How do I create a simplified logic circuit of this given [(A’B’)’ + (A’+ B’)’]’ ? and What is the simplified Boolean expression?

Question

I have to draw the simplified logic circuit of this given [(A’B’)’ + (A’+ B’)’]’ and also get the simplified boolean expression

Answer 1

From DeMorgan Theory

     ((A'B')' + (A'+ B')')'=(A+B + AB)'   
     (A+B + AB)'   = (A+B)'. (AB)'
     (A+B)'. (AB)' = (A'.B') .(A'+B')

assume

X=A' ,Y=B'

we can conclude that

(XY)(X+Y) is (XY)  

as both (XY) and (X+Y) has to be 1 to produce 1 at output as seen from truth table of (XY)(X+Y) is (XY) is identical to XY

so as final result

((A'B')' + (A'+ B')')'=(A+B + AB)'  

(A+B + AB)'   = (A+B)'. (AB)'

(A+B)'. (AB)' = (A'.B') .(A'+B') = A'B'

before simplifying

after simplifying

comparing outputs to make sure

Answer 2

(A'B')' = A + B

Using DeMorgan's theorem (AB)' = A' + B'

(A'+B')' = AB

Again, using DeMorgan's theorem (A+B)' = A'B'

Therefore, now we have the expression:

(A+B + AB)'

Taking A+B as X and AB as Y

(X+Y)' = X'Y' = (A+B)'·(AB)'

Now, creating a logic circuit is fairly simple for this expression, inputs A and B are fed to a NOR gate and NAND gate simultaneously whose outputs act as input to an AND gate