GATE CS 2017 Set 2 Q.27 (CS Q.27): Boolean Algebra and Logic Minimization

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

$wx + w(x + y) + x(x + y) = x + wy$

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

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

$(w + y)(wxy + wyz) = wxy + wyz$

Official answer and solution

Solution

Simplify each Boolean expression using the laws of Boolean algebra to identify the incorrect statement.

Evaluate each option:

Option A: LHS $= wx + w(x + y) + x(x + y) = wx + wx + wy + x + xy = x + wy$ . Matches RHS $x + wy$ . Correct.
Option B: LHS $= \overline{w\bar{x}(y + \bar{z})} + \bar{w}x = (\bar{w} + x + \bar{y}z) + \bar{w}x = \bar{w}(1 + x) + x + \bar{y}z = \bar{w} + x + \bar{y}z$ . Matches RHS. Correct.
Option C: LHS $= (w\bar{x}(y + x\bar{z}) + \bar{w}\bar{x})y = (w\bar{x}y + 0 + \bar{w}\bar{x})y = w\bar{x}y + \bar{w}\bar{x}y = (w + \bar{w})\bar{x}y = \bar{x}y$ . But the RHS is $x\bar{y}$ , which is not equivalent to $\bar{x}y$ (they evaluate differently for $x=0, y=1$ ). Thus, statement C is INCORRECT.
Option D: LHS $= (w + y)(wxy + wyz) = w(wxy + wyz) + y(wxy + wyz) = wxy + wyz + wxy + wyz = wxy + wyz$ . Matches RHS. Correct.

Statement C is the incorrect identity, matching option C.

Final answer

Common trap

Confusing the variable complement positions $\bar{x}y$ and $x\bar{y}$ .

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

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