For a member function, are constexpr, const reference return type, and const qualification mutually independent?

Using

g++ -std=gnu++17 -Werror -Wall -Wextra

the following code compiles without warning:

struct Mutable {
  int x;
};

class State {
public:
  constexpr const Mutable &Immutable() const {
    return mutable_;
  }

private:
  Mutable mutable_;
} state_instance;

From what I've read, every const in that signature means something different:

• The constexpr essentially means that the member function can be evaluated at compile time.
• The first const means that the returned reference cannot be used to mutate the object being referenced.
• The second const means that the member function promises not to mutate any member variables.

Are there any redundancies?

This is not the same question as e.g.

• Usecase for non-const constexpr member functions?
• What is the use of a constexpr on a non-const member function?
• What is the purpose of marking the set function (setter) as constexpr?

which do not describe the constness of returned references.

Each of these is completely independent of each other.

A function which can produce a constant expression may modify any of its parameters (including the implicit this). Just because the result of the expression can be determined at compile-time does not mean nothing was changed when computing that result. And return values which are constant expressions can in fact be modified during constant evaluation, depending on what you're doing with them.

There is no inherent connection between any function parameters (including the implicit this) and a function's return value. As such, whether you can modify a returned reference is independent of whether a function will modify a parameter (again, including the implicit this).