So, the disjunctive normal form of E(x,y,z)=xz∨(y∧z)E(x,y,z)=xz\vee (y\wedge z)E(x,y,z)=xz∨(y∧z) is
(xˉ∧y∧z)∨(x∧yˉ∧z)∨(x∧y∧z)(\bar{x} \wedge y \wedge z) \vee(x \wedge \bar{y} \wedge z) \vee(x \wedge y \wedge z)(xˉ∧y∧z)∨(x∧yˉ∧z)∨(x∧y∧z)
Hence, the conjunctive normal form of E(x,y,z)=xz∨(y∧z) is E(x, y, z)=x z \vee(y \wedge z) \text { is }E(x,y,z)=xz∨(y∧z) is (x∨y∨z)∧(x∨y∨zˉ)∧(x∨yˉ∨z)∧(xˉ∨y∨z)∧(xˉ∨yˉ∨z)\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}(x∨y∨z)∧(x∨y∨zˉ)∧(x∨yˉ∨z)∧(xˉ∨y∨z)∧(xˉ∨yˉ∨z)
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