defpred S₁[ Nat] means Lucas $1 = (tau to_power $1) + (tau_bar to_power $1);
A1: S₁[ 0 ]

proof

tau_bar to_power 0 = 1 by POWER:29;
then (tau to_power 0 ) + (tau_bar to_power 0 ) = 1 + 1 by POWER:29
.= 2 ;
hence S₁[ 0 ] by Th11; :: thesis: verum

end;

A2: S₁[1]

proof

( tau_bar to_power 1 = tau_bar & tau to_power 1 = tau ) by POWER:30;
hence S₁[1] by Th11, FIB_NUM:def 1, FIB_NUM:def 2; :: thesis: verum

end;

A3: 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
A4: S₁[k] and
A5: S₁[k + 1] ; :: thesis: S₁[k + 2]
Lucas (k + 2) = ((tau to_power k) + (tau_bar to_power k)) + (Lucas (k + 1)) by A4, Th12
.= (((tau to_power k) + (tau to_power (k + 1))) + (tau_bar to_power k)) + (tau_bar to_power (k + 1)) by A5
.= ((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;

for k being Nat holds S₁[k] from FIB_NUM:sch 1(A1, A2, A3);
hence for n being Nat holds Lucas n = (tau to_power n) + (tau_bar to_power n) ; :: thesis: verum