Ошибка WCF - не Могла найти элемент конечной точки по умолчанию, что ссылки сокращают 'UserService. UserService'

Я попробовал это, также с использованием linq. Просто замените строку своей формой. но строка облегчает понимание алгоритма. Пожалуйста, проверьте комментарии в коде для различных шагов:

         var lists = new List<List<string>>();
        lists.Add(new List<string> { "a", "b", "c" });
        lists.Add(new List<string> { "a", "c" });
        lists.Add(new List<string> { "d", "e" });
        lists.Add(new List<string> { "e", "d" });
        lists.Add(new List<string> { "e", "a" }); // from my comment

        var results = new List<List<string>>();

        foreach (var list in lists)
        {
            // That checks, if for this list, there is already a list, that contains all the items needed.
            if (results.Any(r => r.Count == r.Union(list).Count()))
            {
                continue;
            }

            // get the lists, that contains at least one item of the current "list".
            // This is important, as depending on the amount of elements, there have to be specific further steps.
            var listsWithItemsOfList = results.Where(r => list.Any(x => r.Contains(x)));

            // if not item, then you just have to add the whole content, as non of the colors exist.
            if (!listsWithItemsOfList.Any())
            {
                results.Add(new List<string>(list));
            }
            // if exactly, one, that add all the items, that were missing
            // (it might be, that nothing is added in case list.Except(l) is empty.
            else if(listsWithItemsOfList.Count() == 1)
            {
                var listWithOneItem = listsWithItemsOfList.Single();
                listWithOneItem.AddRange(list.Except(listWithOneItem));
            }
            else
            {
                // if multiple elements, it's getting complicated.
                // It means, that all needed items are currently spreaded over multiple lists, that have now to be merged.
                var newMergedList = listsWithItemsOfList.SelectMany(x => x).Distinct().ToList(); // merge all into one
                results.RemoveAll(x => listsWithItemsOfList.Contains(x)); // remove those lists from results
                results.Add(newMergedList); // just add one new list, containing all.
            }
        }