An Issue about the rule of explicit specialization where mentioned implicit instantiation - c++

Question

#include <iostream>
template<class T> struct A {
enum E : T;
};
template<class T> enum A<T>::E : T { eT };
template<> enum A<char>::E : char { //#1
echar
};
int main(){
}
Consider the above code which is mentioned in temp.expl.spec#6, it's a typically ill-formed code. The rule says:
If a template, a member template or a member of a class template is explicitly specialized then that specialization shall be declared before the first use of that specialization that would cause an implicit instantiation to take place, in every translation unit in which such a use occurs; no diagnostic is required.
At the point #1, where the specialization A<char> will cause implicit instantiation
As a contrast:
#include <iostream>
template<class T> struct A {
struct Test;
};
template<typename T>
struct A<T>::Test{
};
template<>
struct A<int>::Test{ //#2
int c;
};
int main(){
}
Consider this code, At the point #2, where the specialization A<int> also cause implicit instantiation, The difference of the implicit instantiation between such two codes is said as the following:
temp.inst#2
The implicit instantiation of a class template specialization causes the implicit instantiation of the declarations, but not of the definitions, default arguments, or noexcept-specifiers of the class member functions, member classes, scoped member enumerations, static data members, member templates, and friends; and it causes the implicit instantiation of the definitions of unscoped member enumerations and member anonymous unions. However, for the purpose of determining whether an instantiated redeclaration is valid according to [basic.def.odr] and [class.mem], a declaration that corresponds to a definition in the template is considered to be a definition.
Because in the first code, the member enumeration is unscoped, hence the implicit instantiation of A<char> will also implicit instantiated the definition for enum E. As a contrast, in the second, Test is a member class, hence the implicit instantiation of specialization A<int> will only cause implicit instantiation of the declaration for member class Test rather than definition.
As a variant case of the first code:
#include <iostream>
template<class T> struct A {
enum E : T;
};
template<> enum A<char>::E : char {
echar
};
template<> enum A<int>::E : int { //#1
echar
};
int main(){
}
The difference from the first code is that there's no definition for member enumeration E of its enclosing primary class template, within the primary class template A, it's just a declaration.
So, I think the rule in [temp.expl.spec#6] is not clear. As these above codes shown, in each case, the specialization of primary class template all cause the implicit instantiation for class member, The distinction is one expect the implicit instantiation of declaration , the other is definition.
So, I think modified the rule as the following:
If a template, a member template or a member of a class template is explicitly specialized then that specialization shall be declared before the first use of that specialization that would cause an implicit instantiation of the definition that instantiated from the definition for the corresponding entity, in every translation unit in which such a use occurs; no diagnostic is required.
It will be conform the observable result. If I miss something, please correct me.