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3 votes
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The value of the Boolean expression (with usual definitions) $(A’BC’)’ +(AB’C)’$ is

  1. $0$
  2. $1$
  3. $A$
  4. $BC$
in Digital Logic
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4 Answers

2 votes

$(A'BC')'+(AB'C)'\\=\left\{(A')'+B'+(C')'\right\}+\left\{A'+(B')'+C'\right\};~[\text{Applying De Morgan's law}]\\=A+B'+C+A'+B+C';~[\because (X')'=X]\\=(A+A')+(B+B')+(C+C');~[\text{Commutative and Associtive laws}]\\=1+1+1;~[\because X+X'=1 \text{ as the Complement law}]\\=1;~[\because 1+1=1 \text{ in Boolean algebra}]$

 

So the correct answer is B.


edited by
0 votes

Option B) 1

Given $(A’BC’)’ +(AB’C)’$

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

$1$

0 votes
Answer : B

(A′BC′)′+(AB′C)′  =>     (A+B'+C) + (A'+B+C')

(A+A')+(B+B')+(C+C') = 1
0 votes
(A´BC´)´+(AB´C)´=(A´BC´AB´C)´=0´=1 ( By De Morgan theorem)

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