theorem
  LSeg(|[0,0]|,|[1,0]|) misses LSeg(|[0,1]|,|[1,1]|)
proof
  set x = the Element of LSeg(|[0,0]|,|[1,0]|) /\ LSeg(|[0,1]|,|[1,1]|);
  assume
A1: LSeg(|[0,0]|,|[1,0]|) /\ LSeg(|[0,1]|,|[1,1]|) <> {};
  then x in LSeg(|[0,1]|,|[1,1]|) by XBOOLE_0:def 4;
  then
A2: ex p st p = x & p`1 <= 1 & p`1 >= 0 & p`2 = 1 by Th13;
  x in LSeg(|[0,0]|,|[1,0]|) by A1,XBOOLE_0:def 4;
  then ex p2 st p2 = x & p2`1 <= 1 & p2`1 >= 0 & p2`2 = 0 by Th13;
  hence contradiction by A2;
end;
