Bipartite Graph and Menger's Theorem Menger's Theorem can be stated as follows: LINK
Menger's Theorem.  Let $u$ and $v$ be nonadjacent vertices in a graph $G$. The minimum number of vertices in a $u-v$ separating set equals the maximum number of internally disjoint $u-v$ paths in $G$.
I have a bipartite graph $G=(V_1,V_2,E)$. Can I say the following from Menger's Theorem:
Let $v_1, v_2$ are any two vertices from $V_2$. Then there are at most $k$ paths between $v_1,v_2$ such that these are paths do not share any vertex from $V_1$ if and only if at least $k$ vertices from $V_1$ has to remove in order to divide $G$ into two connected component where $v_1, v_2$ are contained in different components.
Is there any other name for this problem? If it is not followed from Menger's Theorem, then any idea how can I show this?
 A: Let $u,v \in V_2$ and denote by $\mathcal{P}$ the set of all $u \leftrightarrow v$ paths. Moreover, let $V_i(P)$ be the set of vertices of $V_i$ that path $P$ uses. Then, you can formalize Menger's theorem like this:
\begin{align}
&\min\Big\{ |V'| \ \Big|\ V' \subseteq V, \  V'\text{ separates }u\text{ and }v\Big\} = \\ 
&\hspace{50pt} = \max\Big\{|\mathcal{P}'|\ \Big|\ \mathcal{P'}\subseteq\mathcal{P},\ \forall P,Q \in \mathcal{P}'.\ V(P)\cap V(Q) = \varnothing\Big\}
\end{align}
What you want to prove is:
\begin{align}
&\min\Big\{ |V_1'| \ \Big|\ V'_1 \subseteq V_1, \  V_1'\text{ separates }u\text{ and }v\Big\} \stackrel{{\color{red}?}}{=} \\ 
&\hspace{50pt}\stackrel{{\color{red}?}}{=}\max\Big\{|\mathcal{P}'|\ \Big|\ \mathcal{P'}\subseteq\mathcal{P},\ \forall P,Q \in \mathcal{P}'.\ V_1(P)\cap V_1(Q) = \varnothing\Big\}.
\end{align}
Although I dislike overformalization very much, here it can help us a bit.
What we can observe immediately is that
\begin{align}
&\min\Big\{ |V_1'| \ \Big|\ V'_1 \subseteq V_1, \  V_1'\text{ separates }u\text{ and }v\Big\} {\color{red}\geq} \\
{\color{red}\geq}&\min\Big\{ |V'| \ \Big|\ V' \subseteq V, \  V'\text{ separates }u\text{ and }v\Big\} = \\ 
= &\max\Big\{|\mathcal{P}'|\ \Big|\ \mathcal{P'}\subseteq\mathcal{P},\ \forall P,Q \in \mathcal{P}'.\ V(P)\cap V(Q) = \varnothing\Big\} {\color{red}\leq} \\
{\color{red}\leq}&\max\Big\{|\mathcal{P}'|\ \Big|\ \mathcal{P'}\subseteq\mathcal{P},\ \forall P,Q \in \mathcal{P}'.\ V_1(P)\cap V_1(Q) = \varnothing\Big\}
\end{align}
only because the conditions in the fist set are stronger than in the second (that is, more restrictive, the smallest-cardinality set $V'$ might not satisfy requirements of the first set), and in the fourth set are weaker than in the third (less requirements implies possibly larger sets). Yet, because the directions of the inequalities doesn't match that does not tell us anything.
In other words, you cannot naively use Menger's theorem to get your claim.
Of course, if you are wondering whether you can use it at all, then technically yes, because if $\alpha \to \beta$ is true, then $(\alpha \land \gamma) \to \beta$ is true independent of what $\gamma$ is, it could be Menger's theorem ;-)
However, what I think you are asking is whether we can use Menger's theorem sensibly, meaning to offload some part of the proof on its proverbial shoulders. Fortunately, the answer here is yes as well, we only need to use a different graph.
Hint:
Define $G'$ by $V' = V_1 \cup \{u,v\}$ and 
$$E' = E[V'] \cup \Big\{\{x,y\}\ \Big|\ x,y \in V_1,\ \mathcal{N}(x)\cap\mathcal{N}(y) \neq \varnothing\Big\},$$
that is, we take vertices of $V_1$ and join them via an edge if in $G$ they shared a vertex from $V_2$, and then we add $u$ and $v$ together with their edges.
In that graph we could use Menger's theorem, to get bounds on the size of separating set and path set in $G'$. It is quite easy to see, that a $u\leftrightarrow v$ separating set in $G'$ works also in $G$, and any such separating set in $G$ which is contained in $V_1$ works also in $G'$. The reason is that, if it doesn't separate $u$ from $v$, then there exists a path, and then you can map that path to the other graph, both from $G$ to $G'$ and from $G'$ to $G$.
The path set is a bit more problematic, because you have to handle vertices of $V_2$ (add or remove them when mapping), but one can do it in a similar manner – I encourage you to handle this part of the proof yourself.
I hope this helps $\ddot\smile$ 
