theorem Th24:
  superior_setsequence B is non-ascending
proof
  now
    let n be Nat;
    (superior_setsequence B).n = (superior_setsequence B).(n+1) \/ B.n by Th22;
    hence (superior_setsequence B).(n+1) c= (superior_setsequence B).n by
XBOOLE_1:7;
  end;
  hence thesis by PROB_2:6;
end;
