[temp.deduct.type] p6 type name includes non-deduced context

[temp.deduct.type] p6 says

When a type name is specified in a way that includes a non-deduced context, all of the types that comprise that type name are also non-deduced. However, a compound type can include both deduced and non-deduced types.

Consider this example

template<class  U ,class T>
struct A{
    A(T){}
};
template<class U>
struct unknow_content{
    using type = U;
};

template<class U, class T>
void fun(A<U, typename unknow_content<T>::type>, T,U){}

int main(){
   fun(0,0,0);  // ill-formed
}

the type name A<U, typename unknow_content<T>::type> includes a non-deduced context typename unknow_content<T>::type, hence the U that comprises the type name A<U, typename unknow_content<T>::type> is also non-deduced?

Start with [temp.deduct.type] p6 and the supplied deduction example, then compare the wording about non-deduced contexts with the intended deduction behavior. Done means resolving whether U is non-deduced and recording an agreed standard-wording change or clarification.