defpred S1[ Element of NAT , set , set ] means ( ( $2 is not PartFunc of X,COMPLEX & $3 = F . $1 ) or ( $2 is PartFunc of X,COMPLEX & ( for F2 being PartFunc of X,COMPLEX st F2 = $2 holds
$3 = F2 + (F . ($1 + 1)) ) ) );
P1:
for n being Element of NAT
for x being set ex y being set st S1[n,x,y]
proof
let n be
Element of
NAT ;
:: thesis: for x being set ex y being set st S1[n,x,y]let x be
set ;
:: thesis: ex y being set st S1[n,x,y]
now assume
x is
PartFunc of
X,
COMPLEX
;
:: thesis: ex y being Element of K21(K22(X,COMPLEX )) st
( ( x is not PartFunc of X,COMPLEX & y = F . n ) or ( x is PartFunc of X,COMPLEX & ( for F2 being PartFunc of X,COMPLEX st F2 = x holds
y = F2 + (F . (n + 1)) ) ) )then reconsider G2 =
x as
PartFunc of
X,
COMPLEX ;
take y =
G2 + (F . (n + 1));
:: thesis: ( ( x is not PartFunc of X,COMPLEX & y = F . n ) or ( x is PartFunc of X,COMPLEX & ( for F2 being PartFunc of X,COMPLEX st F2 = x holds
y = F2 + (F . (n + 1)) ) ) )thus
( (
x is not
PartFunc of
X,
COMPLEX &
y = F . n ) or (
x is
PartFunc of
X,
COMPLEX & ( for
F2 being
PartFunc of
X,
COMPLEX st
F2 = x holds
y = F2 + (F . (n + 1)) ) ) )
;
:: thesis: verum end;
hence
ex
y being
set st
S1[
n,
x,
y]
;
:: thesis: verum
end;
consider IT being Function such that
P3:
( dom IT = NAT & IT . 0 = F . 0 & ( for n being Element of NAT holds S1[n,IT . n,IT . (n + 1)] ) )
from RECDEF_1:sch 1(P1);
then
rng IT c= PFuncs X,COMPLEX
by TARSKI:def 3;
then reconsider IT = IT as Functional_Sequence of X,COMPLEX by P3, RELSET_1:11, FUNCT_2:def 1;
take
IT
; :: thesis: ( IT . 0 = F . 0 & ( for n being Nat holds IT . (n + 1) = (IT . n) + (F . (n + 1)) ) )
for n being Nat holds IT . (n + 1) = (IT . n) + (F . (n + 1))
hence
( IT . 0 = F . 0 & ( for n being Nat holds IT . (n + 1) = (IT . n) + (F . (n + 1)) ) )
by P3; :: thesis: verum