模板化结构应该包含子类型，但编译器说它不包含

Question

我正在尝试使用C ++元编程，一切都进行得很顺利，除非我尝试编译这个重现器：

struct _0 {
  template<typename F, typename X>
  using apply = X;
};

struct Z {
  template<typename F>
  struct impl {
    struct C {
      template<typename X>
      struct closure {
        template<typename V>
        struct impl {
          // Broken out to help with debugging
          using XX = typename X::template apply<X>;
          using XXV = typename XX::template apply<V>;
        };
        template<typename V>
        using apply = typename impl<V>::XXV;
      };
      template<typename X>
      using apply = typename F::template apply<closure<X>>;
    };
    using exec = typename C::template apply<C>;
  };
  template<typename F>
  using apply = typename impl<F>::exec;
};

struct True {
  template<typename Consequent, typename Alternative>
  using apply = Consequent;
};

struct False {
  template<typename Consequent, typename Alternative>
  using apply = Alternative;
};

struct zero {
  struct constantly_false {
    template<typename>
    using apply = False;
  };

  template<typename N>
  using apply = typename N::template apply<constantly_false, True>;
};

struct mintest {
  template<typename F>
  struct closure {
    template<typename N>
    using apply =
      typename zero::apply<N>::template apply<
      _0, typename F::template apply<_0>>;
  };
  template<typename F>
  using apply = closure<F>;
};
using test = Z::apply<mintest>;

using F = test::apply<_0>;

int main() {
  return 0;
};

我最终得到错误：

reproducer.cc: In instantiation of ‘struct Z::impl<mintest>::C::closure<Z::impl<mintest>::C>::impl<_0>’:
reproducer.cc:19:44:   required by substitution of ‘template<class F> template<class X> template<class V> using apply = typename Z::impl<F>::C::closure<X>::impl::XXV [with V = _0; X = Z::impl<mintest>::C; F = mintest]’
reproducer.cc:56:42:   required from ‘struct mintest::closure<Z::impl<mintest>::C::closure<Z::impl<mintest>::C> >’
reproducer.cc:63:15:   required from here
reproducer.cc:16:53: error: no class template named ‘apply’ in ‘using XX = Z::impl<mintest>::C::apply<Z::impl<mintest>::C> {aka struct mintest::closure<Z::impl<mintest>::C::closure<Z::impl<mintest>::C> >}’
           using XXV = typename XX::template apply<V>;

这对我没有意义。 mintest :: closure没有专门化，所以我认为每个实例都应该适用于它。我尝试使用Clang并得到了同样的错误，这是我的线索，我只是听不懂。

谢谢。