
theorem Th12:
  for FT being filled non empty RelStr, IT being Subset of FT st
FT is symmetric holds IT is connected iff (for A, B being Subset of FT st IT =
  A \/ B & A,B are_separated holds A = IT or B = IT)
proof
  let FT be filled non empty RelStr,IT be Subset of FT;
  assume
A1: FT is symmetric;
A2: now
    assume
A3: for A, B being Subset of FT st IT = A \/ B & A,B are_separated
    holds A = IT or B = IT;
    for A,B being Subset of FT st IT = A \/ B & A <> {} & B <> {} & A
    misses B holds A^b meets B
    proof
      let A,B be Subset of FT;
      assume that
A4:   IT = A \/ B and
A5:   A <> {} & B <> {} and
      A misses B;
      now
        assume
A6:     A^b misses B;
        now
          assume A meets (B^b);
          then consider x being object such that
A7:       x in A and
A8:       x in B^b by XBOOLE_0:3;
          consider x2 being Element of FT such that
A9:       x2=x and
A10:      U_FT x2 meets B by A8;
          set y = the Element of U_FT x2 /\ B;
A11:      U_FT x2 /\ B <>{} by A10,XBOOLE_0:def 7;
          then
A12:      y in U_FT x2 by XBOOLE_0:def 4;
          then reconsider y2=y as Element of FT;
          x2 in U_FT y2 by A1,A12;
          then x2 in U_FT y2 /\ A by A7,A9,XBOOLE_0:def 4;
          then U_FT y2 meets A by XBOOLE_0:def 7;
          then
A13:      y2 in A^b;
          y in B by A11,XBOOLE_0:def 4;
          hence contradiction by A6,A13,XBOOLE_0:3;
        end;
        then A,B are_separated by A6;
        then
A14:    A=IT or B=IT by A3,A4;
A15:    A c= A^b by FIN_TOPO:13;
        A^b /\ B={} by A6,XBOOLE_0:def 7;
        then A /\ B ={} by A15,XBOOLE_1:3,26;
        hence contradiction by A4,A5,A14,XBOOLE_1:21;
      end;
      hence thesis;
    end;
    hence IT is connected;
  end;
  now
    assume
A16: IT is connected;
    thus for A, B being Subset of FT st IT = A \/ B & A,B are_separated holds
    A = IT or B = IT
    proof
      let A, B be Subset of FT;
      assume that
A17:  IT = A \/ B and
A18:  A,B are_separated;
A19:  A misses B by A18,Th6;
      now
        assume A<>{};
        then B={} by A18,A16,A17,A19;
        hence thesis by A17;
      end;
      hence thesis by A17;
    end;
  end;
  hence thesis by A2;
end;
