theorem
  r < s implies ].r,s.[ <> [.p,q.]
proof
  assume that
A1: r < s and
A2: ].r,s.[ = [.p,q.];
A3: not r in ].r,s.[ by Th4;
A4: p <= r by A1,A2,Th51;
  s <= q by A1,A2,Th51;
  then r <= q by A1,XXREAL_0:2;
  hence contradiction by A2,A3,A4,Th1;
end;
