Boolean reduction clarification

Thread Starter

snowrei

Joined Sep 21, 2009
2
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:

Ratch

Joined Mar 20, 2007
1,070
snowrei,

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

Now am I getting this incorrect by assuming that that reduces down to just true?
What is your question? What is your input?

It doesn't seem to work out so any help would be appreciated.
Restate your problem more clearly.

Ratch
 

Thread Starter

snowrei

Joined Sep 21, 2009
2
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.
 

Accipiter

Joined Sep 20, 2009
9
...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.
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).
 
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