reserve n,m,k for Element of NAT,
  x,X for set,
  A1 for SetSequence of X,
  Si for SigmaField of X,
  XSeq for SetSequence of Si;
reserve Omega for non empty set,
  Sigma for SigmaField of Omega,
  ASeq for SetSequence of Sigma,
  P for Probability of Sigma;

theorem Th34:
  for D being non empty Subset-Family of Omega holds (for A being
set holds (A in D iff ex B being set st B in Sigma & ex C being thin of P st A
  = B \/ C)) implies D is SigmaField of Omega
proof
  let D be non empty Subset-Family of Omega;
  assume
A1: for A being set holds A in D iff ex B being set st B in Sigma & ex C
  being thin of P st A = B \/ C;
A2: for A1 being SetSequence of Omega st rng A1 c= D holds Union A1 in D
  proof
    let A1 be SetSequence of Omega;
    reconsider F = A1 as sequence of bool Omega;
A3: dom F = NAT by FUNCT_2:def 1;
    assume
A4: rng A1 c= D;
A5: for n holds ex B being set st B in Sigma & ex C being thin of P st F.
    n = B \/ C
    proof
      let n;
      F.n in rng F by NAT_1:51;
      hence thesis by A1,A4;
    end;
    for n holds F.n in COM(Sigma,P)
    proof
      let n;
      ex B being set st (B in Sigma & ex C being thin of P st F.n = B \/
      C) by A5;
      hence thesis by Def5;
    end;
    then for n being object st n in NAT holds F.n in COM(Sigma,P);
    then reconsider F as sequence of COM(Sigma,P) by A3,FUNCT_2:3;
    consider BSeq being SetSequence of Sigma such that
A6: for n holds BSeq.n in ProbPart(F.n) by Th31;
    consider CSeq being SetSequence of Omega such that
A7: for n holds CSeq.n = F.n \ BSeq.n by Th32;
A8: for n being Element of NAT holds BSeq.n in Sigma & BSeq.n c= F.n & F.
    n \ BSeq.n is thin of P
    proof
      let n be Element of NAT;
      BSeq.n in ProbPart(F.n) by A6;
      hence thesis by Def7;
    end;
    for n holds CSeq.n is thin of P
    proof
      let n be Element of NAT;
      F.n \ BSeq.n is thin of P by A8;
      hence thesis by A7;
    end;
    then consider DSeq being SetSequence of Sigma such that
A9: for n holds CSeq.n c= DSeq.n & P.(DSeq.n) = 0 by Th33;
A10: Union A1 = union rng A1 by CARD_3:def 4;
    ex B being set st B in Sigma & ex C being thin of P st Union A1 = B \/ C
    proof
      set B = union rng BSeq;
      take B;
A11:  union rng BSeq c= union rng F
      proof
        let x be object;
        assume x in union rng BSeq;
        then consider Z being set such that
A12:    x in Z and
A13:    Z in rng BSeq by TARSKI:def 4;
        dom BSeq = NAT by FUNCT_2:def 1;
        then consider n being object such that
A14:    n in NAT and
A15:    Z = BSeq.n by A13,FUNCT_1:def 3;
        reconsider n as Element of NAT by A14;
        set P = F.n;
A16:    BSeq.n c= P by A8;
        ex P being set st P in rng F & x in P
        proof
          take P;
          thus thesis by A3,A12,A15,A16,FUNCT_1:def 3;
        end;
        hence thesis by TARSKI:def 4;
      end;
A17:  ex C being thin of P st Union A1 = B \/ C
      proof
        set C = Union A1 \ B;
        Union DSeq in Sigma by PROB_1:17;
        then
A18:    union rng DSeq in Sigma by CARD_3:def 4;
A19:    C c= union rng CSeq
        proof
          let x be object;
          assume
A20:      x in C;
          then x in union rng F by A10,XBOOLE_0:def 5;
          then consider Z being set such that
A21:      x in Z and
A22:      Z in rng F by TARSKI:def 4;
