prove that $A_n$ is the disjunct union of $B_1$ and $B_2$ prove that $A_n$ is the disjunct union of $B_1$ and $B_2$
Suppose we have $B_1 = \left(\cup_{i=1}^{n} A_i\right) \backslash \left(\cup_{i=1}^{n-1} A_i\right)$ and $B_2 = \cup_{i=1}^{n-1} \left(A_i \cap A_n\right)$.
The disjunction is not that hard, but i don't seem to see why the union is equal to $A_n$.
I've tried writing $B_1$ with demorgan's law but i failed :(.
Any hints or tricks for these type of questions?
Kees
 A: If $x\in B_1 \cup B_2$, then either $x\in B_1$ or $x\in B_2$. If $x\in B_1$, then $x \in A_i$ for some $i$, but $x\notin A_j$ for $1 \le j \le n-1$. So $x\in A_n$. If $x \in B_2$, then $x\in A_i \cap A_n$ for some $i$. Since $A_i\cap A_n \subseteq A_n$, $x\in A_n$. Therefore $B_1\cup B_2 \subseteq A_n$. Conversely, if $x\in A_n$, then either $x$ is in $A_i$ for some $i < n$, or $x$ is not in any $A_i$ for $i < n$. If the former case, $x\in B_2$, and in the latter case, $x\in B_1$. So $x\in B_1\cup B_2$. Thus, $A_n \subseteq B_1 \cup B_2$.
A: $$B_1=\left(\cup_{i=1}^{n} A_i\right) \backslash \left(\cup_{i=1}^{n-1} A_i\right)=
\left(A_n \cup (\cup_{i=1}^{n-1} A_i)\right) \backslash \left(\cup_{i=1}^{n-1} A_i\right) = 
A_n \backslash \left(\cup_{i=1}^{n-1} A_i\right)$$ so 
$$B_1\cup B_2 =(A_n \backslash \left(\cup_{i=1}^{n-1} A_i\right)) \cup (\cup_{i=1}^{n-1} \left(A_i \cap A_n\right)) =A_n$$
A: Note that $$B_2 = \bigcup_{i=1}^{n-1} (A_i \cap A_n) = \left(\bigcup_{i=1}^{n-1} A_i \right) \cap A_n$$
and 
$$\begin{align*} B_1 &= \left( \bigcup_{i=1}^n A_i \right) \backslash \left( \bigcup_{i=1}^{n-1} A_i \right) \\ &= \left( \bigcup_{i=1}^{n-1} A_i \cup A_n \right) \backslash \left( \bigcup_{i=1}^{n-1} A_i \right) \\ &= \underbrace{\left[ \left( \bigcup_{i=1}^{n-1} A_i \right) \backslash \bigcup_{i=1}^{n-1} A_i \right]}_{\emptyset} \cup \left[ A_n \backslash \bigcup_{i=1}^{n-1} A_i \right] \\ &= A_n \backslash \left( \bigcup_{i=1}^{n-1} A_i \right). \end{align*}$$
Hence,
$$A_n = \left( A_n \backslash \bigcup_{i=1}^{n-1} A_i \right) \cup \left( A_n \cap \bigcup_{i=1}^{n-1} A_i \right) = B_1 \cup B_2.$$
