View Full Version : laws of boolean algebra

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?

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.


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

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

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

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)

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.