Solving problems related to Marginal Contribution Nets
I got the following answer:


$v(\emptyset)=0$
$v(\{a\}) = 0$
$v(\{b\}) = 2-2 = 0$
$v(\{a,b\}) = 5+2-2 = 5$
$v(\{a,b,c\}) = 5+2+4 = 11$


Here's a general approach to doing this calculation.
Consider a set $A\subseteq \{a,b,c\}$. Define a map $I_A:\{a,b,c\}\to\{\text{True},\text{False}\}$  as follows:
$$I_A(x) = \left\{ \begin{array}{l@{~~}l} \text{True} &amp; \mbox{if}~x\in A \\
\text{False} &amp; \mbox{otherwise}
\end{array}\right.$$
This is the so-called indicator function of set $A$.

Now take each element of the marginal contribution net and evaluate whether it is true or false given $I_A$. If it is true, add the contribution. If not, add no contribution. More explicitly, given element $\psi\to c$, add $c$ only if $I_A\models\psi$ holds.

Here the notation $I_A\models\psi$ simply means evaluate the truth of the formula $\psi$ given the assignment of the variables $\{a,b,c\}$ to true or false.

Consider $v(\{a,b\})$. This corresponds to the map 
$I_{\{a,b\}}=\{a \mapsto\text{True},b\mapsto\text{True},x\mapsto\text{False}\}$. Given this assignment of truth values, we can conclude that:


$a\wedge b$ is true, so $5$ is contributed.
$b$ is true, so $2$ is contributed.
$c$ is false, so nothing is contributed.
$b\wedge\neg c$ is true, so $-2$ is contributed.


Therefore the total contribution is $5+2-2=5$.