Overriding method that has no access modifier with protected access modifier

Question

When you override methods you are not allowed to reduce the visibility of the inherited method. According to the following table, protected is more accessible than no modifier:

            | Class | Package | Subclass | World
————————————+———————+—————————+——————————+———————
public      |  y    |    y    |    y     |   y
————————————+———————+—————————+——————————+———————
protected   |  y    |    y    |    y     |   n
————————————+———————+—————————+——————————+———————
no modifier |  y    |    y    |    n     |   n
————————————+———————+—————————+——————————+———————
private     |  y    |    n    |    n     |   n

y: accessible
n: not accessible

But when I try to override f() (see SubClass) then I get the error:

Cannot reduce the visibility of the inherited method from MyInterface.

The method in MyInterface has no access modifier and the one in SubClass is protected, so more accessible. What am I missing here?

public interface MyInterface {
  void f();
}

public abstract class MyClass {
  protected abstract void f();
}

public class SubClass extends MyClass implements MyInterface{
   protected void f() { }
}

Answer 1

Methods in interfaces implicitly have the access modifier of public. So when you implement it with protected, it is a weaker access modifier.

Answer 2

Methods in interfaces are implicitly marked public and not default