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

Let us simplify the LHS of each option using Boolean algebra:

• Option A:

  $$\text{LHS} = wx + w(x + y) + x(x + y) = wx + wx + wy + x + xy = wx + wy + x = x(1 + w) + wy = x + wy = \text{RHS}$$

  (Correct identity)

• Option B:

  $$\text{LHS} = \overline{w\bar{x}(y + \bar{z})} + \bar{w}x = \bar{w} + x + \overline{y + \bar{z}} + \bar{w}x = \bar{w}(1 + x) + x + \bar{y}z = \bar{w} + x + \bar{y}z = \text{RHS}$$

  (Correct identity)

• Option C:

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

  Since $\bar{x}x = 0$:

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

  Here, $\text{LHS} = \bar{x}y$, but $\text{RHS} = x\bar{y}$. Hence, Option C is INCORRECT.

• Option D:

  $$\text{LHS} = (w + y)(wxy + wyz) = w(wxy + wyz) + y(wxy + wyz) = wxy + wyz + wxy + wyz = wxy + wyz = \text{RHS}$$

  (Correct identity)