let n be Nat; :: thesis: for R being Relation st n <> 0 holds
( dom (iter (R,n)) c= dom R & rng (iter (R,n)) c= rng R )
let R be Relation; :: thesis: ( n <> 0 implies ( dom (iter (R,n)) c= dom R & rng (iter (R,n)) c= rng R ) )
defpred S₁[ Nat] means ( dom (iter (R,($1 + 1))) c= dom R & rng (iter (R,($1 + 1))) c= rng R );
assume n <> 0 ; :: thesis: ( dom (iter (R,n)) c= dom R & rng (iter (R,n)) c= rng R )
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 (R,(k + 1))) c= dom R and
rng (iter (R,(k + 1))) c= rng R ; :: thesis: S₁[k + 1]
( iter (R,((k + 1) + 1)) = R * (iter (R,(k + 1))) & iter (R,((k + 1) + 1)) = (iter (R,(k + 1))) * R ) by Th68, Th70;
hence S₁[k + 1] by RELAT_1:25, RELAT_1:26; :: thesis: verum

end;

A3: S₁[ 0 ] by Th69;
for k being Nat holds S₁[k] from NAT_1:sch 2(A3, A2);
hence ( dom (iter (R,n)) c= dom R & rng (iter (R,n)) c= rng R ) by A1; :: thesis: verum