Pattern, correct form, and zero packet length rule

Question

Pattern, correct form, and zero packet length rule

From the accepted answer of the previous question, I found a rule. I did not know about templates and the correct form.

The program is poorly formed, diagnostics are not required if:
[...]
Any actual specialization of a variational template requires an empty package of template parameters or
[...]

According to this rule (if I understand correctly), the following template function is poorly formed

template <typename ... Ts>
int foo (std::tuple<Ts...> const &)
 { return std::get<sizeof...(Ts)>(std::tuple<int>{42}); }

since only a valid specialization and an empty parameter package are required Ts....

But (maybe because I do not know English very well). I'm not sure I understand this rule in the case of a template with two packages of additional packages.

I mean ... the following function foo()

#include <tuple>
#include <iostream>

template <typename ... Ts, typename ... Us>
int foo (std::tuple<Ts...> const &, std::tuple<Us...> const &)
 { return std::get<sizeof...(Ts)+sizeof...(Us)-1U>(std::tuple<int>{42}); }

int main ()
 {
   auto t0 = std::tuple<>{};
   auto t1 = std::tuple<int>{0};

   //std::cout << foo(t0, t0) << std::endl; // compilation error
   std::cout << foo(t0, t1) << std::endl;   // print 42
   std::cout << foo(t1, t0) << std::endl;   // print 42
   //std::cout << foo(t1, t1) << std::endl; // compilation error
 }