# Thread: Set - Distributive Law

1. ## Set - Distributive Law

I've came across an example...

By using A-B=A n B', prove that (A u B)-(A n B)=[A-(A n B)] u [B-(A n B)]

The solution they gave is...

(A u B)-(A n B)=(A u B) n (A n B)'
=[A n (A n B)'] u [B n (A n B)'] <---Distributive Law
=[A-(A n B)] u [B-(A n B)]

What I wanna know is what happen at the 2nd step? How to a that by using distributive law?

2. Hello, cloud5!

They "multiplied from the right".

$\displaystyle \underbrace{(A \cup B)}_{(a+b)} \underbrace{\cap}_{\times} \underbrace{ (A \cap B)' }_{c}\;=\;\underbrace{\bigg[A \cap (A \cap B)'\bigg]}_{a\times c} \underbrace{\cup}_{+}$ $\displaystyle \underbrace{\bigg[B \cap (A \cap B)'\bigg]}_{b\times c} \quad\leftarrow\text{ Distributive Law}$

See it?

3. Originally Posted by cloud5
I've came across an example...

By using A-B=A n B', prove that (A u B)-(A n B)=[A-(A n B)] u [B-(A n B)]

The solution they gave is...

(A u B)-(A n B)=(A u B) n (A n B)'
=[A n (A n B)'] u [B n (A n B)'] <---Distributive Law
=[A-(A n B)] u [B-(A n B)]

What I wanna know is what happen at the 2nd step? How to a that by using distributive law?
$\displaystyle A-B=A\cup B'$

$\displaystyle (A\cup B) - ( A\cap B) = \left[A-(A\cap B) \right] \cup [B-(A\cap B) ]$

$\displaystyle (A\cup B ) - (A\cap B)=({\color{red}A}{\color{green}\cup} {\color{blue}B})\cap (A\cup B)'$

$\displaystyle \left[{\color{red}A}\cap (A\cup B)'\right] {\color{green}\cup} \left[{\color{blue}B}\cap (A\cup B)' \right]$

4. Originally Posted by Soroban
Hello, cloud5!

They "multiplied from the right".

$\displaystyle \underbrace{(A \cup B)}_{(a+b)} \underbrace{\cap}_{\times} \underbrace{ (A \cap B)' }_{c}\;=\;\underbrace{\bigg[A \cap (A \cap B)'\bigg]}_{a\times c} \underbrace{\cup}_{+}$ $\displaystyle \underbrace{\bigg[B \cap (A \cap B)'\bigg]}_{b\times c} \quad\leftarrow\text{ Distributive Law}$

See it?

Oh... Now I get it... Thanks for the help~