let S1, S2 be sequence of X; :: thesis: ( ( for n being Nat holds S1 . n = (S . n) * a ) & ( for n being Nat holds S2 . n = (S . n) * a ) implies S1 = S2 )
assume that
A5: for n being Nat holds S1 . n = (S . n) * a and
A6: for n being Nat holds S2 . n = (S . n) * a ; :: 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 = (S . n) * a by A5;
hence S1 . n = S2 . n by A6; :: thesis: verum

end;

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