let f1, f2 be Function of NAT,NAT; :: thesis: ( ( for n being natural number holds f1 . n = n |^ k ) & ( for n being natural number holds f2 . n = n |^ k ) implies f1 = f2 )
assume A2: for n being natural number holds f1 . n = n |^ k ; :: thesis: ( ex n being natural number st not f2 . n = n |^ k or f1 = f2 )
assume A3: for n being natural number holds f2 . n = n |^ k ; :: thesis: f1 = f2
for x being set st x in NAT holds
f1 . x = f2 . x

proof

let x be set ; :: thesis: ( x in NAT implies f1 . x = f2 . x )
assume x in NAT ; :: thesis: f1 . x = f2 . x
then reconsider n = x as natural number ;
f1 . n = n |^ k by A2
.= f2 . n by A3 ;
hence f1 . x = f2 . x ; :: thesis: verum

end;

hence f1 = f2 by FUNCT_2:12; :: thesis: verum