358 lines
		
	
	
		
			11 KiB
		
	
	
	
		
			Prolog
		
	
	
	
	
	
			
		
		
	
	
			358 lines
		
	
	
		
			11 KiB
		
	
	
	
		
			Prolog
		
	
	
	
	
	
| % This file has been included as an YAP library by Vitor Santos Costa, 1999
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| 
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| %   File   : ORDSET.PL
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| %   Author : R.A.O'Keefe
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| %   Updated: 22 May 1983
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| %   Purpose: Ordered set manipulation utilities
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| 
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| %   In this module, sets are represented by ordered lists with no
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| %   duplicates.  Thus {c,r,a,f,t} would be [a,c,f,r,t].  The ordering
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| %   is defined by the @< family of term comparison predicates, which
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| %   is the ordering used by sort/2 and setof/3.
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| 
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| %   The benefit of the ordered representation is that the elementary
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| %   set operations can be done in time proportional to the Sum of the
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| %   argument sizes rather than their Product.  Some of the unordered
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| %   set routines, such as member/2, length/2, select/3 can be used
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| %   unchanged.  The main difficulty with the ordered representation is
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| %   remembering to use it!
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| 
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| :- module(ordsets, [
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| 	list_to_ord_set/2,	%  List -> Set
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| 	merge/3,		%  OrdList x OrdList -> OrdList
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| 	ord_add_element/3,	%  Set x Elem -> Set
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| 	ord_del_element/3,	%  Set x Elem -> Set
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| 	ord_disjoint/2,		%  Set x Set ->
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| 	ord_insert/3,		%  Set x Elem -> Set
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|         ord_member/2,           %  Set -> Elem
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| 	ord_intersect/2,	%  Set x Set ->
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| 	ord_intersect/3,	%  Set x Set -> Set
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| 	ord_intersection/3,	%  Set x Set -> Set
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| 	ord_intersection/4,	%  Set x Set -> Set x Set
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| 	ord_seteq/2,		%  Set x Set ->
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| 	ord_setproduct/3,	%  Set x Set -> Set
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| 	ord_subset/2,		%  Set x Set ->
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| 	ord_subtract/3,		%  Set x Set -> Set
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| 	ord_symdiff/3,		%  Set x Set -> Set
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|         ord_union/2,            %  Set^2 -> Set
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| 	ord_union/3,		%  Set x Set -> Set
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| 	ord_union/4,		%  Set x Set -> Set x Set,
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| 	ord_empty/1,             % -> Set
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| 	ord_memberchk/2             % Element X Set
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|       ]).
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| 
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| /*
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| :- mode
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| 	list_to_ord_set(+, ?),
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| 	merge(+, +, -),
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| 	ord_disjoint(+, +),
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| 	ord_disjoint(+, +, +, +, +),
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| 	ord_insert(+, +, ?),
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| 	ord_insert(+, +, +, +, ?),
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| 	ord_intersect(+, +),
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| 	ord_intersect(+, +, +, +, +),
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| 	ord_intersect(+, +, ?),
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| 	ord_intersect(+, +, +, +, +, ?),
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| 	ord_seteq(+, +),
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| 	ord_subset(+, +),
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| 	ord_subset(+, +, +, +, +),
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| 	ord_subtract(+, +, ?), 
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| 	ord_subtract(+, +, +, +, +, ?),
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| 	ord_symdiff(+, +, ?),
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| 	ord_symdiff(+, +, +, +, +, ?),
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| 	ord_union(+, +, ?),
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| 	ord_union(+, +, +, +, +, ?).
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| */
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| 
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| 
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| %   list_to_ord_set(+List, ?Set)
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| %   is true when Set is the ordered representation of the set represented
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| %   by the unordered representation List.  The only reason for giving it
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| %   a name at all is that you may not have realised that sort/2 could be
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| %   used this way.
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| 
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| list_to_ord_set(List, Set) :-
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| 	sort(List, Set).
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| 
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| 
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| %   merge(+List1, +List2, -Merged)
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| %   is true when Merged is the stable merge of the two given lists.
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| %   If the two lists are not ordered, the merge doesn't mean a great
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| %   deal.  Merging is perfectly well defined when the inputs contain
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| %   duplicates, and all copies of an element are preserved in the
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| %   output, e.g. merge("122357", "34568", "12233455678").  Study this
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| %   routine carefully, as it is the basis for all the rest.
