theorem
  a > 1 & b < 0 implies a #R b < 1
proof
  assume that
A1: a>1 and
A2: b<0;
  -b>0 by A2;
  then a #R (-b) > 1 by A1,Th86;
  then 1 / a #R b > 1 by A1,Th76;
  then 1 / (1 / a #R b) < 1 by XREAL_1:212;
  hence thesis;
end;
