Re: Full adder

Posted by cadet1620 on Feb 06, 2013; 1:33pm
URL: http://nand2tetris-questions-and-answers-forum.52.s1.nabble.com/Full-adder-tp4026235p4026240.html

joobcode wrote

C(A'B+AB') + A(BC'+BC) good
Given that x.x' = 1 I can reduce the 1st term:
C  + A(BC'+BC) bad, am I miss applying here?

The problem is that (~A)B does not equal ~(A(~B)) so X + ~X = 1 does not apply.

The easiest way to simplify Boolean expressions is using Karnaugh maps. http://www.facstaff.bucknell.edu/mastascu/elessonshtml/Logic/Logic3.html is a good introduction.

--Mark