human
Solve the following python programming problem given the following code:
@ not_implemented_for ('undirected') 
@ not_implemented_for ('multigraph') 
def is_reachable(G, s, t) : 
    'Decides whether there is a path from `s` to `t` in the
    tournament.

    This function is more theoretically efficient than the reachability
    checks than the shortest path algorithms in
    :mod:`networkx.algorithms.shortest_paths`.

    The given graph **must** be a tournament, otherwise this function\'s
    behavior is undefined.

    Parameters
    ----------
    G : NetworkX graph
        A directed graph representing a tournament.

    s : node
        A node in the graph.

    t : node
        A node in the graph.

    Returns
    -------
    bool
        Whether there is a path from `s` to `t` in `G`.

    Notes
    -----
    Although this function is more theoretically efficient than the
    generic shortest path functions, a speedup requires the use of
    parallelism. Though it may in the future, the current implementation
    does not use parallelism, thus you may not see much of a speedup.

    This algorithm comes from [1].

    References
    ----------
    .. [1] Tantau, Till.
           "A note on the complexity of the reachability problem for
           tournaments."
           *Electronic Colloquium on Computational Complexity*. 2001.
           <http://eccc.hpi-web.de/report/2001/092/>

    ' 
    def two_neighborhood(G, v) : 
        'Returns the set of nodes at distance at most two from `v`.

        `G` must be a graph and `v` a node in that graph.

        The returned set includes the nodes at distance zero (that is,
        the node `v` itself), the nodes at distance one (that is, the
        out-neighbors of `v`), and the nodes at distance two.

        ' 
        return { x for x in G if ((x == v) or (x in G [v]) or any ((is_path (G, [v, z, x]) for z in G))) } 
def is_closed(G, nodes) : 
        'Decides whether the given set of nodes is closed.

        A set *S* of nodes is *closed* if for each node *u* in the graph
        not in *S* and for each node *v* in *S*,...

    
    

    Fix the buggy line: return all (((not (is_closed (neighborhoods, S) and (s in S) and (t not in S))) for S in neighborhoods))
assistant
#Here is Python code to fix the buggy line: return all (((not (is_closed (neighborhoods, S) and (s in S) and (t not in S))) for S in neighborhoods))
.

return all (((not (is_closed (G, S) and (s in S) and (t not in S))) for S in neighborhoods))