reserve a,b,n for Element of NAT;

theorem Th21: :: Binet Formula for Lucas Numbers
  for n being Nat holds Lucas(n) = tau to_power n + tau_bar to_power n
proof
  defpred P[Nat] means Lucas($1) = tau to_power $1 + tau_bar to_power $1;
  tau_bar to_power 0 = 1 by POWER:24;
  then (tau to_power 0) + (tau_bar to_power 0) = (1 + 1) by POWER:24
    .= 2;
  then
A1: P[0] by Th11;
A2: for k being Nat st P[k] & P[k+1] holds P[k+2]
  proof
    let k be Nat;
    assume that
A3: P[k] and
A4: P[k+1];
    Lucas(k+2)=(tau to_power k + tau_bar to_power k)+Lucas(k+1) by A3,Th12
      .=tau to_power k + tau to_power (k+1) + tau_bar to_power k + tau_bar
    to_power (k+1) by A4
      .=tau to_power (k+2) + tau_bar to_power k + tau_bar to_power (k+1) by Th9
      .=tau to_power (k+2) + (tau_bar to_power k + tau_bar to_power (k+1))
      .=tau to_power (k+2) + tau_bar to_power (k+2) by Th10;
    hence thesis;
  end;
  tau_bar to_power 1 = tau_bar & tau to_power 1 = tau by POWER:25;
  then
A5: P[1] by Th11,FIB_NUM:def 1,def 2;
  for k being Nat holds P[k] from FIB_NUM:sch 1 (A1, A5, A2);
  hence thesis;
end;
