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

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,

$$A_n = \left( A_n \backslash \bigcup_{i=1}^{n-1} A_i \right) \cup \left( A_n \cap \bigcup_{i=1}^{n-1} A_i \right) = B_1 \cup B_2.$$