在Scala中的通用curried方法中键入推断

鉴于下面的定义,它成功编译：

case class SequentialHistory[+E](events: Seq[E])

trait Event

case class SampleEvent(value: String) extends Event

def addEvent[E,E2 >: E](event: E)(history:SequentialHistory[E2]):
 SequentialHistory[E2] = history.copy(events = event +: history.events)

我想了解为什么类型推断不能正常工作并编译：

val hist = SequentialHistory(Seq.empty[Event])
//hist: SequentialHistory[Event] = SequentialHistory(List())

val histWithEvent = addEvent(SampleEvent("Does not compile"))(hist)
//error

导致编译错误：

Error:(21,62) type mismatch;
               found   : SequentialHistory[Event]
               required: SequentialHistory[SampleEvent]
               addEvent(SampleEvent("Does not compile"))(hist)
                              ^

但是,如果我在addEvent方法定义中交换参数列表：

def addEvent2[E,E2 >: E](history: SequentialHistory[E2])(event: E):
 SequentialHistory[E2] = history.copy(events = event +: history.events)

这将解决问题,正确推断类型和下面的代码片段编译：

val hist2 = SequentialHistory(Seq.empty[Event])
//hist2: SequentialHistory[Event] = SequentialHistory(List())

val histWithEvent2 = addEvent2(hist)(SampleEvent("Compiles"))
//histWithEvent1: SequentialHistory[Event] =
// SequentialHistory(List(SampleEvent1(Compiles)))

为什么Scala编译器无法在第一版addEvent中正确推断类型？