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solomon_13000
04-27-2007, 01:22 PM
Proof that (x.y) + (x'.z) + (x'.y.z') = y + (x'.z) using laws of boolean algebra:

My solution:

(x.y) + (x'.z) + (x'.y.z') = y + (x'.z)
(x.y) + (x'.y.z') + (x'.z)
y(x+x'.z') + (x'.z) Distributive law
y(1.z') + (x'.z) Laws of exclude middle
y(1) + (x'.z) Identity law

Is this correct?

stapel
04-27-2007, 03:25 PM
I will guess that you are using the decimal point to stand for multiplication (instead of the "*" or the ""). But what is meant by the "prime" notation?

Also, should we assume multiplication and addition are the usual operations, or are there special rules for this exercise?

Thank you.

Eliz.

daon
04-27-2007, 03:52 PM
1*z'=1 by an identity axiom?

Isn't the point of an "identity" to leave things alone?

solomon_13000
04-27-2007, 06:10 PM
. represents AND
+ represents OR

solomon_13000
04-27-2007, 06:44 PM
If I put it this way it seems that my solution indicates that they are not equal.

(x.y) + (x~.z) + (x~.y.z~) = y + (x~.z)
(x.y) + (x~.y.z~) + (x~.z)
y(x+x~.z~) + (x~.z) Distributive law
y(1.z~) + (x~.z) Laws of exclude middle
y(z~) + (x~.z)

daon
04-29-2007, 10:54 PM
Its definitely true, I just drew a Venn Diagram and verified the identity.

In fact it is true that yz<sup>c</sup>+x<sup>c</sup>z = y + x<sup>c</sup>z.
However your step in the original proof was incorrect.

I haven't taken a class nor read any books on boolean algebra so I'm not sure which identities would be of help.