edited by
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39 votes
39 votes

If $w, x, y, z$ are Boolean variables, then which one of the following is INCORRECT?

  1. $wx+w(x+y)+x(x +y) = x+wy$
  2. $\overline{w \bar{x}(y+\bar{z})} + \bar{w}x = \bar{w} + x + \bar{y}z$
  3. $(w \bar{x}(y+x\bar{z}) + \bar{w} \bar{x}) y = x \bar{y}$
  4. $(w+y)(wxy+wyz) = wxy+wyz$
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4 Answers

Best answer
34 votes
34 votes

Let us try to simplify (minimize) the expression given in each option

Option - A: $wx+w(x+y)+x(x+y)=x+wy$

$wx + wx + wy + x$
$wx + wy + x$    
$x (1+w) + wy$                  
$x  + wy$

Option - B:  $\overline{w \bar{x}(y+\bar{z})} + \bar{w}x = \bar{w} + x + \bar{y}z$

$\overline{w\bar{x}} + \overline{(y+\bar{z})} + \bar{w}x$
$\bar{w} + x + \bar{y}z + \bar{w}x$
$\bar{w} + \bar{w}x + x + \bar{y}z$
$\bar{w} + x + \bar{y}z$

Option - D: $(w+y)(wxy+wyz)=wxy+wyz$

$wxy + wyz + wxy + wyz$
$wxy + wyz$

Option A, B, D are matching fine.

Hence, Option - C is the answer

edited by
15 votes
15 votes
Option C-(wx'(y+xz')+w'.x')y= (wx'y+wx'xz'+w'x')y=
xx'=0 ..so
Wx'y+w'x'y=x'y(w+w')=x'y
6 votes
6 votes

LHS = (wx'(y + xz') + w'x')y = x'y

RHS =xy'

Option C-   Put y=0 and see LHS !=RHS

edited by
–1 votes
–1 votes
option c is incorrect as we will get ~xy+~xy~w on simplification
Answer:

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