC, but there is also TLAPS, and it is specified expressly independent of any resolution strategy.)

Therefore, for someone who is wondering why Prolog programs should be written in terms of something so concrete as lists, when first-order logic rarely utilizes them, it is best to think of Prolog not as a language for reasoning about logical relations (a "logic language"), but as a language for search and unification of structured terms.

[tempguy9999]

The problem is to me that you're imprecise in your claims, which is OK (edit: changed my mind, it's not ok because it's too close to ex falso (sequitur) quodlibet), and won't admit you're wrong when you are, which is bothering people.

You say "Prolog isn't based on first-order logic or anything", I point out you're wrong via wikipedia, you start fudging. You could have said it's not proper in your view because of the resolution strategy (depth first, typically) and allowing (red) cuts and we'd be with you all the way.

You say "You seem to have had no trouble understanding what I wrote", ducking the point made, viz. "That is not a First Order Logic statement", and that's the crucial point (and I'd agree, quantification over predicates isn't 1st order any more).

You evidently know your stuff but equally so do others, why are you burning bridges instead of trying to work out why there's disagreement. You could say prolog just isn't that impressive and there's better stuff available now, so in your view you don't consider prolog worthy any more, but you start denying it's a logic language at all.

[YeGoblynQueenne]

>> Thank you in advance.

I think you read this comment as condencension, but that was definitely not my intent. I meant it in all seriousness that I would be grateful if you could reduce the confrontational tone of your comments.

Anyway I'm sorry that we failed to communicate and have a productive conversation.

[JadeNB]

> I'm pretty sure quantification exists implicitly, but I'm too rusty.

Indeed, as in most informal logic, all free variables are implicitly universally quantified. I think the thing that makes it seem like it's not so is that people see a clause like:

    p(X) :- q(X).

and think "oh, that X isn't really universally quantified, because p(X) isn't always true", but it really is universally quantified; what we're saying is:

    ∀X, p(X) ← q(X).

i.e., for all X values, in order to prove p(X), it suffices to prove q(X). See, for example, https://en.wikipedia.org/wiki/Horn_clause#Definition , which says

> In the non-propositional case, all variables[note 2] in a clause are implicitly universally quantified with the scope being the entire clause.

Prolog's resolution exactly follows this proof strategy; see, for example, https://en.wikipedia.org/wiki/SLD_resolution .

[tempguy9999]

Appreciated. Knew ∀ was in there somewhere... Will check out SLD