1. basic question on sets

How come (A' n B') n [(A u C) n (B' n C')] = (A' n B') n (A u C) n (B' n C') ?

Is it by the commutative laws ? THese brackets and parantheses are really confusing me .

Isnt that by distributive

(A' n B') n [(A u C) n (B' n C')] = [(A' n B')n(A u C)] n [(A' n B') n (B' n C')]

and it doesn't equal to (A' n B') n (A u C) n (B' n C') from the above .

I really thank you for your time and effort to explain this to me .

2. Originally Posted by thereddevils
How come (A' n B') n [(A u C) n (B' n C')] = (A' n B') n (A u C) n (B' n C') ?

Is it by the commutative laws ? THese brackets and parantheses are really confusing me .

Isnt that by distributive

(A' n B') n [(A u C) n (B' n C')] = [(A' n B')n(A u C)] n [(A' n B') n (B' n C')]

and it doesn't equal to (A' n B') n (A u C) n (B' n C') from the above .

I really thank you for your time and effort to explain this to me .
Its because the Intersection operator is associative:

$\displaystyle A\cap\left(B\cap C\right)=A\cap B\cap C=\left(A\cap B\right)\cap C$

You only distribute in cases like $\displaystyle A\cap\left(B\cup C\right)$

3. Originally Posted by Chris L T521
Its because the Intersection operator is associative:

$\displaystyle A\cap\left(B\cap C\right)=A\cap B\cap C=\left(A\cap B\right)\cap C$

You only distribute in cases like $\displaystyle A\cap\left(B\cup C\right)$

Oh thanks a lot !

So A n (B n C) doesnt equal to (A n B) n (A n C) as distributive laws don work on cases like this ??

4. Originally Posted by thereddevils
Oh thanks a lot !

So A n (B n C) doesnt equal to (A n B) n (A n C) as distributive laws don work on cases like this ??
Well it works, but $\displaystyle A\cap\left(B\cap C\right)=\left(A\cap B\right)\cap\left(A\cap C\right)=A\cap B\cap A\cap C=A\cap A\cap B\cap C$ and $\displaystyle A\cap A=A$ so you get $\displaystyle \left(A\cap B\right)\cap \left(A\cap C\right)=A\cap B\cap C$.

Its just extra work.