Inclusion-exclusion Principle for three different sets

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)

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

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 $$B$$.

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

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