theorem Th8:
  p is dominated_by_0 implies 2 * Sum (p|(2*n+1)) < 2*n+1
proof
  assume p is dominated_by_0;
  then
A1: 2 * Sum (p|(2*n+1)) <= 2 * n + 1 by Th2;
  assume 2 * Sum (p|(2*n+1))>= 2 * n + 1;
  then 2 * Sum (p|(2*n+1))= 2 * n + 1 by A1,XXREAL_0:1;
  then 2 * (Sum (p|(2*n+1))-n)=1;
  hence thesis by INT_1:9;
end;
