% deepCount(L,N).

% try writing the solution first in english, declaratively
%
% count of an emply list is 0
% count of a non-empty list adds N to the count of the tail of the list.
%    N: if the head is a list then recursively count the head
%    N: if the head is not a list then add 1

dc([],0).
dc([H|T],N) :- is_list(H), dc(H,N1), dc(T,N2), N is N1+N2.
dc([H|T],N) :- not(is_list(H)), dc(T,M), N is M+1.


% debag(B,S).
%
% an empty list no change
% a list singleton no change
% if the head of a list is in the tail, leave the head out.

% this one finds many solutions that are sub optimal
db([],[]).
db([H|T],L) :- member(H,T), db(T,L).
db([H|T],L) :- db(T,M), append([H],M,L).

% this one uses cut
debag([],[]).
debag([H|T],L) :- member(H,T), !, debag(T,L).
debag([H|T],L) :- debag(T,M), append([H],M,L).

% sum a list of integers
sum([Nth],Nth).
sum( [Ith | Rest], SumAll ) :- sum(Rest,SumRest), SumAll is SumRest+Ith .