Some vertex of degree $5$ must be adjacent to $4$ other vertex of degree $5$ i need help with this proof:
$G$ is an $(,)$-graph if $>()$ and $>()$, where () is the maximum number of points of  that can be chosen so that no two are joined by an edge and () is the maximum number of points in any complete subgraph of G. (in other word,  is a graph that does not have cliques of size  neither independent sets of size )
DEF:  $e(x, y, n)$ is the minimum number of edges possible in an $(x, y)$-graph on $n$ points.
Theorem: $e(3,6,17)\geq 40$
proof: For proof that $e(3,6,17)\geq 38$, use the simplex method, where exist $8$ vertex of degree $5$ and $9$ vertex of degree $4$.
for $e(3,6,17)\geq 40$, first proof that there can not be a vertex of degree $3$
Now, Let $G$ be a $(3, 6)$-graph on $17$ points with $39$ edges; $G$ must then have $7$ vertex of degree $4$ and $10$ vertex of degree $5$. 
then he concludes that every vertex of degree $4$ in $G$ is adjacent to two or more vertex of degree $4$.

Now, the author makes a statement that I do not understand
From this it follows that some vertex of degree $5$ must be adjacent to $4$ other vertex of degree $5$.
There are many possibilities, how can I prove that there is at least one vertex as the statement says?
COMPUTATION L, pp: $144$ of - Some Graph Theoretic Results
Associated with Ramsey's Theorem (JACK E. GRAVER AND JAMES YACKEL)
 A: We have $7$ vertices of degree $4$ and $10$ vertices of degree $5$ in the graph (and no other vertices).
Each vertex of degree $4$ has at least $2$ edges to other vertices of degree $4$, and therefore at most $2$ edges to the vertices of degree $5$. Overall, there can be at most $2\cdot7 = 14$ edges between the degree-$4$ vertices and the degree-$5$ vertices.
If every degree-$5$ vertex had at least $2$ degree-$4$ neighbors, this would give us at least $2\cdot10 = 20$ edges between the degree-$4$ vertices and the degree-$5$ vertices, contradicting the above. So some degree-$5$ vertex must have at most $1$ degree-$4$ neighbor, and therefore at least $4$ degree-$5$ neighbors.
The claim in the paper is that some vertex has exactly $4$ degree-$5$ neighbors, and deduces that $H_2$ has $15$ edges when that vertex is preferred, by applying Corollary 2. The argument above only shows at least $4$ degree-$5$ neighbors. But if a vertex had $5$ degree-$5$ neighbors, then Corollary 2 would tell us that $H_2$ must have $14$ edges when that vertex is preferred, and this contradicts $e(3,5,12) \ge 15$ (computation (7) on p. 131).
