EX-NOR:
F=(A'B + AB')' [EX-NOR=Complement of EX-OR]
(A'B + AA' + BB' + AB') [XX'=0 and X + 0 =X]
( ( ( A'(A+B) )' )' + ( ( B'(A+B) )' )' )' [Take double complement as (X')'=X]
( ( A + (A+B)' )' + ( B + (A+B)' )' )' [De-Morgan's]
- NOR for - (A+B)'
- NOR for - ( A + (A+B)' )'
- NOR for - ( B + (A+B)' )'
- NOR for - ( ( A + (A+B)' )' + ( B + (A+B)' )' )'
Hence, For EX-NOR, required number of gates=4
Since EX-OR=(EX-NOR)' , required number of gates= 4+ 1(for complementing the output of EX-NOR) =5