reserve N for Cardinal;
reserve M for Aleph;
reserve X for non empty set;
reserve Y,Z,Z1,Z2,Y1,Y2,Y3,Y4 for Subset of X;
reserve S for Subset-Family of X;
reserve x for set;
reserve F,Uf for Filter of X;
reserve S for non empty Subset-Family of X;
reserve I for Ideal of X;
reserve S,S1 for Subset-Family of X;
reserve FS for non empty Subset of Filters(X);
reserve X for infinite set;
reserve Y,Y1,Y2,Z for Subset of X;
reserve F,Uf for Filter of X;
reserve x for Element of X;
reserve X for set;
reserve M for non limit_cardinal Aleph;
reserve F for Filter of M;
reserve N1,N2,N3 for Element of predecessor M;
reserve K1,K2 for Element of M;
reserve T for Inf_Matrix of predecessor M, M, bool M;
reserve M for Aleph;

theorem Th40:
  M is measurable implies M is strong_limit
proof
  assume M is measurable;
  then consider F being Filter of M such that
A1: F is_complete_with M and
A2: F is non principal being_ultrafilter;
  assume not M is strong_limit;
  then consider N being Cardinal such that
A3: N in M and
A4: not exp(2,N) in M;
A5: M c= exp(2,N) by A4,CARD_1:4;
  then M c= card Funcs(N,2) by CARD_2:def 3;
  then consider Y being set such that
A6: Y c= Funcs(N,2) and
A7: card Y = M by Th36;
  N is infinite
  proof
    M c= exp(2,card N) by A5;
    then
A8: M c= card bool N by CARD_2:31;
    assume N is finite;
    hence contradiction by A8;
  end;
  then reconsider N1 = N as Aleph;
  Y,M are_equipotent by A7,CARD_1:def 2;
  then consider f being Function such that
A9: f is one-to-one and
A10: dom f = M and
A11: rng f = Y by WELLORD2:def 4;
  defpred P[set,set] means f"{h where h is Function of N1,{{},1}: h in Y & h.
  $1 = $2} in F;
A12: for A being Element of N1 ex i being Element of {{},1} st P[A,i]
  proof
    let A be Element of N1;
    set Y1 = {h where h is Function of N1,{{},1}: h in Y & h.A = 1};
    reconsider Inv1 = f"Y1 as Subset of M by A10,RELAT_1:132;
    set Y0 = {h where h is Function of N1,{{},1}: h in Y & h.A = {}};
    reconsider Inv0 = f"Y0 as Subset of M by A10,RELAT_1:132;
A13: for g1 being Function of N1,{{},1} st g1 in Y holds g1 in Y1 or g1 in Y0
    proof
      let g1 be Function of N1,{{},1} such that
A14:  g1 in Y;
      per cases;
      suppose
        g1 in Y0;
        hence thesis;
      end;
      suppose
        not g1 in Y0;
        then not g1.A = {} by A14;
        then g1.A = 1 by TARSKI:def 2;
        hence thesis by A14;
      end;
    end;
    Y \ Y0 = Y1
    proof
      thus Y \ Y0 c= Y1
      proof
        let X be object such that
A15:    X in Y \ Y0;
        X in Y by A15;
        then consider g1 being Function such that
A16:    g1=X and
A17:    dom g1=N1 & rng g1 c= {{},1} by A6,CARD_1:50,FUNCT_2:def 2;
        reconsider g2=g1 as Function of N1,{{},1} by A17,FUNCT_2:def 1
,RELSET_1:4;
        not X in Y0 by A15,XBOOLE_0:def 5;
        then g2 in Y1 by A13,A15,A16;
        hence thesis by A16;
      end;
      let X be object;
      assume X in Y1;
      then consider h being Function of N1,{{},1} such that
A18:  X=h & h in Y and
A19:  h.A = 1;
      not h in Y0
      proof
        assume h in Y0;
        then
