$e(3,6,17)\geq 40$, minimum number of edges possible in an $(3, 6)$-graph on $17$ points the article is Some Graph Theoretic Results
Associated with Ramsey's Theorem for JACK E. GRAVER AND JAMES YACKEL pp: 144-145: https://core.ac.uk/download/pdf/82034211.pdf
I'm studying graph theory and the author makes a statement that I can not understand

$G$ is an $(x, y)$-graph if $x > C(G)$ and $y > I(G)$, where $I(G)$ is the
maximum number of points of $G$ that can be chosen so that no two are
joined by an edge and $C(G)$ is the maximum number of points in any complete subgraph of G. (in other word, $G$ is a graph that does not have cliques of size $x$ neither independent sets of size $y$)
$e(x, y, n)$ is the minimum number of edges possible in an $(x, y)$-graph on $n$ points.

Theorem: $e(3,6,17)\geq  40$
previously it is proved that $e(3,6,17)\geq  38$ using the simplex method in python. 

the author proves that is not possible that there are points with valence $3$ in a $(3,6)$-graph $G$ on $17$ points and $39$ edges.
Now the author affirms that there must be a $ 0- $ point in $ H_2 $ to then extend the graph to $ 39 $ edges joining $p$ and the $0$-point. Let $G$ be a $(3, 6)$-graph on $17$ points with $38$ edges. It is easy to
show that if $p$ is a preferred point of valence $4$ then $H_2$ must contain
a $0$-point.

Notation:
where: a point p of $H_2$ which has exactly $j$ edges to $H_1$
will be called a $j$-point.
Let $G$ be a graph and $p$ a point of $G$. By $H_1$ we will mean
the graph spanned by all points of $G$ which are joined to $p$ by an edge.
$H_2$ will denote the graph spanned by all points different from $p$ which are
not joined to $p$ by an edge. In the proof to Lemma $2$ we proved that if
$G$ is an $(x, y)$-graph then $H_1$ is an $(x -1, y)$-graph and $H_2$ is an $(x, y-1)$-graph. We will indicate this disection of $G$ by stating that $p$ is the preferred point.

for more than I do accounts I can not find why there must necessarily be a $0$ -point in $ H_2 $, if i suppose there is not $ 0 $ -point, 
I do not come to any contradiction.
 A: I don't know if my argument is the same as the authors' "easy" argument, but here is a proof.
(Note that in modern terminology, "points" are called "vertices", we say "degree" rather than "valence", and probably other terminology in the paper is similarly outdated. I will stick to the paper's terminology.)
We assume for the sake of contradiction that $H_2$ has no $0$-points.
First, we show that there are many edges between $H_1$ and $H_2$. 
If a point in $H_1$ is adjacent to three or more $1$-points in $H_2$, then we find an independent set of size $6$: those three $1$-points, together with the other three points of $H_1$. This is impossible. So a point in $H_1$ can be adjacent to at most two $1$-points in $H_1$, and the total number of $1$-points is at most $8$. If there are no $0$-points, then the remaining $4$ points of $H_2$ are at least $2$-points, giving at least $1 \cdot 8 + 2 \cdot 4 = 16$ edges between $H_1$ and $H_2$.
Second, we show that there cannot be many edges between $H_1$ and $H_2$.
The graph $H_2$ is a $(3,5)$-graph, and it is mentioned earlier in the paper that a $(3,5)$-graph must have at least $20$ edges. There are $4$ more edges between $p$ and $H_1$. If there are $38$ edges total in the graph, this leaves at most $38 - 20 - 4 = 14$ edges between $H_1$ and $H_2$.
There cannot simultaneously be at least $16$ and at most $14$ edges between $H_1$ and $H_2$, contradiction.
