Is using a value that's been subjected to std::move undefined behaviour?

#include <iostream>

void print(int&& value) {
    std::cout << value << std::endl;
}

int main() {
    int value = 20;
    print(std::move(value));
    value = 15;
    std::cout << value << std::endl;
    return 0;
}

I'm trying to get my head around some of the fundamentals of move semantics.

In the case of the code above, my understanding is that std::move has "attempted" to perform a move on value.

Is any attempt to use it after this undefined? Or is something more complex going on, potentially silently degrading performance?