Let $G$ be a graph $6-regular$ order $n=10$ and $u,v \in V(G)$. Show that $G, G-v $ and $G-u-v$ are hamiltonian Hi I have this problem and I would like to know if I explain it well:
Let $G$ be a graph $6-regular$ order $n=10$ and $u,v \in V(G)$. Show that $G, G-v $ and $G-u-v$ are hamiltonian


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*$G$:
$d(v) = 6$ $ \forall v \in V(G)$ then  $d(v) \geq \frac{n}{2}$ for all $v$ then $G$ is hamiltonian

*$G - v$ 
$v$ has 6 neighbors in $G$ all of them with degree 6, once $v$ is removed then its degree is 5, and the others 3 vertex has degree 6.
Now $n= 9$ and $d(v) = {6,5}$ $ \forall v \in V(G-v)$ besides $\frac{n}{2} = 4$  and $d(v) \geq \frac{n}{2} \rightarrow  {6,5} \geq 4$ then $G-v$ is hamiltonian

*$G-v-u$
Here two things can happen:
1.$N_G(u) = N_G(v)$ ($N_G(v)$ is the neighborhood of the vertex $v$ in $G$) now, al the vertex with degree 5 lose one edge and $d(v) = {6,4}$ $ \forall v \in V(G-v-u)$ and $n= 8$ 
$d(v) \geq \frac{n}{2} \rightarrow  {6,4} \geq 4$ then $G-v-u$ is hamiltonian


*$N_G(u) \neq N_G(v)$ then at least one vertex of degree 6 is Neighbour of $u$ in $G$ and as the vertex of degree 6 are 3 in $G-v$ then more than one vertex of degree 5 in $G-v$ is Neighbour of $u$ so, one $u$ is removed the degrees will be $d(v) = {6,5,4}$ $ \forall v \in V(G-v-u)$ and $n= 8$ 


$d(v) \geq \frac{n}{2} \rightarrow  {6,5,4} \geq 4$ then $G-v-u$ is hamiltonian
is it ok? or am i wrong? Thanks!!

It looks OK, although the $G-u-v$ case looks like it could be simpler, since it's apparent that the minimum degree that could be found in the reduced graph is $4$ without the need to split into cases, and that is adequate to show that the reduced $8$-vertex graph is Hamiltonian.
