Does anyone know of any references to successful proofs for ternary or discrete multi-state logic?
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There are two seminar papers by Magnus Hestenes that should give you a good start:\n\n"A  Ternary Algebra with Applications to Matrices and Linear Transformation"\n\n"On a Ternary Algebra"
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well if your looking for books in logic, a book 'beginning logic' by Lemmon is a fantastic introduction, it goes from basic to fairly advanced Logic concepts
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No, nobody really can prove it, with assuming several axioms\nfirst. There are certain things you have to assume are true in your universe, without necessary being able to check them. From those axioms, you can derive conclusions (theorems) that are always true, but only in that particular\nuniverse. The discussed operation of addition is one such theorem. If you assume that the axioms of natural numbers are true, then this addition becomes quite trivival. If you assume those axioms of natural numbers to be false, then the above operation of addition might become false with a different set of supposition. In other words, only with the facts given, one cannot prove, nor disprove, the fact of addition. It all depends on the set of assumptions you would have made before adding the two ones.
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Below you will find the specifications for the AES (Rijndael) and Three-DES algorithms.  \n\nAs for the writing of the programs themselves, I'm almost done with my CSE degree--sounds like your pain is just beginning :-D
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If you can draw it like a circuit it should be a little easier to see.\nClick on this site and go through the 'prev page' and 'next pages' for samples of Boolean logic. \nhttp://computer.howstuffworks.com/boolean2.htm\nHope this helps.\nTry using the output as an input to keep track of state
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Check out the below link. Its good. These are some interview questions on logic design that was online.\n\nhttp://ceaspub.eas.asu.edu/cse517/interview.html
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assigning the values will be fool proof
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I wont give you the entire proof, just give you the key arguments.\n\nknowing the joint distibution of (X, Y) allows to compute the marginal densities of X and Y.\nIndeed, let f_X be the marginal density of X, then:\n\nf_X(x) = &#92;int f(x, y) d_y\n\nf_Y(y) = &#92;int f(x, y) d_x\n\nNow, you might compute E(Y| X = x) = &#92;int y f_{Y|x} (y) d_y\n\nwhere f_{Y| x} is the conditionnal density of Y knowing the event X = x\n\nrecall that f_{Y| x}  = f(x, y) / f_X(x), and you just have to compute the resulting integrals\n\nregards,\n\njfp
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This is a better link that the previous one posted here!! try it! \n\nhttp://pubs.logicalexpressions.com/Pub0009/LPMArticle.asp?ID=305
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http://www.wpi.edu/Pubs/ETD/Available/etd-0430102-120529/unrestricted/gaubatz.pdf\n\nmentions a VHDL implementation and generation of test vectors using Maple V.
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bool function(int a, int b, int c)\n{\nreturn a || b || c;\n}
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http://www.eas.asu.edu/~holbert/ece201/kvl.html
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The links below have good info. The first briefly addresses your first question, and the second is a good brief history of the problem which provides some insight as to the difficulties encountered. I actually played around with this problem a little on an elementary level and it is interesting. Cheers.\n\nhttp://www.sciam.com/print_version.cfm?articleID=000DBA55-B698-1C71-9EB7809EC588F2D7\n\nhttp://www.geocities.com/fermatnow/flt/
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1. yes.\n\n2. one\n\n3. 2 equal segments (this one's a theorem)
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here's an output step by step:\n\nthis is the initial state of deque D\nD :: first -> <- last\n\naddFirst(3)\nD :: first -> 3 <- last\n\naddLast(8)\nD :: first -> 3 8 <- last\n\naddLast(9)\nD :: first -> 3 8 9 <- last\n\naddFirst(5)\nD :: first -> 5 3 8 9 <- last\n\nremoveFirst()\nD :: first -> 3 8 9 <- last\n\nremoveLast()\nD :: first -> 3 8 <- last\n\nfirst()\nfirst = 3\n\naddLast()\nD :: first -> 3 8 <- last\n\nremoveFirst()\nD :: first -> 8 <- last\n\nlast()\nlast = 8\n\nremoveLast()\nD :: first -> <- last\n\nhope this helps...
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i am intrigued by your idea of LOGIC!!! interesting
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i think its true\nim f is isomorphical  with G/(ker f) so\n|G| = |ker f|*|G/(ker f )| = |ker f| *|im f|.\n i applied two well known theorems.