Confusion about the proof of Menger's Theorem in "Introduction to Graph Theory" by Douglas West The proof of Menger's Theorem in the book "Introduction to Graph Theory" by Douglas West (2nd Edition; Page 167) has been divided into two cases. 
The second case assumes that 

"Every minimum $x,y$-cut is $N(x)$ or $N(y)$",

where $N(x)$ denote the set of neighbors of $x$. 
However, it seems that the graph for Case 2 (see below) in the illustration does not satisfy this assumption. What is going on here?

 A: The illustrations will match up with the cases, if we change the descriptions of the cases to:
Case 1'. $G$ has a minimum $x,y$-cut $S$ not contained in $N(x) \cup N(y)$.
Case 2'. Every minimum $x,y$-cut is contained in $N(x) \cup N(y)$.
If we follow these descriptions, then the proof still works, because the only way we use the assumption in case 2 is to say that every vertex outside $\{x\} \cup N(x) \cup N(y) \cup \{y\}$ is in no minimum $x,y$-cut, and this is still true in case 2'. (Since case 1' is a subcase of case 1, there is nothing to worry about there.)
In general, whenever $G$ falls under both case 1 and case 2' (that is, every minimum $x,y$-cut is contained in $N(x) \cup N(y)$, but there is some minimum $x,y$-cut $S$ not equal to $N(x)$ or $N(y)$) then we can handle $G$ by the argument from either case, which is where this flexibility comes from.

Pedagogical note: when I taught this proof last year, I began by considering the case where $N(x) \cap N(y) = \varnothing$ and $V(G) = \{x\} \cup N(x) \cup N(y) \cup \{y\}$, which falls under case 2' and is the case where we can apply König-Egerváry. Then I dealt with the three possibilities below:


*

*$v \in N(x) \cap N(y)$, which is handled in the case 2 proof. (Delete $v$, reducing $\kappa(x,y)$ by $1$.)

*$v \notin \{x\} \cup N(x) \cup N(y) \cup \{y\}$, but $v$ is not part of any minimum $x,y$-cut, which is also handled in the case 2 proof. (Delete $v$, not changing $\kappa(x,y)$.)

*There is a minimum $x,y$-cut $S$ not contained in $N(x) \cup N(y)$. (This is case 1', and we can apply the case 1 proof.)


In some sense, this is the logical progression: we apply König-Egerváry for some cases, and then show that all other cases can be reduced to smaller ones.
