On Fri, Jan 15, 2021 at 10:54 AM Yves Bailly via Std-Proposals <std-proposals@lists.isocpp.org> wrote:

First, the "stronger type alias":
using U = new T;

As you've realized (good!), the problems are going to be with T's existing customization points:
- std::swap(u, u)
- std::hash<T>(u)
With this declaration:
- Any implicit conversion to T (e.g. from constructor) is also an implicit conversion to U.
- There's an implicit conversion from U to T.
- But there's no implicit conversion from T to U.
- Any function which takes a T can also take a U, unless explicitly deleted.
- But any function which takes a U can't take a T.

T's copy constructor is an "implicit conversion to T" — I mean, if T is std::string and U is Name, then
    T t = "Yves";
    void tfoo(T);
    tfoo(t);  // OK, implicitly calls T(const T&)
    void ufoo(U);
    ufoo(t);  // OK, would have implicitly called T(const T&), so there must be a U(const T&) by the first bullet point quoted above

            using Name = new std::string;
            void Store(std::string the_name); // (1)
            void Store(Name the_name);        // (2)
/* error */ Store("Your name here"); // ambiguous, ill-formed

FWIW, this is mildly surprising; this would be a place where `std::is_base_of_v<T, U>` and yet the signature `void(U)` is not considered more-specialized-than `void(T)`.
But I think this is just an indication that your intuition about std::is_base_of is wrong. Types with no inheritance relationships (no base classes) definitely should not claim to have is_base_of relationships.

/* ok    */ using Names_Income = std::unordered_map<Name, double>;
            // std::hash<std::string> used because of
            // implicit cast from Name to std::string

If there's an implicit conversion from Name to std::string, then why did you say
    std::string s;
    s = name;
didn't work?