**Consider a Boolean expression Ex,y,z=xzv(y^z). Find disjunctive and conjunctive normal form.**

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### Solution:

So, the disjunctive normal form of $E(x,y,z)=xz\vee (y\wedge z)$ is

$(\bar{x} \wedge y \wedge z) \vee(x \wedge \bar{y} \wedge z) \vee(x \wedge y \wedge z)$

Hence, the conjunctive normal form of $E(x, y, z)=x z \vee(y \wedge z) \text { is }$$\begin{aligned} & \\(x \vee y \vee z) \wedge(x \vee y \vee \bar{z}) \wedge(x \vee \bar{y} \vee z) \wedge(\bar{x} \vee y \vee z) \wedge(\bar{x} \vee \bar{y} \vee z)& \end{aligned}$