The statement of Lemma 2.2 can be interpreted as follows:

Let $v_0$ be a node in an oriented graph $G$, where "minimum out-degree" means that $v_0$ has the smallest out-degree (3 in this case) compared to the other nodes. The lemma suggests a condition where at least one of the neighbors of $v_0$ has out-degree 0 in the subgraph induced by the neighbors of $v_0$ (i.e., no outgoing edges among those neighbors).

Explanation:

1. Node $v_0$: In the original graph $G$, $v_0$ has out-degree 3. This means that $v_0$ has three directed edges to other nodes.
2. Neighbor-induced subgraph: Consider the subgraph formed only by the neighbors of $v_0$ (denoted as $N(v_0)$), i.e., the vertices that are connected directly to $v_0$ by directed edges. In this subgraph, at least one neighbor has out-degree 0, meaning that this neighbor does not have any outgoing edges to the other nodes in the subgraph.

Claim of the Lemma:

The lemma claims that, under these conditions, the out-degree of $v_0$ will at least double when considering the square of the graph $G^2$. In other words, in $G^2$, the node $v_0$ will have an out-degree of at least 6.

Why would the out-degree of $v_0$ double in $G^2$?

The graph $G^2$ (or "the square of the graph") is constructed such that an edge is directed from node $u$ to node $w$ if there is a path of length 2 from $u$ to $w$ in the original graph $G$. This means that in $G^2$, $v_0$ will have directed edges not only to its original neighbors but also to all nodes that can be reached from $v_0$ via one intermediary node.

Given the condition that at least one neighbor of $v_0$ has out-degree 0 in the neighbor-induced subgraph, this implies that:

• The neighbors of $v_0$ cannot send edges to each other (as at least one of them has out-degree 0).
• However, since $v_0$ has at least 3 neighbors, and these neighbors can have additional neighbors in $G$, there will be new nodes reachable from $v_0$ through these intermediate nodes in $G^2$, significantly increasing its out-degree.

In particular, since $v_0$ already has an out-degree of 3, and some of its neighbors can lead to new nodes in $G^2$, the lemma suggests that $v_0$’s out-degree will at least double to 6.

Conclusion:

In summary, the lemma indicates that under the given conditions (a node $v_0$ with out-degree 3, and at least one neighbor having out-degree 0 in the neighbor-induced subgraph), the out-degree of $v_0$ will increase significantly in $G^2$, doubling from 3 to at least 6.

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Would you like a detailed proof of this lemma, or would you like more information on how graph squares work?

Here are five questions for further exploration:

1. How is the square of a graph formally defined?
2. Why does having a neighbor with out-degree 0 in the induced subgraph affect the out-degree of $v_0$ in $G^2$?
3. Could the out-degree of $v_0$ more than double in $G^2$?
4. What happens if all neighbors of $v_0$ have non-zero out-degree in the induced subgraph?
5. How does the structure of the original graph $G$ affect the properties of $G^2$?

Tip: In oriented graphs, understanding the paths and how they extend in higher powers of the graph is key to determining changes in node degrees.