set mc = addcomplex ;
consider f being FinSequence of COMPLEX such that
A1:
f = F
and
A2:
Sum F = addcomplex $$ f
by RVSUM_1:def 11;
set g = [#] (f,(the_unity_wrt addcomplex));
defpred S1[ Element of NAT ] means addcomplex $$ ((finSeg F),([#] (f,(the_unity_wrt addcomplex)))) is integer ;
A3:
for k being Element of NAT st S1[k] holds
S1[k + 1]
proof
let k be
Element of
NAT ;
( S1[k] implies S1[k + 1] )
A4:
([#] (f,(the_unity_wrt addcomplex))) . (k + 1) is
integer
assume
S1[
k]
;
S1[k + 1]
then reconsider a =
([#] (f,(the_unity_wrt addcomplex))) . (k + 1),
b =
addcomplex $$ (
(finSeg k),
([#] (f,(the_unity_wrt addcomplex)))) as
integer number by A4;
not
k + 1
in Seg k
by FINSEQ_3:9;
then addcomplex $$ (
((finSeg k) \/ {.(k + 1).}),
([#] (f,(the_unity_wrt addcomplex)))) =
addcomplex . (
(addcomplex $$ ((finSeg k),([#] (f,(the_unity_wrt addcomplex))))),
(([#] (f,(the_unity_wrt addcomplex))) . (k + 1)))
by SETWOP_2:4
.=
b + a
by BINOP_2:def 3
;
hence
S1[
k + 1]
by FINSEQ_1:11;
verum
end;
Seg 0 = {}. NAT
;
then A5:
S1[ 0 ]
by BINOP_2:1, SETWISEO:40;
A6:
for n being Element of NAT holds S1[n]
from NAT_1:sch 1(A5, A3);
consider n being Nat such that
A7:
dom f = Seg n
by FINSEQ_1:def 2;
A8:
addcomplex $$ f = addcomplex $$ ((findom f),([#] (f,(the_unity_wrt addcomplex))))
by SETWOP_2:def 2;
n in NAT
by ORDINAL1:def 13;
hence
Sum F is integer
by A2, A8, A7, A6; verum