10,439 views
31 votes
31 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. $P.\bar{R} + Q$

4 Answers

Best answer
46 votes
46 votes
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 QR $

$= P \bar Q $
edited by
6 votes
6 votes

(p+q')(p.q'+p.r)(p'r'+q')

=( 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) )'

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

=(p'+q)'

=p.q'

 

Answer:

Related questions

37 votes
37 votes
5 answers
1
38 votes
38 votes
2 answers
3
Kathleen asked Sep 11, 2014
12,688 views
In the IEEE floating point representation the hexadecimal value $0\text{x}00000000$ corresponds toThe normalized value $2^{-127}$The normalized value $2^{-126}$The normal...