I have a bipartite graph $(V_1\cup V_2, E)$ on which we define $\sigma: V_1\rightarrow\mathbb{Z}$ such that $\sigma(W)=|W|-|\Gamma W|$, where $\Gamma$ gives us the points (from $V_2$, since its a bipartite graph) adjacent to at least one point is $W$.
Now I am asked to prove $\sigma(W_1\cap W_2) + \sigma(W_1\cup W_2) \geq \sigma(W_1) + \sigma(W_2)$.
I have done the following:
$\sigma(W_1\cap W_2) + \sigma(W_1\cup W_2)$
$= |W_1 \cap W_2| - |\Gamma (W_1\cap W_2)| + |W_1 \cup W_2| - |\Gamma (W_1\cup W_2)|$
$= |W_1| + |W_2| - |\Gamma (W_1\cap W_2)| - |\Gamma (W_1\cup W_2)|$ using that $|W_1 \cap W_2| + |W_1 \cup W_2| = |W_1| + |W_2|$ for this last equality.
Now I think that all I have left to prove is that $- |\Gamma (W_1\cap W_2)| - |\Gamma (W_1\cup W_2)| \geq - |\Gamma (W_1)| - |\Gamma (W_2)|$, but I'm not sure how to go about this.
First note that if $A\subseteq B$ then $\Gamma (A)\subseteq\Gamma(B)$. This gives us $$\Gamma (W_1\cap W_2)\subseteq \Gamma(W_1)\cap \Gamma(W_2)$$ and $$\Gamma(W_1)\cup\Gamma(W_2)\subseteq \Gamma(W_1\cup W_2).$$ But if $v\in \Gamma(W_1\cup W_2)$, then $v$ is adjacent to at least one vertex in $W_1\cup W_2$ and hence $v$ is adjacent to a vertex in $W_1$ which implies $v\in \Gamma (W_1)$, OR $v$ is adjacent to a vertex in $W_2$ which implies $v\in \Gamma(W_2)$. This gives $v\in \Gamma(W_1)\cup\Gamma (W_2)$ and therefore $$\Gamma(W_1)\cup\Gamma(W_2)=\Gamma(W_1\cup W_2).$$
Hence $$|\Gamma(W_1)|+|\Gamma(W_2)|=|\Gamma(W_1)\cup\Gamma(W_2)|+|\Gamma(W_1)\cap\Gamma(W_2)|$$$$=|\Gamma(W_1\cup W_2)|+|\Gamma(W_1)\cap\Gamma(W_2)|\geq |\Gamma(W_1\cup W_2)|+|\Gamma(W_1\cap W_2)|.$$