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,g being Relation, A,B being set st A c= B & f|B = g|B holds f|A = g|A
proof
  let f,g be Relation, A,B be set;
  assume that
A1: A c= B and
A2: f|B = g|B;
A3: A = B /\ A by A1,XBOOLE_1:28;
  hence f|A = (f|B)|A by Th65
    .= g|A by A2,A3,Th65;
end;
