Set Theory Proof: Proving Identity (A U B) ∩ (B U C) ∩ (C U A)

AI Thread Summary
The discussion focuses on proving the set theory identity (A U B) ∩ (B U C) ∩ (C U A) = (A ∩ B) U (B ∩ C) U (C ∩ A). The original poster struggled with the proof and attempted to simplify the left side incorrectly using Venn diagrams. Participants advised on the distribution of union over intersection, suggesting to apply set theory rules iteratively. The original poster successfully completed the proof, albeit in a more complex manner than necessary. This exchange highlights the importance of understanding set operations and their properties in solving set theory problems.
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Hi, I have been trying for a very long time to prove the following set theory identity

(A union B) intersect (B union C) intersect (C union A) = (A intersect B) union (B intersect C) union (C intersect A).

I thought that I could simplify (A U B) intersect (B U C) intersect (C U A)
as [B U (A intersect C) ]intersect (C U A), but venn diagrams show that this is not correct.

Thanks
 
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Do you know how to distribute union over intersection and vice versa?
(AuB)^(BuC) = (A^(BuC))u(B^(BuC))
Since B^(BuC) = B you can simplify that a little.
Just keep applying those rules and the result should drop out.
If you get stuck, post your working so far.
 
Thank you, it worked. I probably did it in a more convoluted manner than required, but I was able to do it in around 12 lines.

I didn't realize that you could distribute like that at first.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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