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:

Your blog is a treasure trove of valuable information – thank you!I’ve submitted my email for subscription but haven’t received a confirmation email yet

Thanks for sharing. I read many of your blog posts, cool, your blog is very good.

Your point of view caught my eye and was very interesting. Thanks. I have a question for you.