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${P(A}{\cup}{B}{\cup}{C)}={P(A)+P(B)+P(C)}-{P(A{\cap}B)}-{P(A{\cap}C)}-{P(B{\cap}C)}+{P(A{\cap}B{\cap}C)}$

Proof:

${P(A{\cup}B{\cup}C)}={P(A{\cup}(B{\cup}C))}$ Associativity

${P(A{\cup}(B{\cup}C))}={P(A)+P(B{\cup}C)-P(A{\cap}(B{\cup}C))}$ Well-known theorem

${P(A{\cap}(B{\cup}C)=P(A{\cap}B){\cup}P(A{\cap}C)}$ Distributive

Now I should apply the well-known theorem again.

${P(A{\cap}B){\cup}P(A{\cap}C)=P(A{\cap}B)+P(A{\cap}C)-P((A{\cap}B){\cap}P(A{\cap}C))}$

I'd be finished if I could show this:

${P((A{\cap}B){\cap}P(A{\cap}C))}=P(A{\cap}B{\cap}C)$

Which I could easily justify using a Venn Diagram (which is not acceptable). How else could I do this?

EDIT: The duplicate question doesn't help me. The answer says "and since (A∩C)∩(B∩C)=A∩B∩C, the result is proven." but this is not explained and I DONT want to use a Venn Diagram.

Sydney
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1 Answers1

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https://en.wikipedia.org/wiki/Algebra_of_sets

If you indeed want to play around with the commutative laws and associative laws, here we go:

$$ \begin{align} (A \cap C) \cap (B \cap C) &= A \cap (C \cap (B \cap C)) \tag{Associative Law} \\ &= A \cap ((B \cap C) \cap C) \tag{Commutative Law} \\ &= A \cap (B \cap (C \cap C)) \tag{Associative Law} \\ &= A \cap (B \cap C) \\ \end{align}$$

BGM
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    It is interesting that you think you need to rewrite your entire proof, I do not see why. Maybe your concept start to go wrong at line $3$: $P(A \cap (B \cup C)) = P((A \cap B) \cup (A \cap C))$ (distributive law). At first glance I thought you are just a typo to type the extra $P$ inside, but now it seems that it is a conceptual mistake you have made. Also note that the union operator is usually applied on set, but not numbers. – BGM Jan 22 '19 at 06:34