Answer to Question #276378 in Electric Circuits for skeoskgt

Question

Find the complements of 

[(A'+B)C']' =?

[(AB'+C)D'+E']' =?

[A+((BC')'+D)]" =?

Answer 1

(i) Using de Morgan's theorem

$\begin{bmatrix} (A'+B) &C'\\ \end{bmatrix}=(A'+B)'+C''$

Applying involution law

$=(A'+B)'+C$

Applying de Morgan's theorem

$=A''.B'+C$

Applying involution law

$=A.B'+C$

(ii)

Applying de Morgan's theorem

$=\begin{bmatrix} (AB'+C)D' \end{bmatrix}'.E''$

Applying involution law

$=\begin{bmatrix} a (AB'+C) & D' \end{bmatrix}'.E$

Applying de Morgan's law

$=\begin{bmatrix} (AB'+C)'+D'' \end{bmatrix}.E$

Applying double negation

$\begin{bmatrix} (AB'+C)'+D \end{bmatrix}.E$

Applying de Morgan's theorem

$=\begin{bmatrix} (AB')'.C'+D \end{bmatrix}.E$

$=\begin{bmatrix} (A'+B'').C'+D \end{bmatrix}.E$

$=\begin{bmatrix} (A'+B).C'+D \end{bmatrix}.E$

$=\begin{bmatrix} A'C'+BC'+D \end{bmatrix}.E$

$=E(D+C').(B+D+A')$

Applying involution law

$=\begin{bmatrix} A+((BC')'+D) \\ \end{bmatrix}''$

Using involution law

$=A+((BC')'+D)$

Applying de Morgan's theorem

$=A+(B'+C'')+D$

Applying involution law

$=A+B'+C+D$