Reduce the following Boolean equation.

shamieh

Active member
Is this correct?

The following Boolean equation can be reduced any farther than it already is? Write out the simplified formula.

$$\displaystyle f(w,x,y,z) = (x + z) * (x + \bar{z}) * (x + \bar{y})$$

My answer: $$\displaystyle (x)(\bar{y})$$ ?

Evgeny.Makarov

Well-known member
MHB Math Scholar
$$\displaystyle f(w,x,y,z) = (x + z) * (x + \bar{z}) * (x + \bar{y})$$

My answer: $$\displaystyle (x)(\bar{y})$$ ?
No, the first expression is true when x = y = z = 1, but the second expression is false.

We have\begin{align*}(x + z)(x + \bar{z})(x + \bar{y}) &= (x+z\bar{z})(x+\bar{y})&& \text{by distributivity of disjunction over conjunction}\\ &=x(x+\bar{y})&& \text{since }z\bar{z}=0\\ &=x+x\bar{y}&& \text{by distributivity of conjunction over disjunction}\\ &=x(1+\bar{y})&&\text{since } x=x\cdot1\\ &=x\cdot1&&\text{since }1+\bar{y}=1\\ &=x\end{align*}

shamieh

Active member
When you say

$$\displaystyle (x + z)(x + \bar{z})(x + \bar{y}) = (x + z\bar{z})(x + \bar{y})$$
<-- How are you getting this result? I am confused.

Are you saying: $$\displaystyle (xx + x\bar{z} + xz + z\bar{z})$$ <-- This is just FOILing the first two. What about the one in the last parenthesis? Can you please show details because I am very confused.

Or are you saying

$$\displaystyle xx + xz! + xx + xy! + xz + zz! + xz + zy!$$ ?

Evgeny.Makarov

Well-known member
MHB Math Scholar
$(x+z)(x+\bar{z})=(x+z\bar{z})$ is distributivity of disjunction over conjunction, which I described in a parallel thread.

shamieh

Active member
oh wait I see.

(x + z)(x + z!)(x + y!)

so:

(xx + zz!)(x + y!) = x(x + y!) = then Factor x(1 + y!) and 1 +y! = 1 thus x*1=x. Wow I had to stare at it for a couple minutes. Thanks for the Help, sorry for the hassle lol.

Evgeny.Makarov

Well-known member
MHB Math Scholar
(x + z)(x + z!)(x + y!)

so:

(xx + zz!)(x + y!)
You are writing (x + z)(x + z!) = (xx + zz!) as if you multiplied the first terms (i.e., x and x) and then the second terms (i.e., z and z!) in the two sets of parentheses. There is no law that allows you to do this. Please take a minute to read my description of distributivity of disjunction over conjunction in the other thread and be sure you understand why it is called distributivity and how the two types of distributivity in Boolean algebra are similar.

shamieh

Active member
You are writing (x + z)(x + z!) = (xx + zz!) as if you multiplied the first terms (i.e., x and x) and then the second terms (i.e., z and z!) in the two sets of parentheses. There is no law that allows you to do this. Please take a minute to read my description of distributivity of disjunction over conjunction in the other thread and be sure you understand why it is called distributivity and how the two types of distributivity in Boolean algebra are similar.
You aren't multiplying x and x together but you ARE multiplying z and $$\displaystyle \bar{z}$$ to get $$\displaystyle z\bar{z} = 0$$ correct? If that is was you are doing then I made a mistake and understand how to do it now, if something else is going on then I will need to re-read the definition.

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like for example if i had $$\displaystyle (a + b)(a + \bar{b})(a + c)$$

i would get by, disjunc over conjunc.

$$\displaystyle (a + b\bar{b})(a + c) = a(a + c) = a(1 + c) = a (1+c = 1) = a(1) = a$$

Evgeny.Makarov

Well-known member
MHB Math Scholar
You aren't multiplying x and x together but you ARE multiplying z and $$\displaystyle \bar{z}$$ to get $$\displaystyle z\bar{z} = 0$$ correct?
Yes.

In general, distributivity of some operator % over some other operator # says the following:

x % (y # z) = (x % y) # (x % z).

If % is disjunction (+) and # is multiplication (*), then this amounts to

x + (y * z) = (x + y) * (x + z),

or

x + yz = (x + y)(x + z).

In the right-hand side, one term in both sets of parentheses is the same: x. The other two terms (y and z) are multiplied and the result is added to x to produce the left-hand side.

shamieh

Active member
Great Explanation! Thanks!