          consider n being object such that
A23:      n in NAT and
A24:      Z = F.n by A3,A22,FUNCT_1:def 3;
          reconsider n as Element of NAT by A23;
A25:      not x in union rng BSeq by A20,XBOOLE_0:def 5;
          not x in BSeq.n
          proof
            dom BSeq = NAT by FUNCT_2:def 1;
            then
A26:        BSeq.n in rng BSeq by FUNCT_1:def 3;
            assume x in BSeq.n;
            hence thesis by A25,A26,TARSKI:def 4;
          end;
          then
A27:      x in F.n \ BSeq.n by A21,A24,XBOOLE_0:def 5;
          ex Z being set st x in Z & Z in rng CSeq
          proof
            take CSeq.n;
            dom CSeq = NAT by FUNCT_2:def 1;
            hence thesis by A7,A27,FUNCT_1:def 3;
          end;
          hence thesis by TARSKI:def 4;
        end;
        for A being set holds A in rng DSeq implies P.A = 0
        proof
          let A be set;
A28:      dom DSeq = NAT by FUNCT_2:def 1;
          assume A in rng DSeq;
          then ex n being object st n in NAT & A = DSeq.n
           by A28,FUNCT_1:def 3;
          hence thesis by A9;
        end;
        then
A29:    P.(union rng DSeq) = 0 by Th8;
        union rng CSeq c= union rng DSeq
        proof
          let x be object;
          assume x in union rng CSeq;
          then consider Z being set such that
A30:      x in Z and
A31:      Z in rng CSeq by TARSKI:def 4;
          dom CSeq = NAT by FUNCT_2:def 1;
          then consider n being object such that
A32:      n in NAT and
A33:      Z = CSeq.n by A31,FUNCT_1:def 3;
          reconsider n as Element of NAT by A32;
          n in dom DSeq by A32,FUNCT_2:def 1;
          then
A34:      DSeq.n in rng DSeq by FUNCT_1:def 3;
          CSeq.n c= DSeq.n by A9;
          hence thesis by A30,A33,A34,TARSKI:def 4;
        end;
        then C c= union rng DSeq by A19;
        then
A35:    C is thin of P by A29,A18,Def4;
        Union A1 = C \/ union rng F /\ union rng BSeq by A10,XBOOLE_1:51
          .= B \/ C by A11,XBOOLE_1:28;
        then consider C being thin of P such that
A36:    Union A1 = B \/ C by A35;
        take C;
        thus thesis by A36;
      end;
      Union BSeq in Sigma by PROB_1:17;
      hence thesis by A17,CARD_3:def 4;
    end;
    hence thesis by A1;
  end;
  for A being Subset of Omega holds A in D implies A` in D
  proof
    let A be Subset of Omega;
    assume
A37: A in D;
    ex Q being set st Q in Sigma & ex W being thin of P st Omega \ A = Q \/ W
    proof
      consider B being set such that
A38:  B in Sigma and
A39:  ex C being thin of P st A = B \/ C by A1,A37;
      consider C being thin of P such that
A40:  A = B \/ C by A39;
      reconsider B as Subset of Omega by A38;
      set H = Omega \ B;
      consider G being set such that
A41:  G in Sigma and
A42:  C c= G and
A43:  P.G = 0 by Def4;
      set Q = H \ G;
A44:  Omega \ A = H \ C by A40,XBOOLE_1:41;
A45:  ex W being thin of P st Omega \ A = Q \/ W
      proof
        set W = H /\ (G \ C);
        W c= H by XBOOLE_1:17;
        then reconsider W as Subset of Omega by XBOOLE_1:1;
        reconsider W as thin of P by A41,A43,Def4;
        take W;
        thus thesis by A42,A44,Lm1;
      end;
      take Q;
      B` in Sigma by A38,PROB_1:def 1;
      hence thesis by A41,A45,PROB_1:6;
    end;
    hence thesis by A1;
  end;
  hence thesis by A2,Th4;
end;
