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Given three set $A$, $S$, and $L$. How to prove that $$|A\cap S'\cap L'|=|A|-|A\cap S|-|A\cap L| + |A\cap S\cap L|$$ by using inclusion exclusion principle ? (without the aid of Venn Diagram)

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  • $\begingroup$ Do you know the laws for manipulating unions and intersections? The principle itself? $\endgroup$ – Parcly Taxel May 21 at 16:08
  • $\begingroup$ Can you explain briefly to me about it? $\endgroup$ – Ling Min Hao May 21 at 16:14
  • $\begingroup$ Did you mean to write $|A \cap S \cap L|$ on the Left hand side? $\endgroup$ – OneAndOnlyDaniel May 21 at 16:35
  • $\begingroup$ No, it is $|A\cap S'\cap L'|$ where $S'$ is the complement for $S$. Same goes to $L'$. $\endgroup$ – Ling Min Hao May 21 at 16:38
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Hint:

It is well known that: $|A|=|A\cap B|+|A\cap B'|$

Now, try to use this with $(S'\cap L')$ in place of $B$.

$\begin{array}{rl}|A|&= |A\cap (S'\cap L')| + |A\cap (S'\cap L')'|\\&=|A\cap S'\cap L'| +|A\cap (S\cup L)|\\&=|A\cap S'\cap L'|+|(A\cap S)\cup (A\cap L)|\\&\vdots\end{array}$

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Step 1: Remove the members of $A\cap S$ from $A.$ You now have a set $T$ with $|A|-A\cap S|$ members.

Step 2: To get from $T$ to $A\cap S'\cap L'$ you must remove, from $T,$ the members of $A$ that belong to $L$ that were not removed from $A$ in Step 1. That is , remove the members of $A\cap L$ that do not belong to $S$. That is, remove the members of $A\cap L$ that do not belong to $(A\cap L)\cap S.$ There are $|A\cap L|-|A\cap L\cap S|$ of these.

Therefore $$|A\cap S'\cap L'|=|T|-(\,|A\cap L|-|A\cap L\cap S| \,)=$$ $$=|A|-|A\cap S|-(\,|A\cap L|-|A\cap L\cap S|\,)=$$ $$=|A|-|A\cap S|-|A\cap L|+|A\cap L\cap S|.$$

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