let f be FinSequence of NAT ; :: thesis: for s being Element of NAT st len f >= 3 holds
( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) & Sum (Prefix ((<*s*> ^ f),3)) = (s + (f /. 1)) + (f /. 2) & Sum (Prefix ((<*s*> ^ f),4)) = ((s + (f /. 1)) + (f /. 2)) + (f /. 3) )

let s be Element of NAT ; :: thesis: ( len f >= 3 implies ( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) & Sum (Prefix ((<*s*> ^ f),3)) = (s + (f /. 1)) + (f /. 2) & Sum (Prefix ((<*s*> ^ f),4)) = ((s + (f /. 1)) + (f /. 2)) + (f /. 3) ) )
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;
reconsider x3 = f /. 2 as Element of INT by INT_1:def 2;
reconsider x4 = f /. 3 as Element of INT by INT_1:def 2;
assume A1: len f >= 3 ; :: thesis: ( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) & Sum (Prefix ((<*s*> ^ f),3)) = (s + (f /. 1)) + (f /. 2) & Sum (Prefix ((<*s*> ^ f),4)) = ((s + (f /. 1)) + (f /. 2)) + (f /. 3) )
then consider n being Nat such that
A2: len f = 3 + n by NAT_1:10;
reconsider n = n as Element of NAT by ORDINAL1:def 12;
A3: len <*s*> = 1 by FINSEQ_1:39;
then A4: len (<*s*> ^ f) = 1 + (len f) by FINSEQ_1:22
.= 4 + n by A2 ;
then consider p4, q4 being FinSequence of NAT such that
A5: len p4 = 4 and
len q4 = n and
A6: <*s*> ^ f = p4 ^ q4 by FINSEQ_2:23;
f /. 3 = f . 3 by
.= (<*s*> ^ f) . (1 + 3) by ;
then A7: p4 . 4 = f /. 3 by ;
Seg 4 = dom p4 by ;
then A8: p4 = Prefix ((<*s*> ^ f),4) by ;
A9: 1 <= len f by ;
hence ( Sum (Prefix ((<*s*> ^ f),1)) = s & Sum (Prefix ((<*s*> ^ f),2)) = s + (f /. 1) ) by Th20; :: thesis: ( Sum (Prefix ((<*s*> ^ f),3)) = (s + (f /. 1)) + (f /. 2) & Sum (Prefix ((<*s*> ^ f),4)) = ((s + (f /. 1)) + (f /. 2)) + (f /. 3) )
len (<*s*> ^ f) = 3 + (1 + n) by A4;
then consider p3, q3 being FinSequence of NAT such that
A10: len p3 = 3 and
len q3 = 1 + n and
A11: <*s*> ^ f = p3 ^ q3 by FINSEQ_2:23;
A12: f /. 1 = f . 1 by
.= (<*s*> ^ f) . (1 + 1) by ;
then A13: p4 . 2 = f /. 1 by ;
A14: 2 <= len f by ;
then A15: f /. 2 = f . 2 by FINSEQ_4:15
.= (<*s*> ^ f) . (1 + 2) by ;
then A16: p4 . 3 = f /. 2 by ;
A17: p3 . 2 = f /. 1 by ;
Seg 3 = dom p3 by ;
then A18: p3 = Prefix ((<*s*> ^ f),3) by ;
A19: p3 . 3 = f /. 2 by ;
A20: (<*s*> ^ f) . 1 = s by FINSEQ_1:41;
then p3 . 1 = s by ;
then p3 = <*s,(f /. 1),(f /. 2)*> by ;
hence Sum (Prefix ((<*s*> ^ f),3)) = (s + (f /. 1)) + (f /. 2) by ; :: thesis: Sum (Prefix ((<*s*> ^ f),4)) = ((s + (f /. 1)) + (f /. 2)) + (f /. 3)
p4 . 1 = s by ;
then p4 = <*s,(f /. 1),(f /. 2),(f /. 3)*> by ;
hence Sum (Prefix ((<*s*> ^ f),4)) = ((x1 + x2) + x3) + x4 by
.= ((s + (f /. 1)) + (f /. 2)) + (f /. 3) ;
:: thesis: verum