Relation between XOR and XNOR gates:
As exclusive or gate gives 1 when input variables have odd number of 1’s and equivalence gate gives 1 when there are even number of zeros in input variables
So when we have 2 variables then exclusive or gate gives 1 for x=0, y=1 & x=1, y=0 and it gives 0 for x=0, y=0 & x=1, y=1 while for equivalence gate we have 0 for x=0, y=1 & x=1, y=0 and it gives 1 for x=0, y=0 & x=1, y=1 as shown in the truth tables shown above. So from here we can see that Equivalence operation is compliment of Exclusive OR gate for 2 variables which can also be proved as follow:
A XOR B = A’B + AB’
COMPLIMENT OF A xor B = (A XOR B)’ = (A’B + AB’)’ = (A + B’) (A’ + B) = AA’ + AB + A’B’ + BB’ = AB + A’B’ = A XNOR B
Hence (A’B + AB’)’ = AB + A’B’
This gives (AB)’ =AB
When we have 3 variables we’ll have the following truth tables for equivalence gate and exclusive or gate:
