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solomon_13000

Member

Registered: 2007-04-27

Posts: 6

laws of boolean algebra

I tried to proof that (x.y) + (x~.z) + (x~.y.z~) = y + (x~.z)  but im getting an invalid answer:

(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)

Is my steps correct?

Also I tried this:

(x.y.z) + (x~+z~) = (x.z) + (x~.z~)
y.(x.z) + (x~+z~)  Associative law
1.(x.z) + (x~+z~)  Identity law
(x.z) +  (x~+z~)

and I manage to get a valid answer.

Is my steps correct?

Regards.

Last edited by solomon_13000 (2007-04-27 10:46:11)

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: laws of boolean algebra

This equation is invalid: (x.y.z) + (x~+z~) = (x.z) + (x~.z~)   I checked it with Karnaugh map.

This equation is valid: (x.y) + (x~.z) + (x~.y.z~) = y + (x~.z)  I checked it with Karnaugh mapping.

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