Jan Burse, erstellt 17. Aug 2019
* This module provides ordered sets. The ordered sets are
* represented by lists [x1, .., xn]. The lists must be ordered
* and duplicate free. If this precondition is violated the
* behaviour of the predicates is undefined.
* ?- ord_union([2,3,4],[1,2,4,5],X).
* X = [1,2,3,4,5]
* ?- ord_union([1,2,4,5],[2,3,4],X).
* X = [1,2,3,4,5]
* The realization uses a membership check based on (==)/2 and
* lexical ordering based on (@<)/2. As a result the predicates
* are safe to be used with non-ground terms. On the other hand,
* since this comparison is not arithmetical, 1 and 1.0 are for
* example considered different.
* An unordered set can be converted into an ordered set by
* using the ISO predicate sort/2. Also there is no need for
* predicate permutation/2 here, since equality of ordered sets
* can be tested via the ISO predicate ==/2, provided the elements
* are sufficiently normalized.
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* ord_contains(E, O):
* The predicate succeeds when the set O contains the element E.
% ord_contains(+Elem, +OrdSet)
* ord_difference(O1, O2, O3):
* The predicate succeeds when O3 unifies with the difference of O1 by O2.
% ord_difference(+OrdSet, +OrdSet, -OrdSet)
* ord_intersection(O1, O2, O3):
* The predicate succeeds when O3 unifies with the intersection of O1 and O2.
% ord_intersection(+OrdSet, +OrdSet, -OrdSet)
* ord_union(O1, O2, O3):
* The predicate succeeds when O3 unifies with the union of O1 and O2.
% ord_union(+OrdSet, +OrdSet, -OrdSet)
* ord_subset(O1, O2):
* The predicate succeeds when O1 is a subset of O2.
% ord_subset(+OrdSet, +OrdSet)