Bellman-Ford parent pointer (?) negative cycle
In the notation here, a vertex $v$ is called a parent pointer of a vertex $u$, if a shortest path to $u$, found so far in the algorithm, has $v$ preceding $u$. So in the graph above, starting at $a$ and moving to $b,c,d$ in that order, the parent pointers are $a,b,c$ respectively for $b,c,d$. 

We say there is a loop in parent pointers if we have vertices $v_1,v_2, \ldots ,v_n$ such that $v_i$ is a parent of $v_{i+1}$ and $v_n$ is a parent of $v_1$. Notice that this is not happening in the earlier example. But if we change the weight of edge $(c,a)$ to $-3$, then we have $a$ as parent of $b$, $b$ parent of $c$ and $c$ as parent of $a$, forming a loop in parent pointers, indicating a negative cycle. 

Proof for why this finds negative cycles is given in the document referred in that earlier post

Edit:

Suppose we have found a cycle $v_1,v_2,\ldots , v_n$ in parent pointers ($v_i$ is parent of $v_{i+1}$ and $v_n$ is parent of $v_1$). Let $P(v)$ denote the shortest path length obtained to vertex $v$ so far. Then, as $v_i$ is parent of $v_{i+1}$, we get $P(v_{i+1}) = P(v_i) + w(v_i,v_{i+1})$, where $w(u,v)$ is weight of the edge from $u$ to $v$. So $P(v_n) = P(v_1) + \sum_{i=1}^{n-1} {w(v_i,v_{i+1})}$. We have reached $v_n$ so far and now looking at its neighbour $v_1$. And $v_n$ is made parent of $v_1$. Which means the shortest path length to $v_1$ ($P(v_1)$) found so far is bigger than the path coming from $v_n$. That is, $P(v_1) &gt; P(v_n) + w(v_n,v_1)$. Hence, 

$$
P(v_1) &gt; P(v_1) + \sum_{i=1}^{n-1} {w(v_i,v_{i+1})}
$$
$$
0 &gt; \sum_{i=1}^{n-1} {w(v_i,v_{i+1})}
$$

So any cycle in parent pointers corresponds to a negative cycle in the graph. 

Note that this is only one half of the proof. We still need to show the converse (whenever there is a negative cycle in the graph, we will get a cycle in parent pointers).