reserve A,B for limit_ordinal infinite Ordinal;
reserve B1,B2,B3,B5,B6,D, C for Ordinal;
reserve X for set;
reserve X for Subset of A;
reserve M for non countable Aleph;
reserve X for Subset of M;
reserve N,N1 for cardinal infinite Element of M;

theorem
  omega in cf M implies for S being non empty Subset-Family of M st (
  card S in cf M & for X being Element of S holds X is closed unbounded ) holds
  meet S is closed unbounded
proof
  assume
A1: omega in cf M;
  let S be non empty Subset-Family of M such that
A2: card S in cf M and
A3: for X being Element of S holds X is closed unbounded;
  thus meet S is closed by A3,Th21;
  for B1 st B1 in M ex C st C in meet S & B1 c= C
  proof
    let B1;
    assume B1 in M;
    then reconsider B11=B1 as Element of M;
    deffunc Ch(Element of M) = { LBound($1,X) where X is Element of S : X in S
    } /\ [#]M;
    defpred P[set,Element of M,set] means $3 = sup Ch($2) & $2 in $3;
A4: for B being Element of M holds Ch(B) = { LBound(B,X) where X is
    Element of S : X in S }
    proof
      let B be Element of M;
      set ChB= { LBound(B,X) where X is Element of S : X in S };
      ChB c= M
      proof
        let x be object;
        assume x in { LBound(B,X) where X is Element of S : X in S };
        then consider X being Element of S such that
A5:     LBound(B,X)=x and
        X in S;
        X is unbounded by A3;
        then X is non empty by Th7;
        then LBound(B,X) in X;
        hence thesis by A5;
      end;
      hence thesis by XBOOLE_1:28;
    end;
A6: for B being Element of M holds sup Ch(B) in M & B in sup Ch(B)
    proof
      let B be Element of M;
      deffunc f(Subset of M) = LBound(B,$1);
A7:   Ch(B) c= sup Ch(B) by Th2;
      card { f(X) where X is Element of S: X in S } c= card S from
      TREES_2:sch 2;
      then card Ch(B) c= card S by A4;
      then card Ch(B) in cf M by A2,ORDINAL1:12;
      hence sup Ch(B) in M by CARD_5:26;
      set X = the Element of S;
A8:   X is unbounded by A3;
      then X is non empty by Th7;
      then LBound(B,X) in X;
      then reconsider LB=LBound(B,X) as Element of M;
      LBound(B,X) in { LBound(B,Y) where Y is Element of S: Y in S };
      then
A9:   LB in Ch(B) by A4;
      B in LB by A8,Th9;
      hence thesis by A9,A7,ORDINAL1:10;
    end;
A10: for n being Nat for x being Element of M ex y being
    Element of M st P[n,x,y]
    proof
      let n be Nat;
      let x be Element of M;
      reconsider y=sup Ch(x) as Element of M by A6;
      take y;
      thus thesis by A6;
    end;
    consider L being sequence of M such that
A11: L.0 = B11 and
A12: for n being Nat holds P[n,L.n,L.(n+1)] from RECDEF_1:
    sch 2(A10);
    reconsider L1=L as sequence of [#]M;
    take sup rng L;
A13: B1 in rng L by A11,FUNCT_2:4;
    reconsider RNG = rng L as Subset of M by RELAT_1:def 19;
A14: for C1 being Ordinal st C1 in RNG ex C2 being Ordinal st C2 in RNG &
    C1 in C2
    proof
      let C1 be Ordinal;
      assume C1 in RNG;
      then consider y1 being object such that
A15:  y1 in dom L and
A16:  C1 = L.y1 by FUNCT_1:def 3;
      reconsider y=y1 as Element of NAT by A15,FUNCT_2:def 1;
      reconsider L1=L.(y+1) as Ordinal;
      take L1;
      thus L1 in RNG by FUNCT_2:4;
      thus thesis by A12,A16;
    end;
    sup rng L1 in M by A1,Th22;
    then reconsider SUPL = sup RNG as limit_ordinal infinite Element of M by
A13,A14,Th3;
    for X1 being set st X1 in S holds SUPL in X1
    proof
      let X1 be set;
      assume X1 in S;
      then reconsider X=X1 as Element of S;
A17:  X is closed unbounded by A3;
      then
A18:  X is non empty by Th7;
      sup (X /\ SUPL) = SUPL
      proof
        sup (X /\ SUPL) c= sup SUPL by ORDINAL2:22,XBOOLE_1:17;
        then
A19:    sup (X /\ SUPL) c= SUPL by ORDINAL2:18;
        assume sup (X /\ SUPL) <> SUPL;
        then sup (X /\ SUPL) c< SUPL by A19,XBOOLE_0:def 8;
        then consider B3 being Ordinal such that
A20:    B3 in rng L and
A21:    sup (X /\ SUPL) c= B3 by ORDINAL1:11,ORDINAL2:21;
        consider y1 being object such that
A22:    y1 in dom L and
A23:    B3 = L.y1 by A20,FUNCT_1:def 3;
        reconsider y=y1 as Element of NAT by A22,FUNCT_2:def 1;
        LBound((L.y),X) in X by A18;
        then reconsider LBY=LBound((L.y),X) as Element of M;
        LBound((L.y),X) in { LBound((L.y),Y) where Y is Element of S: Y in S };
        then LBound((L.y),X) in Ch(L.y) by A4;
        then LBY in sup Ch(L.y) by ORDINAL2:19;
        then
A24:    LBound((L.y),X) in L.(y+1) by A12;
        L.(y+1) in rng L by FUNCT_2:4;
        then L.(y+1) in SUPL by ORDINAL2:19;
        then LBY in SUPL by A24,ORDINAL1:10;
        then LBound((L.y),X) in X /\ SUPL by A18,XBOOLE_0:def 4;
        then LBY in sup (X /\ SUPL) by ORDINAL2:19;
        then LBY in L.y by A21,A23;
        hence contradiction by A17,Th9;
      end;
      hence thesis by A17;
    end;
    hence sup rng L in meet S by SETFAM_1:def 1;
    B1 in sup rng L by A13,ORDINAL2:19;
    hence thesis by ORDINAL1:def 2;
  end;
  hence thesis by Th6;
end;
