reserve a,b,n for Element of NAT;

theorem
  for n being Element of NAT holds 5 * (Fib(n)) |^ 2 - (Lucas(n)) |^ 2 =
  4 * (-1) to_power (n+1)
proof
  let n be Element of NAT;
  set a = (tau to_power n);
  set b = (tau_bar to_power n);
A1: (Fib(n))|^2=(Fib(n)*Fib(n)) by WSIERP_1:1
    .=(((tau to_power n)-(tau_bar to_power n))/(sqrt 5))*Fib(n) by FIB_NUM:7
    .=(((tau to_power n)-(tau_bar to_power n))/(sqrt 5)) *(((tau to_power n)
  -(tau_bar to_power n))/(sqrt 5)) by FIB_NUM:7
    .=((a - b)*(a - b))/(sqrt 5 * sqrt 5) by XCMPLX_1:76
    .=(a*a -2*a*b + b*b)/5 by Th2;
A2: a*b=(tau*tau_bar)to_power n by Th8
    .=((1*1-(sqrt 5 * sqrt 5))/4) to_power n by FIB_NUM:def 1,def 2
    .=((1-5)/4) to_power n by Th2
    .=(-1)to_power n;
  (Lucas(n))|^2=(Lucas(n)*Lucas(n)) by WSIERP_1:1
    .=(tau to_power n + tau_bar to_power n)*(Lucas(n)) by Th21
    .=(tau to_power n + tau_bar to_power n)* (tau to_power n + tau_bar
  to_power n) by Th21
    .=a*a+2*a*b+b*b;
  then 5*(Fib(n))|^2-(Lucas(n))|^2=4*(-1)*(-1)to_power n by A1,A2
    .=4*((-1)to_power 1)*(-1)to_power n by POWER:25
    .=4*(((-1)to_power 1)*(-1)to_power n)
    .=4*(-1)to_power(n+1) by FIB_NUM2:5;
  hence thesis;
end;
