theorem Th87:
  a >= 1 & b <= 0 implies a #R b <= 1
proof
  assume that
A1: a>=1 and
A2: b<=0;
  a #R (-b) >= 1 by A1,A2,Th85;
  then 1 / a #R b >= 1 by A1,Th76;
  then 1 / (1 / a #R b) <= 1 by XREAL_1:211;
  hence thesis;
end;
