theorem Th4:
  for R being Equivalence_Relation of Y holds (nabla Y)*R = nabla Y
  & R*nabla Y = nabla Y
proof
  let R being Equivalence_Relation of Y;
  (nabla Y) \/ R c= (nabla Y)*R by Th2;
  then nabla Y c= (nabla Y)*R by EQREL_1:1;
  hence (nabla Y)*R = nabla Y;
  (nabla Y) \/ R c= R*nabla Y by Th2;
  then nabla Y c= R*nabla Y by EQREL_1:1;
  hence thesis;
end;
