Why cast expression to rvalue reference to function is lvalue?

At here,cppreference-lvalue,I found that

Cast expression to rvalue reference to function is lvalue.

I was curious, so I carried out the following experiment:

#include <iostream>

using namespace std;


typedef void (&&funr) (int);
typedef void (&funl) (int);

void test(int num){
    cout<<num<<endl;//output:20
}

void foo(funr fun){
    fun(10);
}

void foo(funl fun){
    fun(20);//call this
}

template <typename T> void foo(T&& fun){
    cout<<is_same<T,void(&)(int)>::value<<endl;//true, it is lvalue. 
    cout<<is_same<T,void(int)>::value<<endl;//false, it isn't rvalue.
}
int main()
{

   foo(static_cast<void(&&)(int)>(test));
   return 0;
}

the fact that so.

Why cast expression to rvalue reference to function is lvalue? Is it because a function type is no need to move semantics or something else? Or I understand this word wrong.

Cast expression to rvalue reference to function is lvalue