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Use the distributivity law to prove the following statement:

$$(A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A) = (A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A)$$

I thought about using De'Morgan's Law to prove the LHS but the problem requires I use the distributive law of sets to prove it. How can you solve this?

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    You could prove this with a Venn diagram directly, actually. But I understand you're asking about a specific proof method. – 2017-02-01
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    Use idenpotence. Distribute pairs and cancel like terms. Recall $X=X\cup X$ and $Y\cap Y=Y$ – 2017-02-01

2 Answers 2

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$(A\cap B)\cup (B\cap C)\cup(C\cap A)=$

$(B\cap (A\cup C))\cup(C\cap A) =$

$(B\cup C) \cap (B \cup A) \cap (A\cup C \cup C)\cap (A\cup C \cup A) =$

$(B \cup A) \cap (B \cup C) \cap (A \cup C) \cap (A \cup C) =$

$(A \cup B) \cap (B \cup C) \cap (C\cup A)$

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Distribute pairs at a time, use idempotence and absorption to eliminate redundancy.

Begin thusly: $\quad(A ∩ B) ∪ (B ∩ C) ∪ (C ∩ A) \\ = \big((A\cup B) \cap (A\cup C) \cap B \cap (B\cap C)\big) \cup (C\cap A) \\ = ((A\cup C)\cap B)\cup (C\cap A) \\ = $

Keep going.   You can do it if you try.