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?
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).
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.