theorem
  abs(f1/f2) = abs(f1)/abs(f2)
proof
  thus abs(f1/f2) = abs(f1(#)(f2^)) by Th31
    .= abs(f1)(#)abs((f2^)) by Th24
    .= abs(f1)(#)((abs(f2))^) by Th30
    .= abs(f1)/abs(f2) by Th31;
end;
