how would prove this using for set operations is..... idempotent laws associa........

**npm1** · April 23rd 2010, 07:32 AM

how would i prove this using Set operations
i.e.
the operations i've been given are ........
idempotent laws
associative laws
commutative laws
distributive laws
identity laws
Involution laws
involution laws
complement laws
de morgans laws
$ \emptyset\;=\;((X \cup Y) \cap (X \cup Y')) \cap ((X' \cup Y) \cap (X' \cup Y')) $
using set operations

if not then atleast provide a similar example...............OR
refer me to a website for learning this......

**Soroban** · April 23rd 2010, 08:12 AM

Hello, npm1!

Prove using Set operations:

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

. . $\begin{array}{ccc}\bigg[(X \cup Y) \cap (X \cup Y')\bigg] \cap \bigg[(X' \cup Y) \cap (X' \cup Y')\bigg] & \text{Given} \\ \\ \bigg[X \cup (Y \cap Y')\bigg] \cap \bigg[X' \cup (Y \cap Y')\bigg] & \text{Distr.} \\ \\ \bigg[X \cup\: \emptyset\bigg] \cap \bigg[X' \cup \:\emptyset\bigg] & A \cap A' \:=\:\emptyset \\ \\ X \cap X' & A \cup \emptyset \:=\:A \end{array}$

. . . . . . . . . . . . . . . . $\begin{array}{ccccccc}\emptyset & \qquad\qquad & \qquad\qquad\;\; & A \cap A' \:=\:\emptyset\end{array}$

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