1. ## Sets 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?

2. ## Re: Sets question (Proving)

Try drawing a Venn Diagram.

3. ## Re: Sets question (Proving)

Originally Posted by MichaelLight
Prove that
(A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)
That is just a messy, messy proof. No one likes doing it.
Here is a suggestion. Consider any point from the LHS. Show it must belong to the union on the RHS.
Then reverse the order: RHS to LHS.

4. ## Re: Sets question (Proving)

Originally Posted by MichaelLight
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?
It's known that:

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

and

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

5. ## Re: 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).

6. ## Re: Sets question (Proving)

Originally Posted by MichaelLight
Prove that
(A U B)∩(B U C)∩(C U A) = (A ∩ B)U(A ∩ C)U(B ∩ C)
As I said above , I prefer a 'pick-a-point' proof.
Suppose that $t \in \left[ {(A \cup B) \cap (A \cup C) \cap (B \cup C)} \right]$
That means all of three are true:
$t \in (A \cup B)\; \Rightarrow \;t \in A \vee t \in B.~~\color{blue}[1]$

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

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

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

So suppose that $t\notin A$. The by [1] & [2] we $t\in B\wedge t\in C$.
So it belongs to the RHS.

That is a sample of a different approach.