Let $N(v)$ be the (open) neighbourhood set of a vertex $v$, and let $N[v]$ be the closed neighbourhood set of $v$.


A graph $G$ is called *4-chordal* if $G$ has no induced cycle with five or more vertices.


**Question:** Does every 4-chordal graph $G$ have two vertices $x$ and $y$ such that
$N(x)$ is a subset of $N(y)$ or $N[x]$ is a subset of $N[y]$?


I think the answer is *yes*, but didn't find a proof so far.


The answer is *yes* for chordal graphs due to simplicial vertices. 


But for 4-chordal graphs, I did not see any result in this direction.
There are generalizations of simplicial vertices of $k$-chordal graphs due to Krithika, Mathew, Narayanaswamy and Sadagopan (2013) and Chavatal, Rusu and Sritharan (2002),
but these notions of simplicial vertices do not answer the question as far as I see ...



Existence of neighborhood inclusion for 4-chordal graphs
The answer is **no.** A counterexample is the [triangular prism](http://en.wikipedia.org/wiki/Triangular_prism) graph. Up to symmetry, there is a unique 5-cycle and a unique 6-cycle in the triangular prism and both these cycles have chords. Hence the triangular prism is 4-chordal. On the other hand, it is also easy to see that no (closed) neighbourhood is contained in any other (closed) neighbourhood. 


