let f be FinSequence of NAT ; :: thesis: for s being Element of NAT st len f >= 1 holds
( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) )

let s be Element of NAT ; :: thesis: ( len f >= 1 implies ( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) ) )
set g = <*s*>;
set h = <*s*> ^ f;
reconsider x1 = s as Element of INT by INT_1:def 2;
reconsider x2 = f /. 1 as Element of INT by INT_1:def 2;
assume A1: len f >= 1 ; :: thesis: ( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) )
then consider n being Nat such that
A2: len f = 1 + n by NAT_1:10;
A3: len <*s*> = 1 by FINSEQ_1:39;
then Seg 1 = dom <*s*> by FINSEQ_1:def 3;
hence Sum (Prefix ((<*s*> ^ f),1)) = Sum <*x1*> by FINSEQ_1:21
.= s by FINSOP_1:11 ;
:: thesis: Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1)
len (<*s*> ^ f) = 1 + (len f) by
.= 2 + n by A2 ;
then consider p2, q2 being FinSequence of NAT such that
A4: len p2 = 2 and
len q2 = n and
A5: <*s*> ^ f = p2 ^ q2 by FINSEQ_2:23;
f /. 1 = f . 1 by
.= (<*s*> ^ f) . (1 + 1) by ;
then A6: p2 . 2 = f /. 1 by ;
Seg 2 = dom p2 by ;
then A7: p2 = Prefix ((<*s*> ^ f),2) by ;
(<*s*> ^ f) . 1 = s by FINSEQ_1:41;
then p2 . 1 = s by ;
hence Sum (Prefix ((<*s*> ^ f),2)) = Sum <*x1,x2*> by
.= s + (f /. 1) by RVSUM_1:77 ;
:: thesis: verum