reserve X for set;

theorem Th20:
  for S being SigmaField of X, N,F being sequence of S holds (
F.0 = N.0 & for n being Nat holds F.(n+1) = N.(n+1) \ N.n & N.n c= N
  .(n+1) ) implies union rng F = union rng N
proof
  let S be SigmaField 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) \ N.n & N.n c= N. (n+1);
  thus union rng F c= union rng N
  proof
    let x be object;
    assume x in union rng F;
    then consider Y being set such that
A3: x in Y and
A4: Y in rng F by TARSKI:def 4;
    consider n being object such that
A5: n in dom F and
A6: Y = F.n by A4,FUNCT_1:def 3;
    reconsider n as Element of NAT by A5;
A7: (ex k being Nat st n = k + 1) implies ex Z being set st x in Z & Z in rng N
    proof
      given k being Nat such that
A8:   n = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      Y=N.(k+1) \ N.k by A2,A6,A8;
      then x in N.(k+1) by A3,XBOOLE_0:def 5;
      hence thesis by FUNCT_2:4;
    end;
    n=0 implies ex Z being set st x in Z & Z in rng N by A1,A3,A6,FUNCT_2:4;
    hence thesis by A7,NAT_1:6,TARSKI:def 4;
  end;
  let x be object;
  assume x in union rng N;
  then consider Y being set such that
A9: x in Y and
A10: Y in rng N by TARSKI:def 4;
A11: ex n being object st n in dom N & Y = N.n by A10,FUNCT_1:def 3;
  ex Z being set st x in Z & Z in rng F
  proof
    ex s being Element of NAT st x in F.s
    proof
      defpred P[Nat] means x in N.$1;
A12:  ex k being Nat st P[k] by A9,A11;
      ex k being Nat st P[k] & for r being Nat st P[r] holds k <= r from
      NAT_1:sch 5(A12);
      then consider k being Nat such that
A13:  x in N.k and
A14:  for r being Nat st x in N.r holds k <= r;
A15:  (ex l being Nat st k = l + 1) implies ex s being Element of NAT st
      x in F.s
      proof
        given l being Nat such that
A16:    k = l + 1;
        take k;
        reconsider l as Element of NAT by ORDINAL1:def 12;
A17:    not x in N.l
        proof
          assume x in N.l;
          then l + 1 <= l by A14,A16;
          hence thesis by NAT_1:13;
        end;
        F.(l+1) = N.(l+1) \ N.l by A2;
        hence thesis by A13,A16,A17,XBOOLE_0:def 5;
      end;
      k=0 implies ex s being Element of NAT st x in F.s by A1,A13;
      hence thesis by A15,NAT_1:6;
    end;
    hence thesis by FUNCT_2:4;
  end;
  hence thesis by TARSKI:def 4;
end;
