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$Q(x,y,z)=(y′\vee z′ \vee 0\vee x′)\wedge1\wedge(z\vee x′\vee 0\vee y\vee z)′\wedge(z′\vee x\vee y\vee z′)$

I'm not supposed to use tables but only proprieties like De Morgan ecc.

EDIT: So I simplyfied and reordered using commutativuty, identity and idempotent laws. Now I have this expression:

$Q(x,y,z)=(x′\vee y′ \vee z′)\wedge(x\wedge y'\wedge z')\wedge(x\vee y\vee z′)$

Think I should bring it in a sum of product form but I'm not sure how.

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$Q(x,y,z)=(\neg y\vee \neg z \vee 0\vee \neg x)\wedge1\wedge \neg (z\vee \neg x\vee 0\vee y\vee z) \wedge(\neg z\vee x\vee y\vee \neg z)$

You need:

  • Identity Law: $A \vee 0 = A$ and $A \wedge 1 = A$.
  • Idempotent Law: $A \wedge A = A$.
  • deMorgans: $\neg(A \vee B) = \neg A \wedge \neg B$

Use this to get you started. You will need Distributive to simplify.

Edit your question to show what you have attempted. Someone will help if you demonstrate a willingness to try.

Edit...

$Q(x,y,z)=(\neg y\vee \neg z \vee \neg x)\wedge (x\wedge \neg y \wedge \neg z) \wedge(x\vee y\vee \neg z)$

$Q(x,y,z)=(\neg y\vee \neg z \vee \neg x)\wedge x\wedge \neg y \wedge \neg z \wedge(x\vee y\vee \neg z)$

Now, look for common factors (Distributive: $A \wedge (B \vee C) = (A \vee C) \wedge (A \vee C)$, and use Annulment $A \wedge 1 = A$.

Hint: factor out $x$ or $\neg y$ or $\neg z$. $A \wedge (A \vee B) = A \vee (1 \wedge B) = A$

Again, edit your question. Someone will verify. (And they do get easier...)

Laws and Theorems of Boolean Algebra

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