Why does `T&&` not stick?

Question

Say I have the following functions:

void f2(int&& i)
{
    puts("f2(int&& i)");
}

void f(int&& i)
{
    puts("f(int&& i)");
    f2(i);
}

The compiler error emitted is:

error C2664: 'void f2(int &&)': cannot convert argument 1 from 'int' to 'int &&'

message : You cannot bind an lvalue to an rvalue reference

It works if you cast to int&& again with another call to std::move():

void f(int&& i)
{
    puts("f(int&& i)");
    f2(std::move(i));
}

But that shouldn't be necessary should it? It looks to me that the variable received by f(int&& i) is an xvalue. If f receives int&&, so we should just be able to pass that along, no?

How is the int&& received by f still an lvalue?