I think "value of" allows typedefs, but "I" and "I+0" are not equivalent (although functionally equivalent), so I wouldn't expect "I+0" to work.

@opensdh, the wording here was introduced by your paper, I believe. Any thoughts here?

The type template parameter that is not a pack or pack expansion is itself a typedef-name. Except that, I cannot figure out we can declare a typedef declaration in a declaration. If functionally equivalent is not considered here, why not we directly say it shall be the name of the corresponding parameter.

My assumption is that [temp.arg.general]/2 means "has the value of" to just be "is". If not that, the equivalence invoked in [expr.prim.id.qual]/3.2 is surely the right interpretation, which would reject I+0. (My intuition is really that the strict current-instantiation rules from [temp.dep.type] would apply here, but that doesn't really work in an out-of-class declaration when we don't know what template to search yet.)

@opensdh, the wording here was introduced by your paper, I believe. Any thoughts here?

That wording was merely moved by P1787R6 from [temp.dep.type]/2, where it dated back to C++11.

"has the value of" to just be "is"

Even if it were that, IMHO, the value of I+0 is the value of I. Since the proposition I = I+0, if I ∈ R, this proposition is always true. Moreover, for template type-parameter, even though we obey [temp.dep.type#2], std::remove_reference_t<decltype(std::declval<T>())> is not permitted in the nested-name-specifier of a member declaration outside its enclosing class template definition.

Let alone it seems there is no way to declare a variable in such context to make it equal to the template parameter. Hence, why not directly say the template argument shall be the identifier introduced by the template parameter declaration.