theorem
  X misses Y implies dom (R /\ [:X,Y:]) misses rng (R /\ [:X,Y:])
proof
  assume
A1: X /\ Y = {};
  dom (R /\ [:X,Y:]) c= X by Th1; then
A2: dom (R /\ [:X,Y:]) /\ rng (R /\ [:X,Y:]) c= X /\ rng (R /\ [:X,Y:]) by
XBOOLE_1:26;
  X /\ rng (R /\ [:X,Y:]) c= X /\ Y by Th1,XBOOLE_1:26;
  hence dom (R /\ [:X,Y:]) /\ rng (R /\ [:X,Y:]) = {} by A1,A2,XBOOLE_1:1,3;
end;
