使用函数器提升接受隐式参数的函数(Scalaz7)

Question

刚刚开始学习Scalaz。以下是我的代码

trait Monoid[A] {
     def mappend(a1: A, a2: A): A
     def mzero: A
}

object Monoid {
    implicit val IntMonoid: Monoid[Int] = new Monoid[Int] {
        def mappend(a1: Int, a2: Int): Int = a1 + a2
        def mzero: Int = 0
    }

    implicit val StringMonoid: Monoid[String] = new Monoid[String] {
        def mappend(a1: String, a2: String): String = a1 + a2
        def mzero: String = ""
    }
}

trait MonoidOp[A] {
    val F: Monoid[A]
    val value: A
    def |+|(a2: A): A = F.mappend(value, a2)
}

object MonoidOp{
    implicit def toMonoidOp[A: Monoid](a: A): MonoidOp[A] = new MonoidOp[A]{
        val F = implicitly[Monoid[A]]
        val value = a
    }
}

我已经定义了一个函数(只是为了它)

def addXY[A: Monoid](x: A, y: A): A = x |+| y

我想提升它，这样它就可以使用Option、List等容器。

def addXYOptioned = Functor[Option].lift(addXY)

上面写着error: could not find implicit value for evidence parameter of type scalaz.Monoid[A] def addOptioned = Functor[Option].lift(addXY)

如何提升这些功能？