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- Thread starter shamieh
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- Jan 30, 2012

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No, the first expression is true when x = y = z = 1, but the second expression is false.\(\displaystyle $f$(w,x,y,z) = (x + z) * (x + \bar{z}) * (x + \bar{y})\)

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

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

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

Are you saying: \(\displaystyle (xx + x\bar{z} + xz + z\bar{z})\) <-- This is just

Or are you saying

\(\displaystyle xx + xz! + xx + xy! + xz + zz! + xz + zy!\) ?

- Jan 30, 2012

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- Jan 30, 2012

- 2,525

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

so:

(xx + zz!)(x + y!)

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You aren't multiplying x and x together but youYou 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.

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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.

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

- Jan 30, 2012

- 2,525

Yes.You aren't multiplying x and x together but youmultiplyingAREzand \(\displaystyle \bar{z}\) to get \(\displaystyle z\bar{z} = 0 \)?correct

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.

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