let f1, f2 be sequence of NAT; :: thesis: ( ( for n being Nat holds f1 . n = n |^ k ) & ( for n being Nat holds f2 . n = n |^ k ) implies f1 = f2 )

assume A2: for n being Nat holds f1 . n = n |^ k ; :: thesis: ( ex n being Nat st not f2 . n = n |^ k or f1 = f2 )

assume A3: for n being Nat holds f2 . n = n |^ k ; :: thesis: f1 = f2

for x being object st x in NAT holds

f1 . x = f2 . x

assume A2: for n being Nat holds f1 . n = n |^ k ; :: thesis: ( ex n being Nat st not f2 . n = n |^ k or f1 = f2 )

assume A3: for n being Nat holds f2 . n = n |^ k ; :: thesis: f1 = f2

for x being object st x in NAT holds

f1 . x = f2 . x

proof

hence
f1 = f2
by FUNCT_2:12; :: thesis: verum
let x be object ; :: thesis: ( x in NAT implies f1 . x = f2 . x )

assume x in NAT ; :: thesis: f1 . x = f2 . x

then reconsider n = x as Nat ;

f1 . n = n |^ k by A2

.= f2 . n by A3 ;

hence f1 . x = f2 . x ; :: thesis: verum

end;assume x in NAT ; :: thesis: f1 . x = f2 . x

then reconsider n = x as Nat ;

f1 . n = n |^ k by A2

.= f2 . n by A3 ;

hence f1 . x = f2 . x ; :: thesis: verum