theorem
  Top EqRelLatt Y = nabla Y
proof
  reconsider K = nabla Y as Element of EqRelLatt Y by MSUALG_5:21;
  now
    let a be Element of EqRelLatt Y;
    reconsider a9 = a as Equivalence_Relation of Y by MSUALG_5:21;
    thus K "\/" a = (the L_join of EqRelLatt Y).(K,a) by LATTICES:def 1
      .= nabla Y "\/" a9 by MSUALG_5:def 2
      .= EqCl (nabla Y \/ a9) by MSUALG_5:1
      .= EqCl (nabla Y) by EQREL_1:1
      .= K by MSUALG_5:2;
    hence a "\/" K = K;
  end;
  hence thesis by LATTICES:def 17;
end;
