Let $\{A_1,A_2\dots A_n\}$ be a mutually disjoint set of events. Let $\{B_1,B_2\dots B_n\}$ be another mutually disjoint set of events. I am trying to transform the expression $$C=\bigcap\limits_{i\in\{1,2,\dots,n\}}\left(A_i\cup B_i\right)$$ to something equivalent by somehow taking the union symbol out. I think that for $n=2$ it's $$C=(A_1\cap B_2)\cup(B_1\cap A_2)$$ but I cannot generalize it for any $n$.

  • $\begingroup$ @layman That's true but as I stated $\{A_1,A_2\dots A_n\}$ are mutually disjoint. This holds for $\{B_1,B_2\dots B_n\}$, as well. So $(A_1\cap A_2)=\emptyset$ and $(B_1\cap B_2)=\emptyset$ $\endgroup$
    – mgus
    Nov 2 '17 at 19:44

I renamed your variables $A_{1,i}$ and $A_{2,i}$ $$C=\bigcap\limits_{i\in\{1,2,\dots,n\}}\left(A_{1,i}\cup A_{2,i}\right)$$

By generalized distributivity $$C=\bigcup\limits_{i_1\in\{1,2\}} \cdots \bigcup\limits_{i_n\in\{1,2\}} \left(\bigcap\limits_{j\in \{1,2,\cdots,n\}} A_{i_j,j} \right)$$

Which can be written as

$$C=\bigcup\limits_{(i_1,\cdots,i_n)\in\{1,2\}^n} \left(\bigcap\limits_{j\in \{1,2,\cdots,n\}} A_{i_j,j} \right)$$

This is a very ugly formula which means write your n brackets.

Choose one term either A or B from each bracket and take the intersection of those. Then take the union for every possible choice.


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