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

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