Question #282837

Consider a Boolean expression Ex,y,z=(x^-z)v(y^z). Find disjunctive and conjunctive normal form.

1
Expert's answer
2021-12-27T16:13:08-0500

Let us find the disjunctive normal form of E(x,y,z)=(x¬z)(yz).E(x,y,z)=(x\land\neg z)\lor(y\land z).


E(x,y,z)=(x¬z)(yz)=(xT¬z)(Tyz)=(x(y¬y)¬z)((x¬x)yz)=(xy¬z)(x¬y¬z)(xyz)(¬xyz).E(x,y,z)=(x\land\neg z)\lor(y\land z) =(x\land T \land \neg z)\lor(T \land y\land z) \\=(x\land (y\lor \neg y) \land \neg z)\lor((x\lor\neg x) \land y\land z) \\=(x\land y \land \neg z)\lor (x\land \neg y\land \neg z)\lor(x \land y\land z)\lor(\neg x \land y\land z).


It follows that (xy¬z)(x¬y¬z)(xyz)(¬xyz)(x\land y \land \neg z)\lor (x\land \neg y\land \neg z)\lor(x \land y\land z)\lor(\neg x \land y\land z) is a disjunctive normal form of E(x,y,z).E(x,y,z).


Let us find the conjunctive normal form of E(x,y,z)=(x¬z)(yz).E(x,y,z)=(x\land\neg z)\lor(y\land z).


E(x,y,z)=(x¬z)(yz)=(xy)(xz)(¬zy)(¬zz)=(xyF)(xFz)(Fy¬z)T=(xy(¬zz))(x(¬yy)z)((¬xx)y¬z)=(xy¬z)(xyz)(x¬yz)(xyz)(¬xy¬z)(xy¬z)=(xy¬z)(xyz)(x¬yz)(¬xy¬z).E(x,y,z)=(x\land\neg z)\lor(y\land z) =(x\lor y)\land(x\lor z)\land (\neg z\lor y)\land (\neg z\lor z) \\=(x\lor y\lor F)\land(x\lor F\lor z)\land ( F\lor y\lor \neg z)\land T \\=(x\lor y\lor (\neg z\land z))\land(x\lor (\neg y\land y)\lor z)\land ( (\neg x\land x)\lor y\lor \neg z) \\=(x\lor y\lor \neg z)\land(x\lor y\lor z)\land(x\lor \neg y\lor z)\land(x\lor y\lor z)\land ( \neg x\lor y\lor \neg z)\land ( x\lor y\lor \neg z) \\=(x\lor y\lor \neg z)\land(x\lor y\lor z)\land(x\lor \neg y\lor z)\land ( \neg x\lor y\lor \neg z).


It follows that (xy¬z)(xyz)(x¬yz)(¬xy¬z)(x\lor y\lor \neg z)\land(x\lor y\lor z)\land(x\lor \neg y\lor z)\land ( \neg x\lor y\lor \neg z) is a conjunctive normal form of E(x,y,z).E(x,y,z).


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