Reduce the following Boolean equation.

[shamieh]

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]

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

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]

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]

$(x+z)(x+\bar{z})=(x+z\bar{z})$ is distributivity of disjunction over conjunction, which I described in a https://driven2services.com/staging/mh/index.php?posts/30680/.

 

[shamieh]

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]

(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]

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]

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]

Great Explanation! Thanks!