Question #276378

Find the complements of 

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

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

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


1
Expert's answer
2021-12-13T11:12:49-0500


(i) Using de Morgan's theorem


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


Applying involution law

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


Applying de Morgan's theorem

=A.B+C=A''.B'+C


Applying involution law

=A.B+C=A.B'+C


(ii)

Applying de Morgan's theorem

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


Applying involution law

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


Applying de Morgan's law

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


Applying double negation

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


Applying de Morgan's theorem

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


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


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


=[AC+BC+D].E=\begin{bmatrix} A'C'+BC'+D \end{bmatrix}.E


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



Applying involution law

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


Using involution law

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


Applying de Morgan's theorem

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


Applying involution law

=A+B+C+D=A+B'+C+D


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