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Raithed
China7078 Posts
bubble NAND gate.... ![[image loading]](http://i32.tinypic.com/3492s1d.jpg) 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. | ||
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fight_or_flight
United States3988 Posts
edit: might be the same thing your saying, I'm getting confused over this whole thread. | ||
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Raithed
China7078 Posts
How is it what I drew not a NAND gate? A NAND gate as a bubble at the output, but I just added two bubbles on the input? | ||
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fight_or_flight
United States3988 Posts
![[image loading]](http://img231.imageshack.us/img231/3885/76077200dc8.png) | ||
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fight_or_flight
United States3988 Posts
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berated-
United States1134 Posts
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azndsh
United States4447 Posts
deMorgan's Law | ||
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sigma_x
Australia285 Posts
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? | ||
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Raithed
China7078 Posts
X = (A’ + B’) + (B+C) X = 1? I get (A’ + B’) = (AB)' + (B+C) right? What is the next step? | ||
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fight_or_flight
United States3988 Posts
X = A’ + (B’ + B) + C X = A’ + 1 + C X = 1 | ||
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Raithed
China7078 Posts
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