reserve u,v,x,x1,x2,y,y1,y2,z,p,a for object,
        A,B,X,X1,X2,X3,X4,Y,Y1,Y2,Z,N,M for set;
reserve e for object, X,X1,X2,Y1,Y2 for set;

theorem
  not x in X & not y in X implies X = X \/ {x,y} \ {x,y}
proof
A1: (X \/ {x,y}) \ {x,y} = X \ {x,y} by XBOOLE_1:40;
  assume ( not x in X)& not y in X;
  hence thesis by A1,Th119;
end;
