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+18 votes

If $P, Q, R$ are Boolean variables, then

$(P + \bar{Q}) (P.\bar{Q} + P.R) (\bar{P}.\bar{R} + \bar{Q})$ simplifies to

  1. $P.\bar{Q}$

  2. $P.\bar{R}$

  3. $P.\bar{Q} + R$

  4. $P.\bar{R} + Q$

asked in Digital Logic by Veteran (59.4k points) | 1.3k views

2 Answers

+22 votes
Best answer
Ans is (A) $ P \bar Q $

$ (P + \bar Q)(P \bar Q + PR)(\bar P \bar R + \bar Q) $

$= (PP \bar Q + PPR + P \bar Q + P \bar QR)(\bar P \bar R + \bar Q) $

$= (P \bar Q + PR +  P \bar Q + P \bar QR)(\bar P \bar R + \bar Q) $

$= P \bar Q + P \bar Q \bar R + P \bar QR $

$= P \bar Q + P \bar Q(\bar R + R) $

$= P \bar Q + P \bar Q $

$= P \bar Q $
answered by Loyal (6.1k points)
selected by
Sir how you get PQ'R'  in third step?
For those who don't like POS form to do simplification
Take complement->simplify in sop->take complement again to get the answer.

For simplifying quickly, properties will play a $\color{RED}{key}$ role here.
In boolean algebra, $+ \ and \ . $ is commutative, associative as well as distributive.

$\Rightarrow (P\bar{P}\bar{R}+\bar{Q})(P.(\bar{Q}+R))$
$\Rightarrow \bar{Q}.P.(\bar{Q}+R)$
$\Rightarrow \bar{Q}.P$

+5 votes


=( p'(q')' )'.{( (pr)'.(pq')' )'.( (p'r')'(q')' )'}

={(p'q)'.( (pr)'.(pq') )'} + (p'r')'q )'

=( p'q+(pr)'.(pq')' + (p'r')'q )'

=( p'q+ (p'+r')(p'+q)+(p+r)q )'

=(p'q +p'+ p'r'+qp'+qr'+pq+qr)'

=( p'(q+1) +p'r+ q(p'+p)+q(r'+r) )'





answered by (75 points)
simple method

assume p= q=r=1 calculate function value

check for same given option

some options will automatically eliminate .if same put another value
can you please elaborate it
take p=q=r=1

funcion will be 0

now put this p,q,r into given options c and d will not satisfy (eliminate )

for option a and b re evaluate the function by p=1 q= 0 r=1  function value= 1

put these into option a and b

it will eliminate b
No it will not eliminate @set2018 there will be conflict btw option (a) and (b)

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