Perfect matching in a graph Assuming I have a bipartite graph with the following property:
for each subgroup of nodes $s \subseteq {V} $ : 
$$
  \sum_{v\epsilon N(S),z\epsilon N(N(S)) }{} {(v,z) \geq 2\left \| S \right \|}
$$
Where $N(S)$ is the neighbourhood of $A$.
i.e. if you go over each of $S$ neighbors and count the edges coming out from each one. the number of edges is grater than or equal to twice the size of the subgroup S. 
How can I show that this graph has a double matching, where double matching is a graph with 2 different perfect matching.
I know it is true by intuition but I can't seem to find a way to prove it formally.
EDIT:
I tried to use Hall's theorem, but it doesn't seem to be correct, because I am counting the number of edges and as a result I am not getting the size of the |N(S)|, because some of the vertices counted more than once.
assuming I have the graph $V= \{v_1, v_2 , u_1, u_2\}$. and $v_1$ is connected to $u_1$ and $u_2$, and $v_2$ is connected to $u_1$ and $u_2$. this graph has the property I mentioned above, and it's pretty obvious that there is a double matching in here.
But if you look at the size of the neighborhood of $S=\{v_1, v_2\}$, it will be $|{u1,u2}| = 2$
and not $2|S|= 4$ as Hall's theorem requires (because when you are counting the vertices you count $u_1$ and $u_2$ twice. once as $v_1$1 neighbors, and once as $v_2$ neighbors).
Also if I try to use Hall's theorem I nned to 1. choose a first perfect matching 2. remove it, and see that there is still a perfect matching (i.e. the is a double matching)
but in this case we can choose a perfect matching that will leave us without a choice for the second matching.
Please advice..
 A: Edit:
Ok, I was wrong, I think that your conjecture is false. As mentioned in comments, I assume the following property
$$\sum_{v\in N(S),\ z\in N(N(S)) }(v,z) \geq 2 |S| \tag{$\spadesuit$}$$
Consider the following bipartite graph (obviously it doesn't even have a single matching):

Let $S_\bullet$ be the part of $S$ that contains black vertices, and $S_\circ$ the rest. If $S_\bullet$ is not empty, then sum contains at least all the blue edges, hence $8 \geq 2 |S_\bullet|$. Also, if $S_\circ$ contains any of the first two white vertices, then again $8 \geq 2 |S_\circ|$ because of the blue edges. Moreover, if $S_\circ$ does not contain any of the first two white vertices, then $4 \geq 2 |S_\circ|$. Finally, because of how the left-hand side sum is stated, the blue edges induces by $S_\circ$ and $S_\bullet$ will be counted appropriate number of times, that is one for $v \in N(S_\bullet)$ and one for $v \in N(S_\circ)$, hence, condition $(\spadesuit)$ is satisfied.
I hope this helps now ;-)
