theorem
 for n being Element of NAT holds
  (superior_setsequence B).n = ((inferior_setsequence Complement B).n)`
proof let n be Element of NAT;
  reconsider Y = (inferior_setsequence(Complement B)).n as Subset of X;
  Y = ((superior_setsequence(Complement Complement B)).n)` by Th30
    .= ((superior_setsequence B).n)`;
  hence thesis;
end;
