• edited by
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59 59 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. $(w+y)(wxy+wyz) = wxy+wyz$

5 Answers

Best answer
50 50 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
17 17 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 6 votes

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

RHS =xy'

Option C-   Put y=0 and see LHS !=RHS

• edited by
3 3 votes
Just observe option C: on the LHS, there is no y' term, but on the RHS, there is y'. Unless we fully complement the LHS, we won't get y' on the RHS. However, we are not complementing the LHS, so option C is the only false one.
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