theorem Th3:
  Z c= dom (f1/f2) implies for x st x in Z holds ((f1/f2).x) #Z n =
  (f1.x) #Z n/(f2.x) #Z n
proof
  assume
A1: Z c= dom (f1/f2);
  let x;
  assume x in Z;
  then ((f1/f2).x) #Z n = (f1.x/f2.x) #Z n by A1,RFUNCT_1:def 1
    .=(f1.x/f2.x) |^ n by PREPOWER:36
    .=(f1.x) |^ n/(f2.x) |^ n by PREPOWER:8
    .=(f1.x) #Z n/(f2.x) |^ n by PREPOWER:36
    .=(f1.x) #Z n/(f2.x) #Z n by PREPOWER:36;
  hence thesis;
end;
