GATE CS 2017 Set 2 Q27 — Boolean Algebra

To find the incorrect Boolean identity, we can simplify the Left-Hand Side (LHS) of each option and compare it with the Right-Hand Side (RHS).

Option A: $wx+w(x+y)+x(x +y) = x+wy$

• $\text{LHS} = wx + wx + wy + x^2 + xy$
• Since $wx + wx = wx$ and $x^2 = x$:
  $\text{LHS} = wx + wy + x + xy$
• Using absorption law $x + xy = x$:
  $\text{LHS} = wx + wy + x = x(w + 1) + wy = x + wy$
• $\text{LHS} = \text{RHS}$. (Correct)

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

• Applying De Morgan's Law to the first term:
  $\overline{w \bar{x}(y+\bar{z})} = \bar{w} + x + \overline{(y+\bar{z})} = \bar{w} + x + \bar{y}z$
• Now, adding the second term:
  $\text{LHS} = (\bar{w} + x + \bar{y}z) + \bar{w}x$
• Combining $\bar{w} + \bar{w}x = \bar{w}(1+x) = \bar{w}$:
  $\text{LHS} = \bar{w} + x + \bar{y}z$
• $\text{LHS} = \text{RHS}$. (Correct)

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

• Simplify the term $w \bar{x}(y+x\bar{z})$:
  $w \bar{x} y + w \bar{x} x \bar{z} = w \bar{x} y + 0 = w \bar{x} y$   (since $\bar{x}x = 0$)
• Substitute back into the expression:
  $\text{LHS} = (w \bar{x} y + \bar{w} \bar{x}) y = w \bar{x} y^2 + \bar{w} \bar{x} y$
• Since $y^2 = y$:
  $\text{LHS} = w \bar{x} y + \bar{w} \bar{x} y = (w + \bar{w}) \bar{x} y = 1 \cdot \bar{x} y = \bar{x} y$
• Here, $\text{LHS} = \bar{x} y$, but $\text{RHS} = x \bar{y}$. Thus, $\text{LHS} \neq \text{RHS}$. (Incorrect)

Option D: $(w+y)(wxy+wyz) = wxy+wyz$

• $\text{LHS} = w(wxy + wyz) + y(wxy + wyz)$
  $= wwxy + wwyz + wxyy + wyyz$
  $= wxy + wyz + wxy + wyz$
  $= wxy + wyz$
• $\text{LHS} = \text{RHS}$. (Correct)

Thus, only the equation in Option C is incorrect.