[temp.over.link] types involving template parameters are not expressions [CWG2584]

Two declarations correspond if they (re)introduce the same name, both declare constructors, or both declare destructors, unless

• [...]
• both declare function templates with equivalent non-object-parameter-type-lists, return types (if any), template-heads, and trailing requires-clauses (if any), and, if both are non-static members, they have corresponding object parameters.

we define whether two types are equivalent in [temp.over.link] p6, which says

When determining whether types or type-constraints are equivalent, the rules above are used to compare expressions involving template parameters.

Consider this example:

template<class T>
T fun();

template<class T>
T&& fun();

In this example, their return types are T and T&&, which involve template parameters. However, T and T&& are not expressions anyway. According to grammar, they are defined as type-specifier. So, whether two function templates correspond relies on the correct definition of [temp.over.link], the latter seems to be a bit clear in its definition.

Another special case is the conversion function template whose name involves template parameter

struct A{
   template<class T>
   operator T();

   template<class T>
   operator T&&();
};

Such two conversion function templates' names are operator T and operator T&&, respectively. whether they declare the same name or not is unclear in this document. For this issue, I think we may clarify it with the help of equivalent return types.

Start with [temp.over.link] paragraph 6 and the cited rules for corresponding declarations. Compare the two function-template return types and the two conversion function-template names in the examples; done means the draft clearly specifies how these template-parameter-dependent types and names are compared.