        ex h1 being Function of N1,{{},1} st h1=h & h1 in Y & h1.A = {};
        hence contradiction by A19;
      end;
      hence thesis by A18,XBOOLE_0:def 5;
    end;
    then
A20: Inv1 = (f"(rng f)) \ Inv0 by A11,FUNCT_1:69
      .= M \ Inv0 by A10,RELAT_1:134;
    per cases by A2;
    suppose
A21:  Inv0 in F;
      reconsider Z = {} as Element of { {},1} by TARSKI:def 2;
      take Z;
      thus thesis by A21;
    end;
    suppose
A22:  M \ Inv0 in F;
      reconsider O=1 as Element of {{},1} by TARSKI:def 2;
      take O;
      thus thesis by A20,A22;
    end;
  end;
  consider g being Function of N1,{{},1} such that
A23: for A being Element of N1 holds P[A,g.A] from FUNCT_2:sch 3(A12);
  deffunc T(Element of N1) = f"{h where h is Function of N1,{{},1}: h in Y & h
  .$1 = g.$1};
  reconsider f1=f as Function of M,Y by A10,A11,FUNCT_2:1;
  set MEET = meet {T(A) where A is Element of N1: A in N1};
A24: rng (f|MEET) = f.:MEET & f|MEET is one-to-one by A9,FUNCT_1:52,RELAT_1:115
;
  card {T(A) where A is Element of N1 : A in N1} c= card N1 from TREES_2:
  sch 2;
  then card {T(A) where A is Element of N1 : A in N1} c= N1;
  then
A25: card {T(A) where A is Element of N1: A in N1} in M by A3,ORDINAL1:12;
  set B = the Element of N1;
A26: {T(A) where A is Element of N1: A in N1} c= F
  proof
    let X be object;
    assume X in {T(A) where A is Element of N1: A in N1};
    then ex A being Element of N1 st X=T(A) & A in N1;
    hence thesis by A23;
  end;
  T(B) in {T(A) where A is Element of N1: A in N1};
  then
A27: meet {T(A) where A is Element of N1: A in N1} in F by A1,A25,A26;
A28: Y is infinite by A7;
A29: for X being set holds X in meet {T(A) where A is Element of N1: A in N1
  } implies f.X = g
  proof
    let X be set;
    assume
A30: X in meet {T(A) where A is Element of N1: A in N1};
    then reconsider X1=X as Element of M by A27;
    f1.X1 in Y by A28,FUNCT_2:5;
    then consider h1 being Function such that
A31: f1.X1=h1 and
A32: dom h1 = N and
    rng h1 c= 2 by A6,FUNCT_2:def 2;
A33: for Z being object st Z in N1 holds h1.Z = g.Z
    proof
      let Z be object;
      assume Z in N1;
      then reconsider Z1=Z as Element of N1;
      T(Z1) in {T(A) where A is Element of N1: A in N1};
      then X1 in T(Z1) by A30,SETFAM_1:def 1;
      then
      f1.X1 in {h where h is Function of N1,{{},1}: h in Y & h.Z1 = g.Z1}
      by FUNCT_1:def 7;
      then ex h being Function of N1,{{} ,1} st f.X1=h & h in Y & h.Z1 = g.Z1;
      hence thesis by A31;
    end;
    dom g = N1 by FUNCT_2:def 1;
    hence thesis by A31,A32,A33,FUNCT_1:2;
  end;
A34: f.:MEET c= {g}
  proof
    let X be object;
    assume X in f.:MEET;
    then ex x being object st x in dom f & x in MEET & X = f.x
      by FUNCT_1:def 6;
    then X = g by A29;
    hence thesis by TARSKI:def 1;
  end;
  MEET = dom f /\ MEET by A10,A27,XBOOLE_1:28;
  then dom (f|MEET) = MEET by RELAT_1:61;
  then
A35: card MEET c= card {g} by A34,A24,CARD_1:10;
  reconsider MEET as Subset of M by A27;
  F is_complete_with card M by A1;
  then F is uniform by A2,Th23;
  then card MEET = card M by A27;
  hence contradiction by A35;
end;
