C++ Template overriding rules - c++

Question

I'm trying to override a C++ template for a specific type, which implements a member function. It seems to me, that this is not possible with the current language spec.
Overriding just works, if I use specialization for one specific type.
The code always prints World instead of Hello.
Is there any way to tell the compiler to use the other template?
I cant modify the first template of serialize() because it is part of an external library. I can just add additional templates.
#include <iostream>
#include <type_traits>
#include <new>
struct PanelBarcode
{
std::string value;
template<class S> void Serialize(S& in_s)
{
std::cout << value;
}
};
template <typename Archive, typename C>
struct has_serialize
{
private:
template<typename T>
static constexpr auto check(T*)
-> typename
std::is_same<
decltype(std::declval<T>().Serialize(std::declval<Archive&>())),
void
>::type;
template<typename>
static constexpr std::false_type check(...);
typedef decltype(check<C>(0)) type;
public:
static constexpr bool value = type::value;
};
struct X {
X() {}
template<class S> void Serialize(S& in_s)
{ }
};
template<class A, class T>
void serialize(A ar, T& obj, const unsigned int version)
{
std::cout << "World";
}
template<class A, class T, typename std::enable_if_t<has_serialize<A, T>::value> * = nullptr>
void serialize(A ar, T& obj, const unsigned int version)
{
obj.Serialize(ar);
}
struct Y
{
Y() {}
};
int main()
{
PanelBarcode bc;
bc.value = "Hello";
Y y;
serialize<Y, PanelBarcode>(y, bc, 13);
return 0;
}