Any and All Puzzle Solution (over-complicated mode)

AnyAndAll.cs

using System.Collections;
using System.Collections.Generic;
using System.Linq;
using static System.Console;

namespace AnyAndAll
{
    public class Program
    {
        public static void Main(string[] args)
        {
            var sequence = GetSpecialSequence();

            // Prints "True"
            WriteLine(sequence.All(b => b == true));

            // Prints "False"
            WriteLine(sequence.Any(b => b == true));

            ShowBehavior();
        }

        private static IEnumerable<bool> GetSpecialSequence()
        {
            // because implementing this with `yield break;` is boring.
            return new TrueBreakFalseBreak();
        }

        private class TrueBreakFalseBreak : IEnumerable<bool>, IEnumerator<bool>
        {
            private int state = 0;
            public bool Current => state % 4 < 2;
            public bool MoveNext() => state++ % 2 == 0;
            public IEnumerator<bool> GetEnumerator() => this;
            IEnumerator IEnumerable.GetEnumerator() => GetEnumerator();
            object IEnumerator.Current => Current;
            public void Reset() { }
            public void Dispose() { }
        }

        private static void ShowBehavior()
        {
            // Note that the solution relies on that the original
            // test code re-uses the same `sequence` variable.
            // .Any and .All call .Dispose, but we ignore it.
            var sequence = new TrueBreakFalseBreak().GetEnumerator();
            for (int i = 0; i < 16; i++)
            {
                WriteLine($"MoveNext: {sequence.MoveNext()}");
                WriteLine($"Current: {sequence.Current}");
            }
        }
    }
}