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:
(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?
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)
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.
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