|
I think you're mixing that with something else fof, I didn't mean if they're the equivalence of anything but such as that:
bubble NAND gate....
NAND gates are basically (AB)' from the beginning, from that diagram alone, but if the input already gives A' and B' the output is already to be (AB)' wouldn't the NOTs cancel? Because the diagrams you provided have bubbles from input to bubbles of output of equivalent gates.
|
Thats not a NAND gate though. What you do is imagine the output bubble is far away. Then when it changes from that pic you have 2 bubbles on the output which cancel.
edit: might be the same thing your saying, I'm getting confused over this whole thread.
|
|
|
Thats how you solve it using "bubble logic". You can also do it algebraically with boolean algebra.
|
thats some pretty drawing, boolean algebra is so much easier though
|
A' NAND B' = (A' AND B')' = A OR B deMorgan's Law
|
"break the line change the sign" as they say. btw: BC+B=BC+B*1 (identity law) BC+B*1=B(C+1) (distributive law)
p.s. For those using examples from the real numbers to show this, why are you trying to explain properties of boolean algebra/rings using field axioms?
|
Alright, and guys, how does...
X = (A’ + B’) + (B+C) X = 1?
I get (A’ + B’) = (AB)' + (B+C) right? What is the next step?
|
X = (A’ + B’) + (B+C)
X = A’ + (B’ + B) + C
X = A’ + 1 + C
X = 1
|
Thank you fof, I'd like to thank everyone for staying up with me while I continue to do this silly stuff. It's 2AM EST. I appreciate it, thanks again fof.
|
|
|
|