Decide T/F about vertex and edge connectivity Decide, whether it's true, that:
$\forall $ graph $G$ : if $K(G) \neq \lambda(G) $ then $\exists $ vertex $v$ : $deg(v)\geq 4$
(where $K(G)$ is the vertex-connectivity and $\lambda (G)$ is the edge-connectivity)
So, I have 2 solutions:
1, if I negate it right away I get: 
$ \exists G : K(v) = \lambda(G) : \forall v : deg(v) \leq 3 $  - which is absolutely true because of $K_4$ - so if the negation is true, the original problem should be false.
2, by contradiction I say that $\forall v : deg(v) \leq 3 $ while $K(G) \neq \lambda(G)$ still holds
because of the degrees we can say, that $K(G) \leq \lambda(G) \leq 3$ but I know, that $K(G) \neq \lambda(G)$ so I get $K(G) < \lambda(G) \leq 3$ which seems like a contradiction, but I don't know how to prove it. (of course that would mean that the original sentence is true)
Any help is appreciated.
 A: Your negation of the statement is incorrect. The correct negation would be: There exists a graph $G$ so that $K(G) \neq \lambda(G)$ and the maximum degree of $G$ is at most 3. Remember that the negation of ($p \Rightarrow q$) is ($p$ and not $q$).
Your second strategy is a good start. You know $K(G) < \lambda(G) \leq 3$. Therefore $K(G) \in \{0,1,2\}$. If $K(G) = 0$, then $\lambda(G) = 0$, the graph is already disconnected. If $K(G) = 1$, there is a vertex $v \in V(G)$ so that deleting $v$ disconnects the graph. But $\deg_G(v) \leq 3$, so for some component of $G - v$, it must be the case that $v$ sends at most 1 edge $e$ to that component. But then $e$ is a cut edge, and $\lambda=1$. If $K(G) = 2$, you can do something similar - look at the cut vertices, each must send at most 1 edge to one of the components, and those edges together would be an edge cut.
In fact you can use this same strategy without doing the case analysis - just look at any set of cut vertices $V$ in the graph, as long as the maximum degree is at most 3, each sends at most one edge to one of the components of $G-V$, and that set of edges would be an edge cut.
