theorem Th6:
  Y in F implies not (X \ Y) in F
proof
  assume
A1: Y in F;
  assume X \ Y in F;
  then
A2: Y /\ (X \ Y) in F by A1,Def1;
  Y misses (X \ Y) by XBOOLE_1:79;
  then {} in F by A2;
  hence contradiction by Def1;
end;
