I need help, but it is not homework for school. It is homework for myself.

What I am doing now is making a hardwired ROM out of discrete logic gates, and part of my problem is that I am lacking enough knowledge of demorgans theorem.

I know the basics, but when I came to this site:
http://www.generalnumbers.com/demorgans_application1.html

It confused me a bit. It states that

AB + CD = ((AB)' (CD)')'

Now It is easy to prove it by drawing the logic gates out on paper.
But there has to be a way to prove it equation wise.

If so, can someone explain it to me without skipping a step?

 

You may have trouble with the proof but here it is. This guy's site was helpful when I was learning real analysis.
http://pirate.shu.edu/~wachsmut/ira/logic/proofs/demorgan.html

 

De Morgan's law states that: (X+Y+Z)'= X'Y'Z', (X'+Y'+Z')= XYZ,







heres how to solve this:


AB+CD=((AB)'+(CD)')'

we use De Morgans Law so that we can distribute the outside prime


AB+CD=(A'+B')'+(C'+D')'

then we apply De Morgans law again so that

AB+CD= (AB)+(CD)


hope that helps!!!

 

hey mik

here is a proof of your equation:

((AB)' (CD)')'

= ((AB)'' + (CD)'') [by demorgan's theorem]

= ((AB) + (CD)) [by cancellation of double negatives]


for demorgan's theorem itself note that the following are all equivalent:

(XY)' is true
XY is false
one of X,Y is false
one of X',Y' is true
X'+Y' is true

peace
stm