Why to use std::move despite the parameter is an r-value reference

Question

I am confused about using std::move() in below code:

If I uncomment line at (2) the output would be: 1 2 3 but if I uncomment line at (1) output would be nothing which means that move constructor of std::vector was called!

Why do we have to make another call to std::move at (1) to make move constructor of std::vector to be called?

What I understood that std::move get the r-value of its parameter so, why we have to get the r-value of r-value at (1)?

I think this line _v = rv; at (2) is more logical and should make std::vector move constructor to be called without std::movebecause rv itself is r-value reference in the first place.

template <class T>
class A
{
public:
    void set(std::vector<T> & lv)
    {

    }  
    void set(std::vector<T> && rv)
    {           
        //_v = std::move(rv);          (1)
        //_v = rv;                     (2)
    }

private:
    std::vector<T> _v;
};

int main()
{
    std::vector<int> vec{1,2,3};
    A<int> a;

    a.set(std::move(vec));
    for(auto &item : vec)
        cout << item << " ";
    cout << endl;

    return 0;
}