Vertex connectivity question I'm currently learning about vertex connectivity and I'm having a bit of trouble understanding some of the terms/definitions, namely "local connectivity" and "$k$-connected" and "connectivity of $G$" (definition $2$, $3$ and $4$) - 
I tried finding the vertex connectivity of the following graph $G$:

I know that you can also find the vertex connectivity by finding the minimum cardinality of a vertex cut of $G$ which is clearly one, but I want to know how to get this answer by looking at internally disjoint paths.
If the minimum cardinality of a vertex cut of $G$ is one then does that mean there should be one pairwise internally disjoint $(x,y)$-path in $G$ (where $x,y$ are distinct vertices from the graph $G$)?
 A: Let's compute $p(u, v)$ for all pairs of distinct vertices $u$ and $v$:
$$\begin{array}{|c|c|c|c|c|c|c|}
\hline
v\backslash u & a & b & c & d & e & f\\\hline
a & - & 2 & 2 & 1 & 1 & 1 \\\hline
b & 2 & - & 2 & 1 & 1 & 1 \\\hline
c & 2 & 2 & - & 1 & 1 & 1 \\\hline
d & 1 & 1 & 1 & - & 2 & 2 \\\hline
e & 1 & 1 & 1 & 2 & - & 2 \\\hline
f & 1 & 1 & 1 & 2 & 2 & - \\\hline
\end{array}$$
So it is easy to see that $\kappa(G) = \min\{\,p(u, v) \colon u, v \in V(G), u \ne v\,\} = 1$.
A: Yes, if $G$ is not complete then the minimum cardinality of a vertex cut is $k$ if and only if there is some pair of vertices $x,y$ such that $p(x,y)=k$. This is Menger's theorem. (We usually count a set of $n-1$ vertices as a vertex cut for this purpose, so that Menger's theorem also works in the complete graph case; the connectivity of $K_n$ is $n-1$ and any two vertices have exactly $n-1$ vertex-disjoint paths.)
In your case any choice of $x$ from the left-hand side of your graph and $y$ on the right-hand side will work. For example, $p(a,d)=1$ because every path from $a$ to $d$ must pass through $c$, so you can't have two vertex-disjoint paths.
