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Match List-I with List-II:
 List-I (Boolean Expression)List-II (Dual of Boolean Expression) 
 A. $x(y+0)$
 I. $(x+0)(y\cdot z)$
 B. $\bar{x}+(\bar{y}+z)$
 II. $x+y\cdot1$
 C. $\bar{x}(y+0)$
 III. $\bar{x}+\bar{y}\cdot1$
 D. $x+(\bar{y}+z)$

 IV. $(x+0)(y\cdot \bar{z})$






Solution

Dual is obtained by interchanging:


$+$ ↔ $\cdot$

$0$ ↔ $1$


Applying dual rule:


A → III

B → IV

C → II

D → I



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