let f be Function; :: thesis: for n being Element of NAT holds
( dom (iter (f,n)) c= (dom f) \/ (rng f) & rng (iter (f,n)) c= (dom f) \/ (rng f) )
let n be Element of NAT ; :: thesis: ( dom (iter (f,n)) c= (dom f) \/ (rng f) & rng (iter (f,n)) c= (dom f) \/ (rng f) )
defpred S₁[ Element of NAT ] means ( dom (iter (f,$1)) c= (dom f) \/ (rng f) & rng (iter (f,$1)) c= (dom f) \/ (rng f) );
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] )
iter (f,(k + 1)) = f * (iter (f,k)) by Th73;
then A2: dom (iter (f,(k + 1))) c= dom (iter (f,k)) by RELAT_1:44;
iter (f,(k + 1)) = (iter (f,k)) * f by Th71;
then A3: rng (iter (f,(k + 1))) c= rng (iter (f,k)) by RELAT_1:45;
assume ( dom (iter (f,k)) c= (dom f) \/ (rng f) & rng (iter (f,k)) c= (dom f) \/ (rng f) ) ; :: thesis: S₁[k + 1]
hence S₁[k + 1] by A2, A3, XBOOLE_1:1; :: thesis: verum

end;

iter (f,0) = id ((dom f) \/ (rng f)) by Th70;
then A4: S₁[ 0 ] by RELAT_1:71;
for k being Element of NAT holds S₁[k] from NAT_1:sch 1(A4, A1);
hence ( dom (iter (f,n)) c= (dom f) \/ (rng f) & rng (iter (f,n)) c= (dom f) \/ (rng f) ) ; :: thesis: verum