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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]

  1. NOR for - (A+B)'
  2. NOR for - ( A + (A+B)' )'
  3. NOR for - ( B + (A+B)' )'
  4. 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

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To realize  EX-NOR gate using NOR gates = 4 NOR GATES.

To realize  EX-OR gate using NOR gates  = 4(for generating EX-NOR)+1(for complementing it) = 5 NOR GATES.
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for construction of EX-OR gate, a minimum number of 2 input NOR is: 5

NOR Construction
  XOR from NOR.svg

similarly, for construction of 2 input, EX-NOR gate minimum number of 2 input NOR gate is:4

NOR Construction
  XNOR from NOR.svg

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