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$A$,$B$,$C$, are subsets of a set $S$. Prove the following set identity using the laws of set theory (set identities)

$\left(A\cup C\right)\cap[\left(A\cap B\right)\cup\left(C'\cap B\right)]=A\cap B$

I'm not exactly sure how to use the identities to solve this problem, and I'm having trouble understanding how to properly apply the identities. Can you guys help guide me in solving this problem?

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  • $\begingroup$ Where's the identity? $\endgroup$ – user2345215 Oct 27 '14 at 0:03
  • $\begingroup$ @user2345215 updated! $\endgroup$ – hax0r_n_code Oct 27 '14 at 0:04
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First use one of the distributive properties: $$(A\cap B)\cup(C'\cap B) = (A\cup C')\cap B$$ Next use the associative property of intersection: $$(A\cup C)\cap((A\cup C')\cap B) = ((A\cup C)\cap(A\cup C'))\cap B$$ Use the other distributive property: $$(A\cup C)\cap (A\cup C') = A\cup(C\cap C')$$
You should know that $C\cap C'=\varnothing$ and $A\cup\varnothing=A$, combine these steps to yield the result.

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  • $\begingroup$ I'm sorry, I studied your answer and I don't understand how the distributive law is working here. $\endgroup$ – hax0r_n_code Oct 27 '14 at 0:32
  • $\begingroup$ @inquisitor I don't know what's your exact definition, but if the problem is that $B$ acts from the right and not left, you can easily derive that from the left handed distributive property which says $A\cap(B\cup C) = (A\cap B)\cup (A\cap C)$. If you change variables it becomes $B\cap (A\cup C') = (B\cap A) \cup (B\cap C')$, now use the commutativity of $\cap$ in all 3 places to get the first identity. $\endgroup$ – user2345215 Oct 27 '14 at 8:34

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