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| 
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| merge([Head1|Tail1], [Head2|Tail2], [Head2|Merged]) :-
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| 	Head1 @> Head2, !,
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| 	merge([Head1|Tail1], Tail2, Merged).
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| merge([Head1|Tail1], List2, [Head1|Merged]) :-
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| 	List2 \== [], !,
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| 	merge(Tail1, List2, Merged).
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| merge([], List2, List2) :- !.
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| merge(List1, [], List1).
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| 
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| 
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| 
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| %   ord_disjoint(+Set1, +Set2)
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| %   is true when the two ordered sets have no element in common.  If the
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| %   arguments are not ordered, I have no idea what happens.
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| 
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| ord_disjoint([], _) :- !.
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| ord_disjoint(_, []) :- !.
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| ord_disjoint([Head1|Tail1], [Head2|Tail2]) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_disjoint(Order, Head1, Tail1, Head2, Tail2).
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| 
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| ord_disjoint(<, _, Tail1, Head2, Tail2) :-
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| 	ord_disjoint(Tail1, [Head2|Tail2]).
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| ord_disjoint(>, Head1, Tail1, _, Tail2) :-
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| 	ord_disjoint([Head1|Tail1], Tail2).
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| 
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| 
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| 
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| %   ord_insert(+Set1, +Element, ?Set2)
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| %   ord_add_element(+Set1, +Element, ?Set2)
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| %   is the equivalent of add_element for ordered sets.  It should give
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| %   exactly the same result as merge(Set1, [Element], Set2), but a bit
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| %   faster, and certainly more clearly.
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| 
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| ord_add_element([], Element, [Element]).
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| ord_add_element([Head|Tail], Element, Set) :-
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| 	compare(Order, Head, Element),
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| 	ord_insert(Order, Head, Tail, Element, Set).
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| 
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| 
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| ord_insert([], Element, [Element]).
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| ord_insert([Head|Tail], Element, Set) :-
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| 	compare(Order, Head, Element),
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| 	ord_insert(Order, Head, Tail, Element, Set).
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| 
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| 
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| ord_insert(<, Head, Tail, Element, [Head|Set]) :-
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| 	ord_insert(Tail, Element, Set).
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| ord_insert(=, Head, Tail, _, [Head|Tail]).
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| ord_insert(>, Head, Tail, Element, [Element,Head|Tail]).
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| 
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| 
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| 
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| %   ord_intersect(+Set1, +Set2)
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| %   is true when the two ordered sets have at least one element in common.
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| %   Note that the test is == rather than = .
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| 
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| ord_intersect([Head1|Tail1], [Head2|Tail2]) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_intersect(Order, Head1, Tail1, Head2, Tail2).
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| 
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| ord_intersect(=, _, _, _, _).
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| ord_intersect(<, _, Tail1, Head2, Tail2) :-
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| 	ord_intersect(Tail1, [Head2|Tail2]).
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| ord_intersect(>, Head1, Tail1, _, Tail2) :-
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| 	ord_intersect([Head1|Tail1], Tail2).
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| 
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| ord_intersect(L1, L2, L) :-
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| 	ord_intersection(L1, L2, L).
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| 
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| 
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| %   ord_intersection(+Set1, +Set2, ?Intersection)
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| %   is true when Intersection is the ordered representation of Set1
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| %   and Set2, provided that Set1 and Set2 are ordered sets.
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| 
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| ord_intersection([], _, []) :- !.
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| ord_intersection([_|_], [], []) :- !.
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| ord_intersection([Head1|Tail1], [Head2|Tail2], Intersection) :-
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| 	( Head1 == Head2 ->
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| 	    Intersection = [Head1|Tail],
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| 	    ord_intersection(Tail1, Tail2, Tail)
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| 	;
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| 	    Head1 @< Head2 ->
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| 	    ord_intersection(Tail1, [Head2|Tail2], Intersection)
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| 	;
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| 	    ord_intersection([Head1|Tail1], Tail2, Intersection)
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| 	).
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| 
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| %   ord_intersection(+Set1, +Set2, ?Intersection, ?Difference)
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| %   is true when Intersection is the ordered representation of Set1
