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
  for f being Relation, X being set st X misses dom f holds f|X = {}
proof
  let f be Relation, X be set such that
A1: X /\ dom f = {};
  thus f|X = (f|dom f)|X
    .= f|({} qua set) by A1,Th65
    .= {};
end;
