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;

theorem Th37:
  for M holds not ex F st F is uniform being_ultrafilter & F is_complete_with M
proof
  let M;
  given F such that
A1: F is uniform being_ultrafilter and
A2: F is_complete_with M;
  Frechet_Ideal(M) c= dual F
  proof
    let X be object such that
A3: X in Frechet_Ideal(M);
    reconsider X1=X as Subset of M by A3;
    assume not X in dual F;
    then X1 in F by A1,Th22;
    then
A4: card X1 = card M by A1;
    card X1 in card M by A3,Th19;
    hence contradiction by A4;
  end;
  then consider S being Subset-Family of M such that
A5: card S = M and
A6: for X1 being set st X1 in S holds not X1 in dual F and
A7: for X1,X2 being set st X1 in S & X2 in S & X1 <> X2 holds X1 misses
  X2 by A2,Th12,Th35;
  S is infinite by A5;
  then consider X1 being object such that
A8: X1 in S by XBOOLE_0:def 1;
  S \ {X1} <> {}
  proof
    assume S \ {X1} = {};
    then S c= {X1} by XBOOLE_1:37;
    hence contradiction by A5;
  end;
  then consider X2 being object such that
A9: X2 in S \ {X1} by XBOOLE_0:def 1;
A10: S qua set\{X1} is Subset of S;
  reconsider X1,X2 as set by TARSKI:1;
  not X2 in {X1} by A9,XBOOLE_0:def 5;
  then X2 <> X1 by TARSKI:def 1;
  then X1 misses X2 by A7,A8,A9,A10;
  then
A11: X1 /\ X2 = {};
  reconsider X1,X2 as Subset of M by A8,A9;
A12: for X1 being set st X1 in S holds X1 in F by A1,A6,Th22;
  then
A13: X1 in F by A8;
  X2 in F by A12,A9,A10;
  then {} in F by A11,A13,Def1;
  hence contradiction by Def1;
end;
