Question #282829
  1. Consider a Boolean expression Ex,y,z=xzv(y^z). Find disjunctive and conjunctive normal form.
1
Expert's answer
2022-01-07T07:40:29-0500

Solution:


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

(xˉyz)(xyˉz)(xyz)(\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)=xz(yz) is E(x, y, z)=x z \vee(y \wedge z) \text { is }(xyz)(xyzˉ)(xyˉz)(xˉyz)(xˉyˉz)\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}


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