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| %   and Set2, provided that Set1 and Set2 are ordered sets.
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| 
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| ord_intersection([], L, [], L) :- !.
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| ord_intersection([_|_], [], [], []) :- !.
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| ord_intersection([Head1|Tail1], [Head2|Tail2], Intersection, Difference) :-
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| 	( Head1 == Head2 ->
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| 	    Intersection = [Head1|Tail],
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| 	    ord_intersection(Tail1, Tail2, Tail, Difference)
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| 	;
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| 	    Head1 @< Head2 ->
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| 	    ord_intersection(Tail1, [Head2|Tail2], Intersection, Difference)
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| 	;
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| 	    Difference = [Head2|HDifference],
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| 	    ord_intersection([Head1|Tail1], Tail2, Intersection, HDifference)
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| 	).
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| 
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| 
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| %   ord_seteq(+Set1, +Set2)
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| %   is true when the two arguments represent the same set.  Since they
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| %   are assumed to be ordered representations, they must be identical.
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| 
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| 
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| ord_seteq(Set1, Set2) :-
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| 	Set1 == Set2.
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| 
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| 
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| 
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| %   ord_subset(+Set1, +Set2)
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| %   is true when every element of the ordered set Set1 appears in the
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| %   ordered set Set2.
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| 
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| ord_subset([], _) :- !.
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| ord_subset([Head1|Tail1], [Head2|Tail2]) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_subset(Order, Head1, Tail1, Head2, Tail2).
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| 
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| ord_subset(=, _, Tail1, _, Tail2) :-
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| 	ord_subset(Tail1, Tail2).
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| ord_subset(>, Head1, Tail1, _, Tail2) :-
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| 	ord_subset([Head1|Tail1], Tail2).
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| 
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| 
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| 
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| %   ord_subtract(+Set1, +Set2, ?Difference)
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| %   is true when Difference contains all and only the elements of Set1
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| %   which are not also in Set2.
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| 
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| 
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| ord_subtract(Set1, [], Set1) :- !.
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| ord_subtract([], _, []) :- !.
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| ord_subtract([Head1|Tail1], [Head2|Tail2], Difference) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_subtract(Order, Head1, Tail1, Head2, Tail2, Difference).
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| 
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| ord_subtract(=, _,     Tail1, _,     Tail2, Difference) :-
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| 	ord_subtract(Tail1, Tail2, Difference).
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| ord_subtract(<, Head1, Tail1, Head2, Tail2, [Head1|Difference]) :-
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| 	ord_subtract(Tail1, [Head2|Tail2], Difference).
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| ord_subtract(>, Head1, Tail1, _,     Tail2, Difference) :-
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| 	ord_subtract([Head1|Tail1], Tail2, Difference).
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| 
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| 
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| %   ord_del_element(+Set1, Element, ?Rest)
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| %   is true when Rest contains the elements of Set1
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| %   except for Set1
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| 
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| 
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| ord_del_element([], _, []).
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| ord_del_element([Head1|Tail1], Head2, Rest) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_del_element(Order, Head1, Tail1, Head2, Rest).
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| 
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| ord_del_element(=, _,     Tail1, _, Tail1).
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| ord_del_element(<, Head1, Tail1, Head2, [Head1|Difference]) :-
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| 	ord_del_element(Tail1, Head2, Difference).
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| ord_del_element(>, Head1, Tail1, _, [Head1|Tail1]).
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| 
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| 
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| 
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| %   ord_symdiff(+Set1, +Set2, ?Difference)
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| %   is true when Difference is the symmetric difference of Set1 and Set2.
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| 
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| ord_symdiff(Set1, [], Set1) :- !.
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| ord_symdiff([], Set2, Set2) :- !.
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| ord_symdiff([Head1|Tail1], [Head2|Tail2], Difference) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_symdiff(Order, Head1, Tail1, Head2, Tail2, Difference).
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| 
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| ord_symdiff(=, _,     Tail1, _,     Tail2, Difference) :-
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| 	ord_symdiff(Tail1, Tail2, Difference).
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| ord_symdiff(<, Head1, Tail1, Head2, Tail2, [Head1|Difference]) :-
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| 	ord_symdiff(Tail1, [Head2|Tail2], Difference).
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| ord_symdiff(>, Head1, Tail1, Head2, Tail2, [Head2|Difference]) :-
