theorem Th41:
 for n being Nat holds
  B is non-descending implies (superior_setsequence B).n = (
  superior_setsequence B).(n+1)
proof let n be Nat;
  assume B is non-descending;
  then (superior_setsequence B).(n+1) \/ B.n = (superior_setsequence B).(n+1)
  by Th40,XBOOLE_1:12;
  hence thesis by Th22;
end;
