What is the Possibility of A given B, i.e., $\sigma(A|B)$ in possibility theory?
In this book, definition of possibility condition is explained, but honestly I can't understand it, could somebody explain it by an example?
Let $(\Gamma, Z, \alpha)$ be a possibility space. Given $B$ occurring, we consider the possibility of $A$, i.e., $\sigma(A|B)$. Suppose $\sigma(B)$ and $\sigma(AB)$ are known. They represent possibilities of $B$ and $AB$, respectively. We note $B = (B - AB) \cup AB$ . If $\sigma(B) = \sigma(AB)$, then it can be said that $B$ achieves its realization on $AB$. If $\sigma(B) > \sigma(AB)$, then it can be said that $B$ achieves its realization on $B - AB$ rather than on $AB$. So, given $B$ occurring, if $B$ achieves its realization on $AB$, then $A$ also occurs. In this way there should be $\sigma(A|B) = 1$. If $B$ achieves its realization on $B - AB$, then the occurrence of $B$ makes no difference on occurrence of $AB$. Thus $\sigma(A|B) = \sigma(AB|B) = \sigma(AB)$.
Definition 2.3.5 - Let $(\Gamma, Z, \alpha)$ be a possibility space and $A,B \in Z$. The possibility of $A$ conditional on $B$ is defined as
$$\sigma(A|B)=\begin{cases} 1, & \text{if $\sigma(AB)=\sigma(B)$} \\ \sigma(AB), & \text{if $\sigma(AB)<\sigma(B)$} \end{cases}$$ Here we note that in order to make $\sigma(\bar B|B) = 0$ , there should be $\sigma(B) = 0$.