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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- Jun 15th 2011, 03:23 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? - Jun 15th 2011, 03:48 AMProve ItRe: Sets question (Proving)
Try drawing a Venn Diagram.

- Jun 15th 2011, 03:51 AMPlatoRe: Sets question (Proving)
- Jun 15th 2011, 04:52 AMAlso sprach ZarathustraRe: Sets question (Proving)
- Jun 15th 2011, 05:58 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). - Jun 15th 2011, 06:20 AMPlatoRe: Sets question (Proving)
As I said above , I prefer a 'pick-a-point' proof.

Suppose that $\displaystyle t \in \left[ {(A \cup B) \cap (A \cup C) \cap (B \cup C)} \right]$

That means all of three are true:

$\displaystyle t \in (A \cup B)\; \Rightarrow \;t \in A \vee t \in B.~~\color{blue}[1]$

$\displaystyle t \in (A \cup C)\; \Rightarrow \;t \in A \vee t \in C.~~\color{blue}[2]$

$\displaystyle t \in (B \cup C)\; \Rightarrow \;t \in B \vee t \in C.~~\color{blue}[3]$

If $\displaystyle t\in A\cap B\cap C$ the clearly it is RHS.

So suppose that $\displaystyle t\notin A$. The by [1] & [2] we $\displaystyle t\in B\wedge t\in C$.

So it belongs to the RHS.

That is a sample of a different approach.