assertEvalsToFalse(Predicates.and(FALSE, NEVER_REACHED));
  }

  public void testAnd_equalityBinary() {
    new EqualsTester()
        .addEqualityGroup(Predicates.and(TRUE, NEVER_REACHED), Predicates.and(TRUE, NEVER_REACHED))
        .addEqualityGroup(Predicates.and(NEVER_REACHED, TRUE))
        .addEqualityGroup(Predicates.and(TRUE))
        .addEqualityGroup(Predicates.or(TRUE, NEVER_REACHED))

   * first decrementing to zero, and then removing. putIfAbsent or replace could observe the
   * intermediate zero-state. Ways we could deal with this are:
   *
   * - Don't define any of the ConcurrentMap operations. This is the current state of affairs.
   *
   * - Define putIfAbsent and replace as treating zero and absent identically (as currently

   * all references {@code x}, {@code y}, and {@code z} (any of which may be null):
   *
   * <ul>
   *   <li>{@code equivalent(x, x)} is true (<i>reflexive</i> property)
   *   <li>{@code equivalent(x, y)} and {@code equivalent(y, x)} each return the same result
   *       (<i>symmetric</i> property)
   *   <li>If {@code equivalent(x, y)} and {@code equivalent(y, z)} are both true, then {@code

   * all references {@code x}, {@code y}, and {@code z} (any of which may be null):
   *
   * <ul>
   *   <li>{@code equivalent(x, x)} is true (<i>reflexive</i> property)
   *   <li>{@code equivalent(x, y)} and {@code equivalent(y, x)} each return the same result
   *       (<i>symmetric</i> property)
   *   <li>If {@code equivalent(x, y)} and {@code equivalent(y, z)} are both true, then {@code

   * enclosed} by both this range and {@code other}.
   *
   * <p>For example,
   *
   * <ul>
   *   <li>{@code [2, 4)} and {@code [5, 7)} are not connected
   *   <li>{@code [2, 4)} and {@code [3, 5)} are connected, because both enclose {@code [3, 4)}
   *   <li>{@code [2, 4)} and {@code [4, 6)} are connected, because both enclose the empty range
   *       {@code [4, 4)}

 * remove) are supported. Null keys and values are supported.
 *
 * <p>{@code containsKey(k)}, {@code put(k, v)} and {@code remove(k)} are all (expected and
 * amortized) constant time operations. Expected in the hashtable sense (depends on the hash
 * function doing a good job of distributing the elements to the buckets to a distribution not far

    // Every node is, at a minimum, reachable from itself. Since the resulting transitive closure
    // will have no isolated nodes, we can skip adding nodes explicitly and let putEdge() do it.

    if (graph.isDirected()) {
      // Note: works for both directed and undirected graphs, but we only use in the directed case.
      for (N node : graph.nodes()) {
        for (N reachableNode : reachableNodes(graph, node)) {

   *   <li>If {@code tolerance} is zero, and neither {@code a} nor {@code b} is NaN, then {@code a}
   *       and {@code b} are fuzzily equal if and only if {@code a == b}.
   *   <li>With {@link Double#POSITIVE_INFINITY} tolerance, all non-NaN values are fuzzily equal.
   *   <li>With finite tolerance, {@code Double.POSITIVE_INFINITY} and {@code

   * and https://github.com/jspecify/jspecify/issues/490.
   *
   * (Perhaps it should also have required that its cause was a RuntimeException. However, that
   * would have required that we throw a different kind of exception for wrapping *checked*
   * exceptions in methods like Futures.getUnchecked and LoadingCache.get.)
   */

  /**

 * existing key-value pair, the {@code ArrayListMultimap} will contain entries for both the new
 * value and the old value.
 *
 * <p>Keys and values may be null. All optional multimap methods are supported, and all returned
 * views are modifiable.
 *
 * <p>The lists returned by {@link #get}, {@link #removeAll}, and {@link #replaceValues} all
 * implement {@link java.util.RandomAccess}.
 *