let X be set ; :: thesis: for n being Element of NAT holds iter (id X),n = id X
let n be Element of NAT ; :: thesis: iter (id X),n = id X
defpred S₁[ Element of NAT ] means iter (id X),$1 = id X;
A1: for k being Element of NAT st S₁[k] holds
S₁[k + 1]

proof

let k be Element of NAT ; :: thesis: ( S₁[k] implies S₁[k + 1] )
assume A2: S₁[k] ; :: thesis: S₁[k + 1]
thus iter (id X),(k + 1) = (iter (id X),k) * (id X) by Th71
.= id X by A2, FUNCT_2:23 ; :: thesis: verum

end;

( dom (id X) = X & rng (id X) = X ) by RELAT_1:71;
then id ((dom (id X)) \/ (rng (id X))) = id X ;
then A3: S₁[ 0 ] by Th70;
for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A3, A1);
hence iter (id X),n = id X ; :: thesis: verum