# Boolean reduction clarification

Discussion in 'Homework Help' started by snowrei, Sep 21, 2009.

1. ### snowrei Thread Starter New Member

Sep 21, 2009
2
0
OK so maybe I'm just an idiot but I have a question regarding the following problem.

(x || y || !z) & (!x || !y || z)

Now am I getting this incorrect by assuming that that reduces down to just true? It doesn't seem to work out so any help would be appreciated.

Last edited: Sep 21, 2009
2. ### Ratch New Member

Mar 20, 2007
1,068
4
snowrei,

Why don't you post your problems with notation like this. (x+y+z')(x'+y'+z) ?

Ratch

3. ### snowrei Thread Starter New Member

Sep 21, 2009
2
0
Sorry, I'm trying to reduce the equation down to a minimal form, and if I'm reading the distributive property properly, I can reduce (x+y+z')(x'+y'+z) to xx'+yy'+z'z which reduces to 1.

I don't think that's correct and I'm trying to find the error in my logic.

4. ### Accipiter New Member

Sep 20, 2009
9
0
xx'+yy'+z'z reduces to zero, not one.

Also, I think you have misunderstood the distributive property. You need to distribute each term in (x+y+z') to each term in (x'+y'+z).