# Thread: set operations 2

1. ## set operations 2

use the law of algebra of sets to that emptyset=((X union Y) intersection (X union Ycomplement)) intersect ((Xcomplement union Y) intersection (Xcomplement union Ycomplement)) please specify which you use in the derivation....

ANDANDANDANDANDANDANDANDANDANDANDANDANDANDANDANDAN DANDAND

what does it mean by |A * A| where A={1,2,3,4,5}

small explainations would be helpful

2. Hello, npm1!

I don't know the names of your Rules and Theorems.
I'll make up some names:

. . $\begin{array}{cccccc}\text{Rule} & \text{Na{m}e} \\ \hline \\[-3mm]
A \cap A' \:=\:\emptyset & \text{[C1]} \\ \\[-3mm]
A \cup \emptyset \:=\:A & \text{[C2]} \end{array}$

Use the laws of algebra of sets to prove that:

. . $(X \cup Y) \cap (X \cup Y') \cap (X' \cup Y) \cap (X' \cup Y') \;=\;\emptyset$

$\begin{array}{ccccc}(X \cup Y) \cap (X \cup Y') \cap (X' \cup Y) \cap (X' \cup Y') & \text{Given} \\ \\[-3mm]

\bigg[X \cup (Y \cap Y')\bigg] \cap \bigg[X' \cup (Y \cap Y')\bigg] & \text{Distr.} \\ \\[-3mm]

\bigg[X \cup \emptyset\bigg] \cap \bigg[X' \cup \emptyset\bigg] & \text{[C1]} \\ \\[-3mm]

X \cap X' & \text{[C2]} \\ \\[-3mm]

\emptyset & \text{[C1]}
\end{array}
$

3. ## thanks u

very clear
$((X \cup Y) \cap (X \cup Y')) \cap ((X' \cup Y) \cap (X' \cup Y')) \;=\;\emptyset$ (new to forum and don't how to use the math tags)

the laws i've been given for set operations is.....
idempotent laws
associative laws
commutative laws
distributive laws
identity laws
Involution laws
involution laws
complement laws
de morgans laws