 reserve x,y,z,t for object,X,Y,Z,W for set;
 reserve R,S,T for Relation;

theorem
  X misses Y & R c= [:X,Y:] \/ [:Y,X:] implies R|X c= [:X,Y:]
proof
  assume that
A1: X /\ Y = {} and
A2: R c= [:X,Y:] \/ [:Y,X:];
  R|X = R /\ [:X,rng R:] by RELAT_1:67;
  then R|X c= ([:X,Y:] \/ [:Y,X:]) /\ [:X,rng R:] by A2,XBOOLE_1:26;
  then
A3: R|X c= [:X,Y:] /\ [:X,rng R:] \/ [:Y,X:] /\ [:X,rng R:] by XBOOLE_1:23;
  [:Y,X:] /\ [:X,rng R:] = [:X /\ Y, X /\ rng R:] by ZFMISC_1:100
    .= {} by A1,ZFMISC_1:90;
  hence thesis by A3,XBOOLE_1:18;
end;
