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 Th166:
  dom R c= X implies R \ (R|A) = R|(X\A)
proof
  assume dom R c= X;
  hence R \ (R|A) =  R|X \ R|A by Th62
     .= R|(X\A) by Th74;
end;
