let f be Function; :: thesis: for n being Element of NAT st n <> 0 holds
( dom (iter (f,n)) c= dom f & rng (iter (f,n)) c= rng f )
let n be Element of NAT ; :: thesis: ( n <> 0 implies ( dom (iter (f,n)) c= dom f & rng (iter (f,n)) c= rng f ) )
defpred S₁[ Nat] means ( dom (iter (f,($1 + 1))) c= dom f & rng (iter (f,($1 + 1))) c= rng f );
assume n <> 0 ; :: thesis: ( dom (iter (f,n)) c= dom f & rng (iter (f,n)) c= rng f )
then A1: ex k being Nat st n = k + 1 by NAT_1:6;
A2: for k being Nat st S₁[k] holds
S₁[k + 1]

proof

let k be Nat; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume that
dom (iter (f,(k + 1))) c= dom f and
rng (iter (f,(k + 1))) c= rng f ; :: thesis: S₁[k + 1]
( iter (f,((k + 1) + 1)) = f * (iter (f,(k + 1))) & iter (f,((k + 1) + 1)) = (iter (f,(k + 1))) * f ) by Th71, Th73;
hence S₁[k + 1] by RELAT_1:44, RELAT_1:45; :: thesis: verum

end;

A3: S₁[ 0 ] by Th72;
for k being Nat holds S₁[k] from NAT_1:sch 2(A3, A2);
hence ( dom (iter (f,n)) c= dom f & rng (iter (f,n)) c= rng f ) by A1; :: thesis: verum