let S1, S2 be sequence of X; :: thesis: ( ( for n being Nat holds S1 . n = (1r / (n !)) * (z #N n) ) & ( for n being Nat holds S2 . n = (1r / (n !)) * (z #N n) ) implies S1 = S2 )

assume that

A2: for n being Nat holds S1 . n = (1r / (n !)) * (z #N n) and

A3: for n being Nat holds S2 . n = (1r / (n !)) * (z #N n) ; :: thesis: S1 = S2

for n being Element of NAT holds S1 . n = S2 . n

assume that

A2: for n being Nat holds S1 . n = (1r / (n !)) * (z #N n) and

A3: for n being Nat holds S2 . n = (1r / (n !)) * (z #N n) ; :: thesis: S1 = S2

for n being Element of NAT holds S1 . n = S2 . n

proof

hence
S1 = S2
by FUNCT_2:63; :: thesis: verum
let n be Element of NAT ; :: thesis: S1 . n = S2 . n

S1 . n = (1r / (n !)) * (z #N n) by A2;

hence S1 . n = S2 . n by A3; :: thesis: verum

end;S1 . n = (1r / (n !)) * (z #N n) by A2;

hence S1 . n = S2 . n by A3; :: thesis: verum