let seq be Complex_Sequence; :: thesis: ( ( for n being Element of NAT holds seq . n = 0c ) implies for m being Element of NAT holds (Partial_Sums seq) . m = 0c )
assume A1: for n being Element of NAT holds seq . n = 0c ; :: thesis: for m being Element of NAT holds (Partial_Sums seq) . m = 0c
let m be Element of NAT ; :: thesis: (Partial_Sums seq) . m = 0c
defpred S₁[ Element of NAT ] means seq . $1 = (Partial_Sums seq) . $1;
A2: S₁[ 0 ] by Def7;
A3: 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 A4: S₁[k] ; :: thesis: S₁[k + 1]
thus seq . (k + 1) = 0c + (seq . (k + 1))
.= ((Partial_Sums seq) . k) + (seq . (k + 1)) by A1, A4
.= (Partial_Sums seq) . (k + 1) by Def7 ; :: thesis: verum

end;

for n being Element of NAT holds S₁[n] from NAT_1:sch 1(A2, A3);
then seq = Partial_Sums seq by FUNCT_2:113;
hence (Partial_Sums seq) . m = 0c by A1; :: thesis: verum