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May 21, 2012 at 14:30 comment added celtschk @Anixx: Commutativity should already be accounted for by the line SetAttributes[#,Orderless]&/@{plus,times};. To account for associativity as well, you can change Orderless to {Flat,Orderless} in that line. I don't think there's an attribute for distributivity, though.
May 21, 2012 at 5:58 comment added Anixx But can it automatically simplify the expressions with unknown variables? For example, simplify $(a\otimes b)\oplus (a\otimes c)$ to $a\otimes (b\oplus c)$?
May 21, 2012 at 5:53 vote accept Anixx
May 21, 2012 at 2:49 comment added J. M.'s missing motivation Also, Function[{a, b, c}, a\[CircleTimes](b\[CirclePlus]c) == (a\[CircleTimes]b)\[CirclePlus](a\[CircleTimes]c)] @@@ Apply[Join, Permutations /@ Tuples[{0, 1, Infinity}, 3]]
May 21, 2012 at 2:47 comment added J. M.'s missing motivation @Anixx: okay then... Function[{a, b, c}, a\[CirclePlus](b\[CirclePlus]c) == (a\[CirclePlus]b)\[CirclePlus]c] @@@ Apply[Join, Permutations /@ Tuples[{0, 1, Infinity}, 3]]
May 21, 2012 at 2:43 comment added Anixx I want it do derive all the properties of the new algebra for itself, just taking the tables of addition and multiplication as the input.
May 21, 2012 at 2:42 comment added Anixx the tables I provided are enough for such derivation.
May 21, 2012 at 2:41 comment added Anixx I wonder whether it will derive associativity ans distributivity laws itself.
May 21, 2012 at 2:39 comment added J. M.'s missing motivation @Anixx: why not try it out yourself? BTW: you didn't mention if associativity is a property of your algebra...
May 21, 2012 at 2:38 comment added rm -rf what sort of expressions do you have in mind? Have you tried it with your expressions? You should add such info to the question before posting.
May 21, 2012 at 2:37 history edited rm -rf CC BY-SA 3.0
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May 21, 2012 at 2:37 comment added Anixx Will it simplify complex expressions?
May 21, 2012 at 2:31 history answered rm -rf CC BY-SA 3.0