1. ## Another sets prove

By using the set algebra , prove that , for any sets A and B
A u ( A' n B ) = A u B

Deduce , or using set algebra , show that

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

2. The question test the distributive laws.

A u ( A' n B ) = A u B
from L.H.S : A u ( A' n B )
= ( A U A' ) n ( A u B )
= S n ( A u B )
= A u B ( proven )

You should be able to solve the second part using the law or the procedure listed above

By using the set algebra , prove that , for any sets A and B
A u ( A' n B ) = A u B

Deduce , or using set algebra , show that

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

Deduce , or using set algebra , show that

A u ( A' n B ) u [A' n (A u B') n {B u (B' n C}] = A u B u C
I tried this one but i couldn't get the prove . Can someone pls spot my mistake .

My working :

(A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C

wondering , where is the A .

4. ## Proofs using the Laws of Sets

I tried this one but i couldn't get the prove . Can someone pls spot my mistake .

My working :

(A u B) u (A' n B' ) n ( B u C ) = S n (B u C ) = B u C

wondering , where is the A .
Here's a proof - there might be a quicker one:

$A \cup (A'\cap B)\cup [A'\cap (A\cup B')\cap (B\cup (B'\cap C)]$

$=A\cup B\cup [A'\cap (A\cup B') \cap (B \cup C)]$, using the first result twice

$=A \cup B \cup[\{(A' \cap A) \cup (A'\cap B')\} \cap (B \cup C)]$, Distributive Law

$= A \cup B \cup [\{\oslash \cup (A' \cap B') \} \cap (B \cup C)]$, Complement Law

$= A \cup B \cup [(A' \cap B') \cap (B \cup C)]$, Identity Law

$= A \cup B \cup [(A \cup B)' \cap (B \cup C)]$, De Morgan's Law

$= [(A \cup B) \cup (A \cup B)'] \cap [(A \cup B) \cup (B \cup C)]$, Distributive Law

$= \boldsymbol{\text {U}} \cap [(A \cup B) \cup (B \cup C)]$, Complement Law

$= A \cup (B \cup B) \cup C$, Identity Law, Associative Law

$= A \cup B \cup C$, Idempotent Law