Re: Chap 1 Slide 19: Simplifications of the canonical representation

Posted by xedover on
URL: http://nand2tetris-questions-and-answers-forum.52.s1.nabble.com/Chap-1-Slide-19-Simplifications-of-the-canonical-representation-tp4032566p4032567.html

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.