list_to_set/2 is not required to order

[Jean-Luc-Picard-2021]

I find a commit comment:

diff --git a/ports/fcube/fcube.lgt b/ports/fcube/fcube.lgt
index acdc0ebbf5..d1d46b5951 100644
--- a/ports/fcube/fcube.lgt
+++ b/ports/fcube/fcube.lgt
+ % Paulo Moura: added the next list_to_set/2 to allow using an ordered set representation
list_to_set(NewSet0, NewSet).

list_to_set/2 does not order, only deduplicate.
You can try yourself in SWI-Prolog:

/* SWI-Prolog 8.5.14 */
?- list_to_set([c,a,b,a], X).
X = [c, a, b].
/* Its not ordered, if it were ordered it were [a,b,c] */

In some Prolog systems it does order, but this
is not a guarantee neither a requirement.

[Jean-Luc-Picard-2021]

If you would like to use ordered sets, you need
to use the ISO core standard sort/2.
But its not guaranteed to run faster in any way.

It can also run slower. It would only run
faster if you could eliminate the sort/2. Which
is sometimes possible, like the result of

a ord_XXX operation, doesn't need a sort/2.

P.S.: You see here, its not list_to_set/2, but it would be list_to_ord_set/2,

list_to_ord_set(+List, -OrdSet)
Transform a list into an ordered set. This is the same as sorting the list.
https://www.swi-prolog.org/pldoc/man?predicate=list_to_ord_set/2

It has this implementation:

list_to_ord_set(List, Set) :-
 sort(List, Set). 

https://www.swi-prolog.org/pldoc/doc/_SWI_/library/ordsets.pl?show=raw

[pmoura]

In the same fcube.lgt file:

	:- uses(set, [
		as_set/2 as list_to_set/2, intersection/3, subtract/3, union/3
	]).

The Logtalk sets library is an ordered sets library.

[Jean-Luc-Picard-2021]

Ok, you use the same protocol for ord_XXX.
And it is implemented as follows:

as_set(List, Set) :-
     sort(List, Set).

Is it more performant? Or is it slower than with lists?
Does your lists have a list_to_set/2? Oh you cannot test it,

no union, intersection, etc.. here:
https://github.com/LogtalkDotOrg/logtalk3/blob/master/library/types/list.lgt

Edit 24.08.2022
But your lists have an interesting outlier:

	subtract([], _, []).
	subtract([Head| Tail], List, Rest) :-
		(	memberchk(Head, List) ->
			subtract(Tail, List, Rest)
		;	Rest = [Head| Tail2],
			subtract(Tail, List, Tail2)
		).

[pmoura]

Union, intersection, ... are set operations. Subtract is not a set only operation.

[Jean-Luc-Picard-2021]

What do you mean are subtract is no set operation?
Do you mean no ordered list-set operation or no list-set operation.
Its actually both. I find it also in your set.lgt:

subtract(Set, [], Set) :- !.
subtract([], _, []) :- !.
subtract([Head1| Tail1], [Head2| Tail2], Difference) :-
		compare(Order, Head1, Head2),
		subtract(Order, Head1, Tail1, Head2, Tail2, Difference).

subtract(=, _, Tail1, _, Tail2, Difference) :-
		subtract(Tail1, Tail2, Difference).
subtract(<, Head1, Tail1, Head2, Tail2, [Head1| Difference]) :-
		subtract(Tail1, [Head2| Tail2], Difference).
subtract(>, Head1, Tail1, _, Tail2, Difference) :-
                subtract([Head1| Tail1], Tail2, Difference).

In SWI-Prolog its clearer since there is no name overloading:

https://www.swi-prolog.org/pldoc/doc_for?object=ord_subtract/3

P.S.: Mathematically we would write P \ Q for the set difference.