let rseq be Real_Sequence; :: thesis: ( ( for n being Element of NAT holds rseq . n = 0 ) implies for m being Element of NAT holds (Partial_Sums (abs rseq)) . m = 0 )
defpred S₁[ Element of NAT ] means (abs rseq) . $1 = (Partial_Sums (abs rseq)) . $1;
assume A1: for n being Element of NAT holds rseq . n = 0 ; :: thesis: for m being Element of NAT holds (Partial_Sums (abs rseq)) . m = 0
A2: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A3: S₁[k] ; :: thesis: S₁[k + 1]
thus (abs rseq) . (k + 1) = 0 + ((abs rseq) . (k + 1))
.= (abs 0 ) + ((abs rseq) . (k + 1)) by ABSVALUE:def 1
.= (abs (rseq . k)) + ((abs rseq) . (k + 1)) by A1
.= ((Partial_Sums (abs rseq)) . k) + ((abs rseq) . (k + 1)) by A3, SEQ_1:16
.= (Partial_Sums (abs rseq)) . (k + 1) by SERIES_1:def 1 ; :: thesis: verum

end;

let m be Element of NAT ; :: thesis: (Partial_Sums (abs rseq)) . m = 0
A4: S₁[ 0 ] by SERIES_1:def 1;
for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A4, A2);
hence (Partial_Sums (abs rseq)) . m = (abs rseq) . m
.= abs (rseq . m) by SEQ_1:16
.= abs 0 by A1
.= 0 by ABSVALUE:def 1 ;
:: thesis: verum