Chap 1 Slide 19: Simplifications of the canonical representation

[Bahanet]

Hello !

This is my first post so, first of all, I would like to thank you for making this very interesting course available on internet !

I do not understand how the simplifications of the canonical representation on the slide 19 were made: https://docs.wixstatic.com/ugd/56440f_51f7dbaa5d004686a9576588088458c6.pdf

I do not understand how we went from:
(Not(x) And Not(y) And Not(z)) Or (Not(x) And y And Not(z)) Or (x And Not(y) And Not(z))
to:
(Not(x) And Not(z)) Or (x And Not(y) And Not(z))
and even less to:
(Not(x) And Not(z)) Or (Not(y) And Not(z))

But I do understand the 2 last lines:
(Not(x) And Not(z)) Or (Not(y) And Not(z)) =
Not(z) And (Not(x) Or Not(y))

Thank you !

[xedover]

The first level is easy...
(X'Y'Z') + (X'YZ') + (XY'Z')

In the first two terms, we can factor out a common (X'Z')
--> (X'Z')(Y' + Y) + (XY'Z')

Y' + Y = 1
and (X'Z')1 = (X'Z')
so
--> (X'Z') + (XY'Z')

The next level is much trickier. It uses the law of A + A'B = A + B
(I assume this the same as A' + AB = A' + B ?)

again, start by factoring out a common Z'
--> Z'(X' + XY')

so (X' + XY') = X' + Y' (right?)
--> Z'(X' + Y')

which is the same as the next to the last line (X'Z') + (Y'Z'), but already reduced.


Is that right? I'm a little fuzzy on the second reduction (especially since it skipped straight to the final reduction), but I think that's right.

[xedover]

found this list of laws: https://www.mi.mun.ca/users/cchaulk/misc/boolean.htm

AB' + B = A + B -- redundancy law

[Bahanet]

Hello ! Thank you for the answer. I now understand !

Just a little correction; you wrote :

Y' + Y = 0
and (X'Z')0 = (X'Z')
so
--> (X'Z') + (XY'Z')

It should be:

Y' + Y = 1                   (Complement law)
and (X'Z')1 = (X'Z')     (Identity Law)
so
--> (X'Z') + (XY'Z')

Thanks !

[xedover]

oh...totally right... I'll make the correction above.
thanks.