Possible resolution:

Change in 8.7 [stmt.expand] bullet 5.2 as follows:

• ...
• Otherwise, if S is an iterating expansion statement, S is equivalent to:

    {
      init-statement
      constexpr auto&& decltype(auto) range = (expansion-initializer) ;
      constexpr auto begin = begin-expr;  // see 8.6.5 [stmt.ranged]
      constexpr auto end = end-expr;      // see 8.6.5 [stmt.ranged]
      S0
      .
      .
      .
      SN-1
    }
  

  where N is ...

(From submission #805.)

Consider:

  #include <span>

  constexpr int arr[3] = { 1, 2, 3 };
  consteval std::span<const int> foo() {
    return std::span<const int>(arr);
  }

  int main() {
    int r = 0;
    template for (constexpr auto m : foo())
      r += m;
  }

The expansion of this iterable expansion statement contains:

   constexpr auto &&range = foo();

The deduction yields std::span<const int>&& for the type of range, and the non-const lifetime-extended temporary causes constant evaluation to fail.