Proposed resolution:

This wording is relative to N4988.

1. Modify [forward.list.erasure] as indicated:

   template<class T, class Allocator, class U = T>
     typename forward_list<T, Allocator>::size_type
       erase(forward_list<T, Allocator>& c, const U& value);
   

   -1- Effects: Equivalent to: return erase_if(c, [&](const auto& elem) -> bool { return elem == value; });

2. Modify [list.erasure] as indicated:

   template<class T, class Allocator, class U = T>
     typename list<T, Allocator>::size_type
       erase(list<T, Allocator>& c, const U& value);
   

   -1- Effects: Equivalent to: return erase_if(c, [&](const auto& elem) -> bool { return elem == value; });

[ Wrocław 2024-11-23; Status changed: Voting → WP. ]

[ 2024-08-21; Reflector poll ]

Set status to Tentatively Ready after nine votes in favour during reflector poll.

std::erase for list is specified to return erase_if(c, [&](auto& elem) { return elem == value; }). However, the template parameter Predicate of erase_if only requires that the type of decltype(pred(...)) satisfies boolean-testable, i.e., the return type of elem == value is not necessarily bool.

This means it's worth explicitly specifying the lambda's return type as bool to avoid some pedantic cases (demo):

#include <list>

struct Bool {
  Bool(const Bool&) = delete;
  operator bool() const;
};

struct Int {
  Bool& operator==(Int) const;
};

int main() {
  std::list<Int> l;
  std::erase(l, Int{}); // unnecessary hard error
}