Thread: 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 .

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:



You only distribute in cases like

Originally Posted by Chris L T521

Its because the Intersection operator is associative:



You only distribute in cases like

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 ??

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 and so you get .

Its just extra work.