# XOR gate to XNOR gate boolean algebra

1. Aug 23, 2010

### Ogakor

Xor gate with negated input and negated output.

The expected output is an XNOR gate. I can't get it with boolean algebra.

My initial formula is:
((AB')' + (A'B)')'

The expected output is:
AB + (AB)' right?

2. Aug 23, 2010

### I_am_learning

3. Aug 23, 2010

### Ogakor

No, this was my seatwork and I didn't get the answer.

4. Aug 24, 2010

### xcvxcvvc

must you show it with boolean algebra? I'd recommend, since this has only 2 inputs and 1 output, simply running through the 4 possible inputs and calculating the 4 possible outputs 1 at a time in a truth table.

5. Aug 24, 2010

### vela

Staff Emeritus
Neither of those expressions looks right to me.

6. Sep 18, 2010

### zgozvrm

I'm pretty sure that we're talking about inverting only one of the inputs to the XOR gate.
So, instead of A XOR B, we would have A' XOR B

This would, in fact have the result as A XNOR B

7. Jul 30, 2011

### RonnieRongz

[(AB')+(A'B)]'

By De Morgan's Theorem
(AB')' • (A'B)'

Again by De Morgan's Theorem
(A'+B'') • (A''+B')

Simplify
(A'+B) • (A+B')

Multiply
A'A+A'B'+AB+BB'

*AA' and BB' are equal to 0

Therefore [(AB')+(A'B)]' = A'B'+AB
Hope this helps