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| 	ord_symdiff([Head1|Tail1], Tail2, Difference).
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| 
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| 
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| 
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| %   ord_union(+Set1, +Set2, ?Union)
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| %   is true when Union is the union of Set1 and Set2.  Note that when
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| %   something occurs in both sets, we want to retain only one copy.
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| 
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| ord_union([S|Set1], [], [S|Set1]).
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| ord_union([], Set2, Set2).
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| ord_union([Head1|Tail1], [Head2|Tail2], Union) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_union(Order, Head1, Tail1, Head2, Tail2, Union).
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| 
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| ord_union(=, Head,  Tail1, _,     Tail2, [Head|Union]) :-
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| 	ord_union(Tail1, Tail2, Union).
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| ord_union(<, Head1, Tail1, Head2, Tail2, [Head1|Union]) :-
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| 	ord_union(Tail1, [Head2|Tail2], Union).
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| ord_union(>, Head1, Tail1, Head2, Tail2, [Head2|Union]) :-
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| 	ord_union([Head1|Tail1], Tail2, Union).
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| 
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| 
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| %   ord_union(+Set1, +Set2, ?Union, ?Difference)
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| %   is true when Union is the union of Set1 and Set2 and Difference is the
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| %   difference between Set2 and Set1.
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| 
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| ord_union(Set1, [], Set1, []) :- !.
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| ord_union([], Set2, Set2, Set2) :- !.
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| ord_union([Head1|Tail1], [Head2|Tail2], Union, Diff) :-
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| 	compare(Order, Head1, Head2),
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| 	ord_union(Order, Head1, Tail1, Head2, Tail2, Union, Diff).
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| 
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| ord_union(=, Head,  Tail1, _,  Tail2, [Head|Union], Diff) :-
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| 	ord_union(Tail1, Tail2, Union, Diff).
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| ord_union(<, Head1, Tail1, Head2, Tail2, [Head1|Union], Diff) :-
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| 	ord_union(Tail1, [Head2|Tail2], Union, Diff).
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| ord_union(>, Head1, Tail1, Head2, Tail2, [Head2|Union], [Head2|Diff]) :-
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| 	ord_union([Head1|Tail1], Tail2, Union, Diff).
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| 
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| 
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| 
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| %   ord_setproduct(+Set1, +Set2, ?Product)
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| %   is in fact identical to setproduct(Set1, Set2, Product).
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| %   If Set1 and Set2 are ordered sets, Product will be an ordered
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| %   set of x1-x2 pairs.  Note that we cannot solve for Set1 and
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| %   Set2, because there are infinitely many solutions when
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| %   Product is empty, and may be a large number in other cases.
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| 
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| ord_setproduct([], _, []).
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| ord_setproduct([H|T], L, Product) :-
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| 	ord_setproduct(L, H, Product, Rest),
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| 	ord_setproduct(T, L, Rest).
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| 
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| ord_setproduct([], _, L, L).
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| ord_setproduct([H|T], X, [X-H|TX], TL) :-
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| 	ord_setproduct(T, X, TX, TL).
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| 
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| 
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| ord_member(El,[H|T]):-
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|     compare(Op,El,H),
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|     ord_member(Op,El,T).
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| 
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| ord_member(=,_,_).
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| ord_member(>,El,[H|T]) :-
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|     compare(Op,El,H),
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|     ord_member(Op,El,T).
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| 
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| ord_union([], []).
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| ord_union([Set|Sets], Union) :-
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|     length([Set|Sets], NumberOfSets),
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|     ord_union_all(NumberOfSets, [Set|Sets], Union, []).
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| 
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| ord_union_all(N,Sets0,Union,Sets) :-
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|     (  N=:=1  -> Sets0=[Union|Sets]
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|     ;  N=:=2  -> Sets0=[Set1,Set2|Sets], 
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|                  ord_union(Set1,Set2,Union)
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|     ;  A is N>>1,
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|        Z is N-A,
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|        ord_union_all(A, Sets0, X, Sets1),
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|        ord_union_all(Z, Sets1, Y, Sets),
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|        ord_union(X, Y, Union)
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|     ).
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| 
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| ord_empty([]).
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| 
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| ord_memberchk(Element, [E|_]) :- E == Element, !.
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| ord_memberchk(Element, [_|Set]) :-
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| 	ord_memberchk(Element, Set).
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| 
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