## 3 Characteristics of OS Graphs
Hence P\({}_{1}\)(4) is connected to the 3 vertices--P\({}_{3}\)(3), P\({}_{5}\)(2) and P\({}_{6}\)(2). And P\({}_{2}\)(4) is connected to three vertices say P\({}_{6}\)(2) and P\({}_{3}\)(3), P\({}_{4}\)(3). Now P\({}_{5}\)(2) and P\({}_{6}\)(2) cannot be connected according to theorem 8. Hence according to axiom d at least one of the vertices having V = 2, say P\({}_{5}\)(2), must be connected to P\({}_{3}\)(3) (The mediator between P\({}_{5}\)(2) and P\({}_{1}\)(4) must be one of the vertices already connected to P\({}_{1}\)(4) as all its connections are already allocated. It cannot be P\({}_{6}\)(2) due to Theorem 8 and it cannot be P\({}_{2}\)(4) which has only one V = 2 vertex in its SS structure and this is P\({}_{6}\)(2)). Since P\({}_{2}\)(4) is connected to P\({}_{4}\)(3) (circled in the figure), then according to axiom d they both have to be connected to their mediator, but this is impossible since the structure includes already six vertices and all the other vertices, except P\({}_{4}\)(3), are fully connected in accordance with their valence. We see then that the attempt to design an OS satisfying the above-mentioned conditions (\(m_{2}=m_{3}=m_{4}=2\)) results in contradiction. **Case B**. \(m_{5}=1\). In this case, \(m_{4}\leq 2\) (Theorem 7). Also, \(m_{2}\leq 2\) because the only possible SSs for the vertex having V = 2 are 5,4 and 5,3 (since these vertices must be connected to the vertex having V = 5, and according to Theorem 8, they cannot be connected to each other). Hence, if \(m_{4}=0\), then \(m_{2}=1\), and according to Theorem 3, \(m_{3}=4\). But this result is impossible since, according to Theorems 10 and 12, there are only two possible SSs for the vertices having V = 3; SS = 5,3,3; and SS = 5,3,2. Hence, \(m_{4}\
eq 0\), so that \(m_{4}=1,2\). Let us show now that m\({}_{2}\
eq 0\). Suppose that m\({}_{2}=0\), since m\({}_{4}\leq 2\) (Theorem 7) and m\({}_{3}+\)m\({}_{4}+1=6\) (Theorem 3) -> m\({}_{3}=5-\)m\({}_{4}\); then, m\({}_{3}\geq\)3. The only possible SSs for the vertex having V = 3 are 5,3,3; 5,4,3; and 5,4,4. But the case where SS = 5,3,3 is impossible. To show this, let us try to form an OS satisfying this condition. Let V(P\({}_{1}\)) = 3 and SS(P\({}_{1}\)) = 5,3,3. Let P\({}_{1}\) be connected to P\({}_{2}\), P\({}_{3}\), and P\({}_{4}\), where V(P\({}_{2}\)) = 3, V(P\({}_{3}\)) = 3, and V(P\({}_{4}\)) = 5. P\({}_{2}\) and P\({}_{3}\) have to be connected to P\({}_{4}\) (having V = 5). Since V(P\({}_{2}\)) = 3, and it is connected only to two vertices (P\({}_{1}\) and P\({}_{4}\)), it has to be connected to an additional vertex, P\({}_{5}\) (not to P\({}_{3}\) because, in this case, SS(P\({}_{3}\)) = 5,3,3, the same as SS(P\({}_{1}\))), and P\({}_{5}\) has to be connected to P\({}_{4}\). By the same token, P\({}_{3}\) and P\({}_{4}\) have to be connected to P\({}_{6}\). Now, the only possible values of the V of P\({}_{5}\) and P\({}_{6}\) is 4 (since m\({}_{2}\) = 0, and there are already three vertices having V = 3 and one vertex having V = 5). Therefore, SS(P\({}_{2}\)) = SS(P\({}_{3}\)) = 3,4,5. Hence, P\({}_{2}\) = P\({}_{3}\), in contradiction to axiom e.  
We are left then with the following four possibilities: m\({}_{2}\) = 1,2 and m\({}_{4}\) = 1,2. For each of these cases, m\({}_{3}\) = 5 \(-\)m\({}_{2}\) - m\({}_{4}\), as 3 m\({}_{3}\)+5 m\({}_{5}\) = even. It follows for m\({}_{5}\) = 1 that  
\[3\;\mbox{m${}_{3}$}{= even}-5=\mbox{odd}\]  
Thus, m\({}_{3}\;\
eq\;2\). The remaining cases are given in Table 7.  
We subsequently show that the attempt to design an OS satisfying the conditions of each row in this table always results in the contradiction of one of the axioms, a to e. Pay attention to the fact that, in all the cases discussed below, all the vertices have to be connected to P\({}_{6}\)(5). Let us begin with the first row of Table 7, in which m\({}_{2}\) = 1, m\({}_{3}\) = 3, m\({}_{4}\) = 1, and m\({}_{5}\) = 1. Let us denote the vertices of the possible OS by P\({}_{1}\)(2), P\({}_{2}\)(3), P\({}_{3}\)(3), P\({}_{4}\)(3), P\({}_{5}\)(4), and P\({}_{6}\)(5).