Copy assignment operator for variadic std::tuple - c++

Question

I have a variadic template with a variadic std::tuple as a member.
I declared the default assignment operator.
MSVC complains that the operator is deleted. I can still run the program.
It doesnt give the expected result.
So 3 Questions come to my mind:
Why is the default copy assignment operator not generated even if flagged as default?
Why can I run the programm still with the code posted?
How to declare a assignment operator for std::tuple<T...> template?
#include <string>
#include <iostream>
#include <tuple>
template<typename ...T>
class Variadic
{
public:
explicit Variadic(T... args)
:m_data{ std::move(args)... }
{
}
Variadic& operator=(const Variadic&) = default;
template<typename Type>
Type get_element() const
{
return std::get<Type>(m_data);
}
private:
std::tuple<T...> m_data{};
};
int main()
{
Variadic<int, std::string> tuple1{ 1, "a" };
Variadic<int, std::string> tuple2{ 2, "b" };
std::cout << tuple1.get_element<int>() << '\n'; // 1
std::cout << tuple1.get_element<std::string>() << '\n'; // a
std::cout << tuple2.get_element<int>() << '\n'; // 2
std::cout << tuple2.get_element<std::string>() << '\n'; // b
tuple2 = tuple1; // MSVC2017 gives warning here:
//function "Variadic<T...>::operator=(const Variadic<T...> &) [with T=<int, std::string>]"
//(declared at line 13) cannot be referenced -- it is a deleted function
// still it compiles ???
std::cin.get();
std::cout << tuple1.get_element<int>() << '\n'; // 1
std::cout << tuple1.get_element<std::string>() << '\n'; // a
std::cout << tuple2.get_element<int>() << '\n'; // expect 1 but 2 ?
std::cout << tuple2.get_element<std::string>() << '\n'; // expect b but a ?
}