theorem
  -f/g = (-f)/g & f/(-g) = -f/g
proof
  thus -f/g = (-f)/g by Th32;
  thus f/(-g) = f (#) ((-g)^) by Th31
    .= f (#) ((-1) (#) (g^)) by Lm1,Th28
    .= -(f (#) (g^)) by Th13
    .= -(f/g) by Th31;
end;
