reserve k, l, m, n, i, j for Nat,
  K, N for non empty Subset of NAT,
  Ke, Ne, Me for Subset of NAT,
  X,Y for set;

theorem Th8:
  n is Subset of NAT
proof
  now
    let x be object;
    assume x in {l where l is Nat: l<n};
    then ex l be Nat st x=l & l<n;
    hence x in NAT by ORDINAL1:def 12;
  end;
  then {l where l is Nat:l<n} c= NAT;
  hence thesis by AXIOMS:4;
end;
