Prove that

(A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)

I am so clueless on solving this... can anyone help me?

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- June 15th 2011, 04:23 AMMichaelLightSets question (Proving)
Prove that

(A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)

I am so clueless on solving this... can anyone help me? - June 15th 2011, 04:48 AMProve ItRe: Sets question (Proving)
Try drawing a Venn Diagram.

- June 15th 2011, 04:51 AMPlatoRe: Sets question (Proving)
- June 15th 2011, 05:52 AMAlso sprach ZarathustraRe: Sets question (Proving)
- June 15th 2011, 06:58 AMemakarovRe: Sets question (Proving)
Consider what intersections you get when you apply the first distributive law that Zarathustra gave above to the left-hand side

(A ∪ B) ∩ (B ∪ C) ∩ (C ∪ A).

Each resulting intersection will have one set from each of the three sets of parentheses. Thus, there will be

A ∩ B ∩ C,

A ∩ B ∩ A,

A ∩ C ∩ C,

A ∩ C ∩ A,

and so on. Note that repetitions can be removed because X ∩ X = X for any set X.

Each intersection will have either two sets (A ∩ B or A ∩ C or B ∩ C) or all three sets (A ∩ B ∩ C). Also, each of A ∩ B, A ∩ C, and B ∩ C will be a part of the left-hand side. Finally, note that A ∩ B ∩ C ⊆ A ∩ B, and if X ⊆ Y, then X ∪ Y = Y, so A ∩ B ∩ C can be omitted from the final union.

All together, this implies that the left-hand side simplifies to (A ∩ B) ∪ (A ∩ C) ∪ (B ∩ C). - June 15th 2011, 07:20 AMPlatoRe: Sets question (Proving)