reserve
  A,B,X for set,
  S for SigmaField of X;

theorem Th2:
  for S being non empty Subset-Family of X for F, G being sequence of S
  st (G.0 = F.0 & for n being Nat holds G.(n+1) = F.(n+1) \/
G.n) holds for H being sequence of S st (H.0 = F.0 & for n being Nat
   holds H.(n+1) = F.(n+1) \ G.n) holds union rng F = union rng H
proof
  let S be non empty Subset-Family of X;
  let F, G be sequence of S;
  assume that
A1: G.0 = F.0 and
A2: for n being Nat holds G.(n+1) = F.(n+1) \/ G.n;
  let H be sequence of S;
  assume that
A3: H.0 = F.0 and
A4: for n being Nat holds H.(n+1) = F.(n+1) \ G.n;
  thus union rng F c= union rng H
  proof
    let r be object;
    assume r in union rng F;
    then consider E being set such that
A5: r in E and
A6: E in rng F by TARSKI:def 4;
A7: ex s being object st s in dom F & E = F.s by A6,FUNCT_1:def 3;
    ex s1 being Element of NAT st r in H.s1
    proof
      defpred P[Nat] means r in F.$1;
A8:   ex k being Nat st P[k] by A5,A7;
      ex k being Nat st P[k] & for n being Nat st P[n] holds k <= n from
      NAT_1:sch 5(A8);
      then consider k being Nat such that
A9:   r in F.k and
A10:  for n being Nat st r in F.n holds k <= n;
A11:  (ex l being Nat st k = l + 1) implies ex s1 being Element of NAT st
      r in H.s1
      proof
        given l being Nat such that
A12:    k = l + 1;
        take k;
        reconsider l as Element of NAT by ORDINAL1:def 12;
A13:    not r in G.l
        proof
          assume r in G.l;
          then consider i being Nat such that
A14:      i <= l and
A15:      r in F.i by A1,A2,MEASURE2:5;
          k <= i by A10,A15;
          hence thesis by A12,A14,NAT_1:13;
        end;
        H.(l+1) = F.(l+1) \ G.l by A4;
        hence thesis by A9,A12,A13,XBOOLE_0:def 5;
      end;
      k=0 implies ex s1 being Element of NAT st r in H.s1 by A3,A9;
      hence thesis by A11,NAT_1:6;
    end;
    then consider s1 being Element of NAT such that
A16: r in H.s1;
    H.s1 in rng H by FUNCT_2:4;
    hence thesis by A16,TARSKI:def 4;
  end;
A17: for n being Element of NAT holds H.n c= F.n
  proof
    let n be Element of NAT;
A18: (ex k being Nat st n = k + 1) implies H.n c= F.n
    proof
      given k being Nat such that
A19:  n = k + 1;
      reconsider k as Element of NAT by ORDINAL1:def 12;
      H.n = F.n \ G.k by A4,A19;
      hence thesis;
    end;
    n=0 implies H.n c= F.n by A3;
    hence thesis by A18,NAT_1:6;
  end;
  union rng H c= union rng F
  proof
    let r be object;
    assume r in union rng H;
    then consider E being set such that
A20: r in E and
A21: E in rng H by TARSKI:def 4;
    consider s being object such that
A22: s in dom H and
A23: E = H.s by A21,FUNCT_1:def 3;
    reconsider s as Element of NAT by A22;
A24: F.s in rng F by FUNCT_2:4;
    E c= F.s by A17,A23;
    hence thesis by A20,A24,TARSKI:def 4;
  end;
  hence thesis;
end;
