Lemma 2.2 If a minimum out-degree node v_0 in an oriented graph is has out-degree 3 and at least one of its neighbors has out-degree 0 in its neighbor induced subgraph, then the out-degree of v_0 will at least double in G^2

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.

Understand Lemma 2.2, which discusses how the out-degree of a minimum out-degree node (v_0) in an oriented graph with at least one neighbor having out-degree 0 in the induced subgraph doubles in the square of the graph (G^2). This is crucial in graph theory for analyzing higher-order graph properties.

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