
theorem
  for A being set, b1,b2 being Rbag of A st (for x being set st x in A
  holds b1.x = b2.x) holds Sum b1 = Sum b2
proof
  let A be set, b1,b2 be Rbag of A;
  assume
A1: for x being set st x in A holds b1.x = b2.x;
  then for x being set st x in A holds b2.x <= b1.x;
  then
A2: Sum b2 <= Sum b1 by Th5;
  for x being set st x in A holds b1.x <= b2.x by A1;
  then Sum b1 <= Sum b2 by Th5;
  hence thesis by A2,XXREAL_0:1;
end;
