GATE Overflow Test Series | Digital Logic | Test 1 | Question: 6

Question

The Boolean expression $(A'BC)'+(AB'C')'$ simplifies to which of the following options?

• A. $0$
• B. $B$
• C. $1$
• D. $AC$

Comments

A’ + B’ = (AB)’ . apply this formula.

Answer 1

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

Answer 2

$ F = (A′BC)′+(AB′C′)$

apply compliment

$ F’ = ((A′BC)′+(AB′C′)’)’$

$ F’ = (A′BC)(AB′C′)$(by demorgans law)

$ F’ = 0 $

to get F back take compliment again

$ (F’)’ = 0’ $

$F=1$