Help explain this Boolean equation XOR with NOR gates

Question

I'm just starting Boolean algebra and am following an example given in the text that shows the configuration of NOR gates to create an XOR. I cannot follow the algebraic example and would like to have detailed steps filled in as I cannot match up rules I know to the example:

Example of XOR to NOR algebra

$$\color{silver}{\boxed{\color{black}{\begin{align} A ~\overline B + \overline A ~B ~=&~ \overline{~\overline{~A ~\overline B~}~} + \overline{~\overline{~\overline A ~B~}~} \\[1ex] = & ~ \overline{~\overline A + B~} + \overline{~A + \overline B~} \end{align}}}}$$

I understand that (AB)' = A'+B' but I'm not following how the complements were worked out in the example (see jpg) to get the end equation.

Thank you.

Answer 1

Remember that $\overline{\overline X}=X$ by the law of double negation. Then $$\eqalignno{ A \overline B + \overline A B &= \overline{\overline{\left(A \overline B\right)}} + \overline{\overline{\left(\overline A B\right)}} &{\rm by~double~negation} \cr&= \overline{\left(\overline A + \overline{\overline B{}}\right)} + \overline{\left(\overline{\overline A} + \overline B\right)} &{\rm by~DeMorgan} \cr&= \overline{\left(\overline A + B\right)} + \overline{\left(A + \overline B\right)}. &{\rm by~double~negation} \cr } $$

Answer 2

Hint when its $(AB')'=A'+B$ and $(A+B')'= A'.B$ where $" ' "$ represents bar.

Answer 3

I understand that (AB)' = A'+B' but I'm not following how the complements were worked out in the example

$$\begin{align} A~B'+A'~B ~=&~ \Big(\color{navy}{\big(A~B'\big)'}\Big)'+\Big(\color{navy}{\big(A'~B\big)'}\Big)' & \because X = \Big(\color{navy}{\big(X\big)'}\Big)' \\[1ex] = & ~ \big(\color{navy}{A'+B}\big)' + \big(\color{navy}{A+ B'}\big)' & \because \color{navy}{\big(X~Y'\big)'} = \color{navy}{X'+Y} \end{align}$$