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Proposition 2:

| G | δ ( G ) Δ ( G ) .

Proof: We start by looking at the vertex with the smallest degree, v s . Since for any vertex v , N ( v ) N ( v s ) we can say that every vertex is connected to either v s or a vertex in N ( v s ) . There are δ ( G ) such vertices, having a maximum degree of Δ ( G ) , implying that there are at most δ G Δ ( G ) vertices in the graph.

Proposition 3 (the lonely neighbor property):

For each edge ( u , v ) there exists a vertex w such that exactly one of ( w , v ) or ( w , u ) is in E ( G ) . Assuming ( w , v ) is in E ( G ) then for each vertex y exactly one of ( y , w ) or ( y , u ) is in E ( G ) .

Proof: Let an edge ( u , v ) be given. Because our graph is minimal, there exists a vertex w such that N ( a ) N ( w ) = b for some labels a , b { u , v } . Without loss of generality assign a to u and b to v . Now suppose for the sake of contradiction that ( w , u ) is in E ( G ) . Then { u , v , w } induces a K 3 and so there exists x V ( G ) { u , v , w } such that { u , v , w , x } induces a K 4 , but then x N ( u ) N ( w ) = v a contradiction. Now let a vertex y V ( G ) { u , v , w } be given. Suppose both ( y , w ) and ( y , u ) are in E ( G ) then y N ( u ) N ( w ) = v which is again a contradiction.

Proposition 4:

For any edge ( u , v ) G , at least one of u , v has degree of at least 4.

Proof: Suppose that neither of u nor v have degree of at least 4. It follows from Proposition 1 that d e g ( u ) = d e g ( v ) = 3 . u and v must both be contained in a K 4 subgraph. By the lonely neighbor property, there must exist a vertex w such that either ( u , w ) G or ( v , w ) G . This is a contradiction, as desired.

Proposition 5:

For any induced K 4 subgraph, at most one vertex has degree 3.

Proof: This follows directly from Proposition 4.

Proposition 5:

If G ( V , E ) is a graph in S then there are atleast | E | 6 choices of 4-tuples which induce a K 4 .

Proof: Each edge is in a K 4 and so it must make up at least 1 6 of a K 4 .

Proposition 6:

For any v V ( G ) , v is contained in at least deg( v ) 3 K 4 subgraphs.

Proof: Since any K 4 subgraph that contains v must contain three other vertices, and v must be adjacent to all of these vertices, there must be deg( v ) 3 induced K 4 subgraphs. Since this implies that there may be some uncounted incident edges to v , and that all edges are contained in a K 4 subgraph, it follows that v is contained in at least deg( v ) 3 K 4 subgraphs.

Proposition 7:

In any induced K 4 subgraph, at least one vertex must have a minimum degree of 5.

Proof: Let { v 1 , v 2 , v 3 , v 4 } G induce a K 4 subgraph. Suppose for the sake of contradiction that d e g ( v i ) 4 , for i { 1 , 2 , 3 , 4 } . K 4 is not a minimal 3-cover, so without loss of generality, let v 1 have degree 4. Since all edges are contained in a K 4 subgraph, it follows that two more vertices in our original K 4 subgraph ( { v 2 , v 3 } ) must be part of another K 4 subgraph, along with v 1 , and a fifth vertex, called v 5 . v 4 cannot have degree of 4, since this would imply that at least one of { v 1 , v 2 , v 3 } would have degree more than 5, so it must have degree 3. However, the graph induced by { v 1 , v 2 , v 3 , v 4 , v 5 } is not a minimal 3-cover, so at least one incident edge must be added to v 5 . Call the corresponding adjacent vertex v 6 . Since G is a 3-cover, there must be a vertex that is adjacent to both v 3 and v 6 , which implies that we must add an incident edge to at least one of the vertices in { v 1 , v 2 , v 3 , v 4 } , which leads to a contradiction, as desired.

Conclusion

Future work on this topic could either find a counterexample to the conjecture, or show that there is no counterexample by using bounds to arrive at a contradiction.

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Source:  OpenStax, The art of the pfug. OpenStax CNX. Jun 05, 2013 Download for free at http://cnx.org/content/col10523/1.34
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