reserve A,X,X1,X2,Y,Y1,Y2 for set, a,b,c,d,x,y,z for object;
reserve P,P1,P2,Q,R,S for Relation;

theorem
  X c= Y implies (R \ R|Y)|X = {}
  proof
    assume
A1: X c= Y;
    dom R /\ X c= X by XBOOLE_1:17; then
A2: dom R /\ X c= Y by A1;
    dom(R \ R|Y) = dom R \ Y by Th167;
    then dom((R \ R|Y)|X) =(dom R \ Y) /\ X by Th55
       .= dom R /\ X \ Y by XBOOLE_1:49
       .= {} by A2,XBOOLE_1:37;
    hence (R \ R|Y)|X = {};
  end;
