remdup predicate need hlep!

[vuzu]

Implement a predicate remdup/2 which removes duplicate elements from a list,
indexes the remaining elements (1 to n), and prints out how many elements there are.
For example,
?- remdup([7,2,4,5,6,7,9,4],X).
There are 6 elements in the list.
X = [[1,7],[2,2],[3,4],[4,5],[5,6],[6,9]]

!! the question is this. what I have done is :

remdup(L, L1) :- remdup(L, [], L1).


remdup([H|T], Xnew, List) :-
member(H, Xnew),
!,
remdup(T, Xnew, List).

remdup([H|T], Xnew, List) :-
remdup(T, [H|Xnew], List).

remdup([], L, L):-

write('There'), write(' '), write('are'), write(' '),
write(N), write(' '), write('elements'), write(' '),
write('in'), write(' '), write('the'), write(' '), write('list').


the remaining part is the indexing part. how could I do that.

 

[joel76]

You have to travel through the list and add 1 each time, buiding a pair. Your rule will be like that
numbering(+List, +Number, +BuildingList, -FinalList).