let n be Nat; :: thesis: ( n >= 2 & n is even implies Lucas (n + 1) = [/((tau * (Lucas n)) - 1)\] )
set tn = tau_bar to_power n;
assume A1: ( n >= 2 & n is even ) ; :: thesis: Lucas (n + 1) = [/((tau * (Lucas n)) - 1)\]
A2: Lucas (n + 1) = (tau to_power (n + 1)) + (tau_bar to_power (n + 1)) by FIB_NUM3:21;
A3: tau * (Lucas n) = tau * ((tau to_power n) + (tau_bar to_power n)) by FIB_NUM3:21
.= (tau * (tau to_power n)) + (tau * (tau_bar to_power n))
.= ((tau to_power 1) * (tau to_power n)) + (tau * (tau_bar to_power n))
.= (tau to_power (n + 1)) + (tau * (tau_bar to_power n)) by Th2 ;
A4: (tau * (Lucas n)) - 1 <= Lucas (n + 1)
proof
(tau * (tau_bar to_power n)) - 1 <= tau_bar to_power (n + 1)
proof end;
then (tau to_power (n + 1)) + ((tau * (tau_bar to_power n)) - 1) <= (tau to_power (n + 1)) + (tau_bar to_power (n + 1)) by XREAL_1:6;
hence (tau * (Lucas n)) - 1 <= Lucas (n + 1) by A3, FIB_NUM3:21; :: thesis: verum
end;
((tau * (Lucas n)) - 1) + 1 > Lucas (n + 1)
proof end;
hence Lucas (n + 1) = [/((tau * (Lucas n)) - 1)\] by A4, INT_1:def 7; :: thesis: verum