let f1, f2 be sequence of ({a} *); :: thesis: ( ( for n being Nat holds f1 . n = n |-> a ) & ( for n being Nat holds f2 . n = n |-> a ) implies f1 = f2 )
assume that
A3: for n being Nat holds f1 . n = n |-> a and
A4: for n being Nat holds f2 . n = n |-> a ; :: thesis: f1 = f2

now :: thesis: for k being Element of NAT holds f1 . k = f2 . k

let k be Element of NAT ; :: thesis: f1 . k = f2 . k
thus f1 . k = k |-> a by A3
.= f2 . k by A4 ; :: thesis: verum

end;

hence f1 = f2 ; :: thesis: verum