let S1, S2 be sequence of X; :: thesis: ( ( for n being Nat holds S1 . n = (1 / (n !)) * (z #N n) ) & ( for n being Nat holds S2 . n = (1 / (n !)) * (z #N n) ) implies S1 = S2 )
assume that
A2: for n being Nat holds S1 . n = (1 / (n !)) * (z #N n) and
A3: for n being Nat holds S2 . n = (1 / (n !)) * (z #N n) ; :: thesis: S1 = S2
for n being Element of NAT holds S1 . n = S2 . n

proof

let n be Element of NAT ; :: thesis: S1 . n = S2 . n
S1 . n = (1 / (n !)) * (z #N n) by A2;
hence S1 . n = S2 . n by A3; :: thesis: verum

end;

hence S1 = S2 by FUNCT_2:63; :: thesis: verum