**Q- Implement 2 variable XNOR gate using NOR only in minimum number of gates.**

F= x XNOR y = (x XOR y)’

F’= (x XOR y) = x’y+xy’ = x’y+xy’+xx’+yy’ = (x+y) (x’+y’)

**F’=** x’(x+y) + y’(x+y)

Take compliment

F= (x’(x+y) + y’(x+y))’ = (x’(x+y) )’ . (y’(x+y))’ =(x+(x+y)’). (y + (x+y)’)

Take compliment again

F’= ( (x+(x+y)’). **(**y + (x+y)’) )’ = (x+(x+y)’)’ + (y + (x+y)’)’

Take compliment again

F= [ (x+(x+y)’)’ + (y + (x+y)’)’ ]’

Now we can implement the circuit of XNOR gate using NOR gates

**Hence we need only 4 NOR GATES to implement XNOR gate and we require 5 NOR gates to implement XOR gate with 5 ^{th} NOR gate used as inverter and placed in front of 4^{th} NOR gate.**