laws of boolean algebra

[solomon_13000]

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]

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]

1*z'=1 by an identity axiom?

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

[solomon_13000]

. represents AND
+ represents OR

[solomon_13000]

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]

Its definitely true, I just drew a Venn Diagram and verified the identity.

In fact it is true that yz^c+x^cz = y + x^cz.
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.