let f1, f2 be Function of NAT,({a} *); :: thesis: ( ( for n being Element of NAT holds f1 . n = n |-> a ) & ( for n being Element of NAT holds f2 . n = n |-> a ) implies f1 = f2 )
assume that
A2: for n being Element of NAT holds f1 . n = n |-> a and
A3: for n being Element of NAT holds f2 . n = n |-> a ; :: thesis: f1 = f2

now

let k be Element of NAT ; :: thesis: f1 . k = f2 . k
thus f1 . k = k |-> a by A2
.= f2 . k by A3 ; :: thesis: verum

end;

hence f1 = f2 by FUNCT_2:63; :: thesis: verum