discrete structure ,Graphs 
A graph is called good graph if there are two spanning tree with no common edges between them...
  If $G=(V,E)$ is a good graph and $G$ has two vertices that have more than $\frac{n+2}2$ neighbors, prove that there is an edge or a vertex that can be removed and $G$ would still be a good graph. 

I didn't get advanced a lot in the solution, what I reached is that the there is no vertex with one neighbor... Because if that happen the edge between this vertex and its neighbor is in both of the trees which can't be...
For the two edges with more than $\frac{n+2}2$ neighbors there are more than two common neighbors at least (according to pigeons in holes). I know I need to look at them but I couldn't reach any further... Please help  and thank you.
 A: At first note that if a vertex $u \in V(G)$ has more than $\frac{n + 2}{2}$ neighbours then $\deg u \ge \frac{n + 3}2$ since $\deg u$ is an integer. The graph $G$ is not empty, because it has two vertices with positive degree. Then there is no vertex of degree $0$ or $1$, since there are two edge-disjoint spanning trees. If there is a vertex $v$ of degree $2$ then $v$ is a leaf in both edge-disjoint spanning trees, thus $G - v$ is a good graph. If $\deg v \ge 3$ for all $v \in V(G)$, then taking into account two vertices of degree at least $\frac{n + 3}2$ we get that
$$|E(G)| = \frac12\sum_{v \in V(G)} \deg v \ge \frac12\left((n - 2) \cdot 3 + 2 \cdot \frac{n + 3}2\right) = \frac12 (4n - 3).$$
Since $|E(G)|$ is an integer then $|E(G)| \ge \left\lceil \frac{4n - 3}2\right\rceil = 2n - 1$. Each spanning tree has $n - 1$ edges for $2n - 2 < |E(G)|$ edges in total for the pair of edge-disjoint spanning trees. Therefore there is an edge that doesn't belong to either spanning tree of  the pair. We can remove this edge keeping graph being good.
