reserve a,b,n for Element of NAT;

theorem
  for n being Element of NAT holds Fib(2*n+1) = Fib(n+1) * Lucas(n+1) -
  Fib(n) * Lucas(n)
proof
  let n be Element of NAT;
  Fib(n+1) * Lucas(n+1) - Fib(n) * Lucas(n) = Fib(2*(n+1)) - Fib(n)* Lucas
  (n) by Th28
    .= Fib((2*n)+2) - Fib(2*n) by Th28
    .= Fib((2*n)+1) by FIB_NUM2:29;
  hence thesis;
end;
