Candidate template ignored because template argument could not be inferred

template<typename K>
void foo(const typename A<K>::X& x, const typename A<K>::Y& y) {
    std::cout << "A" << std::endl;
}

K cannot been deduced, since it's in non-deduced context.

n3337 14.8.2.5/4

In certain contexts, however, the value does not participate in type deduction, but instead uses the values of template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

n3337 14.8.2.5/5

The non-deduced contexts are:

— The nested-name-specifier of a type that was specified using a qualified-id.

The argument K in const typename A<K>::X is not deducible. Basically, everything left of a :: is not deducible (if :: separates a nested-name).

It's trivial to see why it makes no sense to ask for deduction by running through this thought experiment:

struct A { typedef int type; }
struct B { typedef int type; }

template <typename T> void foo(typename T::type);

foo(5);   // is T == A or T == B ??

There's no one-to-one mapping from types to nested types: Given any type (such as int), there could be many ambient types of which it is a nested type, or there needn't be any.