reserve FT for non empty RelStr,
  A,B,C for Subset of FT;

theorem
  FT is filled symmetric & ( for A, B being Subset of FT st [#]FT = A \/
B & A <> {}FT & B <> {}FT & A is closed & B is closed holds A meets B) implies
  FT is connected
proof
  assume
A1: FT is filled symmetric;
  assume
A2: for A, B being Subset of FT st [#]FT = A \/ B & A <> {}FT & B <> {}
  FT & A is closed & B is closed holds A meets B;
  assume not FT is connected;
  then not [#]FT is connected;
  then consider P, Q being Subset of FT such that
A3: [#]FT = P \/ Q and
A4: P misses Q and
A5: P,Q are_separated and
A6: P <> {}FT & Q <> {}FT by A1,Th6;
  P is closed & Q is closed by A1,A3,A5,Th27;
  hence contradiction by A2,A3,A4,A6;
end;
