defpred S1[ 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: S1[ 0 ] by Th11;
A2: for k being Nat st S1[k] & S1[k + 1] holds
S1[k + 2]
proof
let k be Nat; :: thesis: ( S1[k] & S1[k + 1] implies S1[k + 2] )
assume that
A3: S1[k] and
A4: S1[k + 1] ; :: thesis: S1[k + 2]
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 S1[k + 2] ; :: thesis: verum
end;
( tau_bar to_power 1 = tau_bar & tau to_power 1 = tau ) by POWER:25;
then A5: S1[1] by Th11, FIB_NUM:def 1, FIB_NUM:def 2;
for k being Nat holds S1[k] from FIB_NUM:sch 1(A1, A5, A2);
 hence for n being Nat holds Lucas n = (tau to_power n) + (tau_bar to_power n) ; :: thesis: verum