How can I find a DNF and Minimal Form for this boolean expression? $Q(x,y,z)=(y′\vee z′ \vee 0\vee x′)\wedge1\wedge(z\vee x′\vee 0\vee y\vee z)′\wedge(z′\vee x\vee y\vee z′)$
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.
 A: $Q(x,y,z)=(\neg y\vee \neg z \vee 0\vee \neg x)\wedge1\wedge \neg (z\vee \neg x\vee 0\vee y\vee z) \wedge(\neg z\vee x\vee y\vee \neg z)$
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
