Confusion std::move, lvalue and x value [duplicate]

This question already has an answer here:

• How does std::move() transfer values into RValues? 2 answers

I was studying the difference between x value and l value from stackoverflow, github and cppreference and yet not convinced about the difference between l value and x value.

One problem is that to understand x value, one should understand std::move and to understand std::move you should know x value [ref]. Circular!

Also, according to cppref example:

#include <iostream>
#include <utility>
#include <vector>
#include <string>

int main()
{
    std::string str = "Hello";
    std::vector<std::string> v;

    // uses the push_back(const T&) overload, which means 
    // we'll incur the cost of copying str
    v.push_back(str);
    std::cout << "After copy, str is \"" << str << "\"\n";

    // uses the rvalue reference push_back(T&&) overload, 
    // which means no strings will be copied; instead, the contents
    // of str will be moved into the vector.  This is less
    // expensive, but also means str might now be empty.
    v.push_back(std::move(str));
    std::cout << "After move, str is \"" << str << "\"\n";

    std::cout << "The contents of the vector are \"" << v[0]
                                         << "\", \"" << v[1] << "\"\n";
}

Using std::move may cause the original value be loosed. To me, it looks like

v.push_back(std::move(str))

causes a new member v[1] being created. Then,

&v[1]=&str

But why should it damage the value in str? It does not make sense.

Therefore,

1- How does std::move harm the original value? (through which mechanism)

2- How does x value differ from l value? Is there any practical example for it? (Not a dummy example)