XOR gate to XNOR gate boolean algebra

[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?

Please help me.

 

[I_am_learning]

Is this your homework?

 

[Ogakor]

Is this your homework?

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

 

[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.

 

[vela]

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?

Please help me.

Neither of those expressions looks right to me.

 

[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

 

[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