theorem Th9:
  (for n holds seq.n <= r) iff seq is bounded_above & upper_bound seq <= r
proof
  thus (for n holds seq.n <= r) implies seq is bounded_above &
  upper_bound seq <= r
  proof
    assume
A1: for n holds seq.n <= r;
    now
      let m;
      seq.m <= r by A1;
      hence seq.m < r + 1 by Lm1;
    end;
    hence
A2: seq is bounded_above by SEQ_2:def 3;
    now
      set r1= (upper_bound seq) - r;
      assume r < upper_bound seq;
      then r1 > 0 by XREAL_1:50;
      then ex k st (upper_bound seq) - r1 < seq.k by A2,Th7;
      hence contradiction by A1;
    end;
    hence thesis;
  end;
  assume that
A3: seq is bounded_above and
A4: upper_bound seq <= r;
  now
    let n;
    seq.n <= upper_bound seq by A3,Th7;
    hence seq.n <= r by A4,XXREAL_0:2;
  end;
  hence thesis;
end;
