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seen Oct 17 '12 at 16:48

Oct
18
awarded  Student
Oct
15
comment Proving if equations are products of XOR
Oh, I see what I did wrong. I get it thanks!
Oct
15
revised Proving if equations are products of XOR
deleted 209 characters in body
Oct
15
comment Proving if equations are products of XOR
$$A⊕B⊕A.B= A+B$$ $$\begin{array}{c|c|c} A&B&A\oplus B&A\cdot B&(A\oplus B)\oplus A\cdot B&A+B\\ \hline 0&0&0&0&0\oplus 0=0&0\\ 0&1&1&0&1\oplus 0=1&0\\ 1&0&1&0&0\oplus 1=1&0\\ 1&1&0&1&1\oplus 1=0&1 \end{array}$$
Oct
15
comment Proving if equations are products of XOR
Following what you demostrated is this what the table is suppose to look like?@Brian M. Scott
Oct
15
comment Proving if equations are products of XOR
Okay thanks Ben.
Oct
15
accepted Proving if equations are products of XOR
Oct
15
revised Proving if equations are products of XOR
added 160 characters in body
Oct
14
asked Proving if equations are products of XOR
Oct
10
comment Boolean simplification
Okay, I get it! Thanks copper.hat.
Oct
10
comment Boolean simplification
Thank you by the way.
Oct
10
accepted Boolean simplification
Oct
10
comment Boolean simplification
Alright excuse my ignorance but what rule is that? Also after each step a variable was dropped. Why is that? i.e. D',C',A. Where do they go? Are they get canceled out?
Oct
10
awarded  Editor
Oct
10
revised Boolean simplification
added 69 characters in body
Oct
10
asked Boolean simplification
Oct
10
asked Boolean Function
Oct
10
comment Boolean Simplification
Thanks Brain for help.
Oct
10
comment Boolean Simplification
Alright that makes sense. My question is now do we always follow those same steps? i.e. de Morgans, distribute laws, absorption laws. Or does each equation go through a different approach, depending how it is presented?
Oct
10
awarded  Scholar