Match List I with List II. List IList II A. x·(y+0) I. (x̄+0)·(ȳ·z) B. x̄·1+(ȳ+z) II. x+y·1 C.…

The dual of a Boolean expression is found by exchanging AND (·) with OR (+) and 0 with 1, leaving the variables and their complements unchanged. Thus $x \cdot (y + 0)$ has the dual $x + y \cdot 1$ (II); $\bar{x} \cdot 1 + (\bar{y} + z)$ has the dual $(\bar{x} + 0) \cdot (\bar{y} \cdot z)$ (I); $\bar{x} \cdot (\bar{y} + 0)$ has the dual $\bar{x} + \bar{y} \cdot 1$ (III); and $x \cdot 1 + (y + \bar{z})$ has the dual $(x + 0) \cdot (y \cdot \bar{z})$ (IV).