clear(print_given).
set(binary_res).
clear(detailed_history).
clear(print_kept).
clear(print_back_sub).
assign(max_seconds,10).
assign(stats_level,0).

list(usable).

% Everybody has a mother.
% forall y exists x mother(x,y).  mother_of is a function.
mother(mother_of(y),y).

% You don't have more than one mother.
% forall x,y (mother(x,y) -> x = mother_of(y)
-mother(x,y) | x = mother_of(y).
 
% You can't be your own mother.
% forall x,y (mother(x,y) -> x != y)
-mother(x,y) | x != y.

% Your mother is in your maternal line.
% forall x,y (mother(x,y) -> maternal(x,y))
-mother(x,y) | maternal(x,y).

% The mother of anyone in your maternal line is in it too.
% forall x,y,z (mother(x,y) & maternal(y,z) -> maternal(x,z))
-mother(x,y) | -maternal(y,z) | maternal(x,z).

end_of_list.

list(sos).

% I want to prove that someone is in your maternal line
% that is not your mother.
%	forall y exists x (maternal(x,y) & -mother(x,y))
% I need to refute the negation of this sentence which is:
% 	exists y forall x (-maternal(x,y) | mother(x,y)
% I use Jane as a Skolem constant.

-maternal(x,Jane) | mother(x,Jane).

end_of_list.