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);

theorem Th16:
  FS is c=-linear implies union FS is Filter of X
proof
A1: for S being set st S in FS holds S c= bool X
  proof
    let S be set;
    assume S in FS;
    then S is Element of Filters(X);
    hence thesis;
  end;
  consider S being object such that
A2: S in FS by XBOOLE_0:def 1;
  assume
A3: FS is c=-linear;
A4: for Y1,Y2 holds (Y1 in union FS & Y2 in union FS implies Y1 /\ Y2 in
  union FS) & ( Y1 in union FS & Y1 c= Y2 implies Y2 in union FS)
  proof
    let Y1,Y2;
    thus Y1 in union FS & Y2 in union FS implies Y1 /\ Y2 in union FS
    proof
      assume that
A5:   Y1 in union FS and
A6:   Y2 in union FS;
      consider S1 being set such that
A7:   Y1 in S1 and
A8:   S1 in FS by A5,TARSKI:def 4;
A9:   S1 is Filter of X by A8,Th15;
      consider S2 being set such that
A10:  Y2 in S2 and
A11:  S2 in FS by A6,TARSKI:def 4;
A12:  S1,S2 are_c=-comparable by A3,A8,A11;
A13:  S2 is Filter of X by A11,Th15;
      per cases by A12;
      suppose
        S1 c= S2;
        then Y1 /\ Y2 in S2 by A7,A10,A13,Def1;
        hence thesis by A11,TARSKI:def 4;
      end;
      suppose
        S2 c= S1;
        then Y1 /\ Y2 in S1 by A7,A10,A9,Def1;
        hence thesis by A8,TARSKI:def 4;
      end;
    end;
    assume that
A14: Y1 in union FS and
A15: Y1 c= Y2;
    consider S1 being set such that
A16: Y1 in S1 and
A17: S1 in FS by A14,TARSKI:def 4;
    S1 is Filter of X by A17,Th15;
    then Y2 in S1 by A15,A16,Def1;
    hence thesis by A17,TARSKI:def 4;
  end;
A18: not {} in union FS
  proof
    assume {} in union FS;
    then consider S being set such that
A19: {} in S and
A20: S in FS by TARSKI:def 4;
    S is Filter of X by A20,Th15;
    hence contradiction by A19,Def1;
  end;
  reconsider S as set by TARSKI:1;
  S is Filter of X by A2,Th15;
  then X in S by Th5;
  then union FS is non empty Subset-Family of X by A1,A2,TARSKI:def 4
,ZFMISC_1:76;
  hence thesis by A18,A4,Def1;
end;
