Question

Merter 11. Prove that: (X n Y)' = (X'U Y')
​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm :\longmapsto\:x \: \in \: (X\cap Y)'$

$\rm \implies\:x \: \cancel \in \: X\cap Y$

$\rm \implies\:x \: \cancel \in \: X \: or \:x\cancel \in \: Y$

$\rm \implies\:x \: \in \: X' \: or \: x \: \in \: Y'$

$\rm \implies\:x \: \in \: X' \:\cup \: Y'$

$\bf\implies \:(X \: \cap \: Y)' \: \sub \: X' \: \cup \: Y' - - - (1)$

Again, Let assume that

$\rm :\longmapsto\:\:x \: \in \: X' \:\cup \: Y'$

$\rm \implies\:x \: \in \: X' \: or \: x \: \in \: Y'$

$\rm \implies\:x \: \cancel \in \: X \: or \:x\cancel \in \: Y$

$\rm \implies\:x \: \cancel \in \: X\cap Y$

$\rm \implies\:x \: \in \: (X\cap Y)'$

$\bf\implies \:X' \: \cup \: Y' \: \sub \: (X \: \cap \: Y)' - - - - (2)$

From equation (1) and (2), we concluded that

$\bf\implies \:X' \: \cup \: Y' \: = \: (X \: \cap \: Y)'$

Hence, Proved

More to know :-

1. Commutative Law

$\boxed{ \tt{ \: X\cup Y = Y\cup X \: }}$

$\boxed{ \tt{ \: X\cap Y = Y\cap X \: }}$

2. Associative Law

$\boxed{ \tt{ \: (X\cup Y)\cup Z = X\cup (Y\cup Z) \: }}$

$\boxed{ \tt{ \: (X\cap Y)\cap Z = X\cap (Y\cap Z) \: }}$

3. Distributive Law

$\boxed{ \tt{ \: X\cup (Y\cap Z) = (X\cup Y)\cap (X\cup Z) \: }}$

$\boxed{ \tt{ \: X\cap (Y\cup Z) = (X\cap Y)\cup (X\cap Z) \: }}$

4. Complement Law

$\boxed{ \tt{ \: U' \: = \: \phi \: }}$

$\boxed{ \tt{ \: \phi \:' \: = \: U }}$