Understanding `std::is_move_constructible`

Types without a move constructor, but with a copy constructor that accepts const T& arguments, satisfy std::is_move_constructible. For example, in the following code:

#include <type_traits>

struct T {
    T(const T&) {}
    //T(T&&) = delete;
};

int main() {
    static_assert(std::is_move_constructible<T>::value, "not move constructible");
    return 0;
}

T will have no implicit move constructor as it has a user-defined copy constructor.

However, if we uncomment the explicit delete of the move constructor, the code no longer compiles. Why is this? I would have expected that the explicit copy constructor would still satisfy std::is_move_constructible.

Does overload play a role, choosing the declared move constructor and then failing because it is deleted?