defpred S1[ Nat] means (2 * (Lucas \$1)) + (Lucas (\$1 + 1)) = 5 * (Fib (\$1 + 1));
A1: 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
A2: S1[k] and
A3: S1[k + 1] ; :: thesis: S1[k + 2]
(2 * (Lucas (k + 2))) + (Lucas ((k + 2) + 1)) = (2 * (Lucas (k + 2))) + (Lucas (k + (2 + 1)))
.= (2 * (Lucas (k + 2))) + ((Lucas (k + 1)) + (Lucas (k + 2))) by Th13
.= ((2 * (Lucas (k + 2))) + (Lucas (k + 1))) + (Lucas (k + 2))
.= ((2 * (Lucas (k + 2))) + (Lucas (k + 1))) + (() + (Lucas (k + 1))) by Th12
.= ((Lucas (k + 2)) + ((2 * (Lucas (k + 1))) + (Lucas (k + 2)))) + ()
.= ((() + (Lucas (k + 1))) + (5 * (Fib (k + 2)))) + () by
.= (5 * (Fib (k + 1))) + (5 * (Fib (k + 2))) by A2
.= 5 * ((Fib (k + 1)) + (Fib (k + (1 + 1))))
.= 5 * (Fib (((k + 1) + 1) + 1)) by PRE_FF:1
.= 5 * (Fib ((k + 2) + 1)) ;
hence S1[k + 2] ; :: thesis: verum
end;
(0 + 1) + 1 = 2 ;
then Fib 2 = 1 by PRE_FF:1;
then A4: S1 by Th11, Th14;
A5: S1[ 0 ] by ;
thus for k being Nat holds S1[k] from FIB_NUM:sch 1(A5, A4, A1); :: thesis: verum