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MendelSchmiedekamp comments on Arrow's Theorem is a Lie - Less Wrong
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Note, according to the wikipedia article listed, Arrow's theorem is valid "if the decision-making body has at least two members and at least three options to decide among". This makes me suspect the Pareto-efficiency counter-example as this assumes we have only 2 options.
It doesn't matter if there are ten thousand other options. If you sum numbers A-1 through A-N, and you sum numbers B-1 through B-N, and A-X > B-X for all X, then A must be larger than B; it doesn't matter how many alternatives there are.
Fair enough. Although in considering the implications of more than two options for the other conditions, I noticed something else worrisome.
The solution you present weakens a social welfare function, after all if I have two voters, and they vote (10,0,5) and (0,10,5) the result is an ambiguous ordering, not a strict ordering as required by Arrow's theorem (which is really a property of very particular endomorphisms on permutation groups).
It seems like a classic algorithmic sacrifice of completeness for power. Was that your intent?