reserve X for set;

theorem Th5:
  for S being non empty Subset-Family of X, N,F being sequence of S
  holds F.0 = N.0 & (for n being Nat holds F.(n+1) = N.(n+1) \/
F.n) implies for r being set for n being Nat holds (r in F.n iff ex
  k being Nat st k <= n & r in N.k)
proof
  let S be non empty Subset-Family of X, N,F be sequence of S;
  assume that
A1: F.0 = N.0 and
A2: for n being Nat holds F.(n+1) = N.(n+1) \/ F.n;
  let r be set, n be Nat;
  defpred P[Nat] means
   (ex k being Nat st k <= $1 & r in N.k) implies r in F.$1;
A3: for n being Nat st P[n] holds P[n+1]
  proof
    let n be Nat;
    assume
A4: (ex k being Nat st k <= n & r in N.k) implies r in F.n;
A5: F.(n+1) = N.(n+1) \/ F.n by A2;
    given k being Nat such that
A6: k <= n+1 and
A7: r in N.k;
    k <= n or k = n + 1 by A6,NAT_1:8;
    hence thesis by A4,A7,A5,XBOOLE_0:def 3;
  end;
  thus r in F.n implies ex k being Nat st k <= n & r in N.k
  proof
    defpred P[Nat] means r in F.$1;
    assume
A8: r in F.n;
    then
A9: ex n being Nat st P[n];
    ex s being Nat st P[s] & for k being Nat st P[k] holds s <= k from
    NAT_1:sch 5(A9);
    then consider s being Nat such that
A10: r in F.s and
A11: for k being Nat st r in F.k holds s <= k;
A12: (ex k being Nat st s = k + 1) implies ex k being Element of NAT st k
    <= n & r in N.k
    proof
      given k being Nat such that
A13:  s = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
A14:  not r in F.k
      proof
        assume r in F.k;
        then s <= k by A11;
        hence thesis by A13,NAT_1:13;
      end;
      take s;
A15:  s in NAT by ORDINAL1:def 12;
      F.s = N.s \/ F.k by A2,A13;
      hence thesis by A8,A10,A11,A14,A15,XBOOLE_0:def 3;
    end;
    s=0 implies ex k being Element of NAT st k <= n & r in N.k
            by A1,A10,NAT_1:2;
    hence thesis by A12,NAT_1:6;
  end;
A16: P[0] by A1,NAT_1:3;
  for n being Nat holds P[n] from NAT_1:sch 2(A16,A3);
  hence thesis;
end;
