Morris Mano Edition 3 Exercise 4 Question 25 (Page No. 151)

Question

prove that $ x’ \oplus y = x \oplus y’ = (x \oplus y)’ = xy+x’y’$

Answer 1

we know that ${\color{Magenta}{{x\oplus y}=x\overline{y}+\overline{x}y}}$

${{\overline{x}\oplus y}=\overline{x}\cdot\overline{y}+\overline{\overline{x}}\cdot y}$   

$\overline{x}\oplus y=\overline{x}\cdot\overline{y}+x\cdot y$             $[\overline{\overline{A}}=A]$

$\overline{x}\oplus y=x\cdot y+\overline{x}\cdot\overline{y}$

 
$x\oplus\overline{y}=x\cdot\overline{\overline{y}}+\overline{x}\cdot \overline{y}$

$x\oplus\overline{y}=x\cdot y+\overline{x}\cdot \overline{y}$

 
$\overline{x\oplus y}=\overline{x\overline{y}+\overline{x}y}=(\overline{x}+\overline{\overline{y}})\cdot(\overline{\overline{x}}+\overline{y})=(\overline{x}+y)\cdot(x+\overline{y})=\overline{x}\cdot\overline{y}+xy=xy+\overline{x}\cdot\overline{y}=x\odot y$       $ [A\cdot \overline{A}=0]$

Hence we can write

${\color{DarkOrange}{\overline{x}\oplus y=x\oplus\overline{y}=\overline{x\oplus y}=xy+\overline{x}\cdot\overline{y}} }$

Answer 2

we can solve using truth table

all of these operation are EX-NOR