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> "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 d

> "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 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 7y ago
>> 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 7y ago
  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 7y ago
    >> 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.

YeGoblynQueenne 7y ago

>> 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 7y ago
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 7y ago
  >> 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.