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The problem you uploaded asks for the following:

1. **Prove**: For any graph \( G \), if \( \overline{G} \) denotes the complement of \( G \), then prove that \( G - x \cong \overline{G} - x \) for any vertex \( x \in G \).

2. **Define**: What is meant by "self-complementary."

3. **Prove**: The number of vertices in a self-complementary graph must be congruent to 0 or 1 (mod 4).

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### 1. **Proof for \( G - x \cong \overline{G} - x \):**
Let's start by examining the first part of the problem.

- Let \( G \) be any graph and \( \overline{G} \) its complement, where for every edge in \( G \), there is no corresponding edge in \( \overline{G} \), and vice versa.
- The operation \( G - x \) denotes removing the vertex \( x \) and all edges incident to it from \( G \). Similarly, \( \overline{G} - x \) removes vertex \( x \) from \( \overline{G} \).

#### Goal: Prove \( G - x \cong \overline{G} - x \).

- Since \( \overline{G} \) is the complement of \( G \), removing the vertex \( x \) from both \( G \) and \( \overline{G} \) results in subgraphs where the relationships between the remaining vertices are preserved in the complement.
  
- Formally, this means that the graph \( G - x \) will have edges between all pairs of vertices that were not adjacent in \( G \) (and thus adjacent in \( \overline{G} \)). Therefore, \( G - x \) and \( \overline{G} - x \) are isomorphic, which proves the first part of the problem:
  \[
  G - x \cong \overline{G} - x.
  \]

### 2. **Definition of Self-Complementary Graph**:
A graph \( G \) is **self-complementary** if \( G \cong \overline{G} \). This means that the graph is isomorphic to its own complement. In other words, the structure of the graph is such that the complement can be rearranged (via some permutation of vertices) to look exactly like the original graph.

### 3. **Proving the Number of Vertices in a Self-Complementary Graph is Congruent to \( 0 \mod 4 \) or \( 1 \mod 4 \)**:

Let \( n \) be the number of vertices in a self-complementary graph.

- If \( G \cong \overline{G} \), then the number of edges in \( G \) must equal half the number of edges in the complete graph on \( n \) vertices. The complete graph \( K_n \) has \( \frac{n(n-1)}{2} \) edges, so the number of edges in a self-complementary graph must be:
  \[
  \text{Number of edges in } G = \frac{n(n-1)}{4}.
  \]
  
- For this to be an integer, \( n(n-1) \) must be divisible by 4. This happens only when \( n \equiv 0 \mod 4 \) or \( n \equiv 1 \mod 4 \) (since both \( n \) and \( n-1 \) need to include factors of 2 for divisibility by 4).

Thus, the number of vertices in a self-complementary graph must be congruent to 0 or 1 modulo 4.

### Would you like any further clarification on this problem?

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Here are five follow-up questions related to this topic:

1. What are some examples of self-complementary graphs with specific numbers of vertices?
2. How does the complement operation affect other graph properties like connectedness?
3. Can a bipartite graph ever be self-complementary? Why or why not?
4. What is the relationship between automorphisms and self-complementary graphs?
5. How does the concept of graph isomorphism generalize to other mathematical structures?

### Tip:
Always verify if a graph's complement is isomorphic by checking vertex permutations and edge mappings explicitly.

For any graph G, if G-bar denotes its complement, then prove that G-x is isomorphic to G-bar-x for any vertex x in G. Define self-complementary. Prove that the number of vertices in a self-complementary graph must be congruent to 0 or 1 (mod 4).

Explore a graph theory problem proving that G-x is isomorphic to G-bar-x for any vertex x, defining self-complementary graphs, and showing that the number of vertices in a self-complementary graph is congruent to 0 or 1 modulo 4.

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