structural types should be more relaxed about protected methods

[scabug]

scala> def f(x: { def clone(): AnyRef }) = x.clone()           
<console>:5: error: method clone cannot be accessed in AnyRef{def clone(): AnyRef}
       def f(x: { def clone(): AnyRef }) = x.clone()
                                             ^

The problem of course is that the access of clone() will be widened to public in any number of AnyRef subclasses, any of which could legally call the method if it would compile.

I can't figure any way to get this signature in. Type alias doesn't get us any closer despite accepting the declaration of a public method:

scala> type FakeClone = { def clone(): AnyRef }
defined type alias FakeClone

scala> def f(x: FakeClone) = x.clone()         
<console>:6: error: method clone cannot be accessed in FakeClone
       def f(x: FakeClone) = x.clone()
                               ^

[scabug]

Imported From: https://issues.scala-lang.org/browse/SI-3197?orig=1
Reporter: @paulp

[scabug]

@paulp said:
See also #1352 and #1994, I see. (But I consider this issue distinct.)

[scabug]

@paulp said:
Here's another related thing to fix.

scala> class A { override def clone(): A = new A  }
defined class A

scala> class C { override def clone(): A = new A  }
defined class C

scala> List(new A, new C)
res0: List[ScalaObject{protected[package lang] def clone(): A}] = List(A@2b619bca, C@153b0106)