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    Because π fixes all elements of pairs in L, π must fix th... — Carmelics
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    Home/Modality & Possibility
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    Supports→A symmetric function defined on a set of pairs cannot be a choice function on that set

    Because π fixes all elements of pairs in L, π must fix the value of f at U

    Modality & PossibilityTruth & Knowledge
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    A symmetric function defined on a set of pairs cannot be a choice function on th...Because π does not fix the members of U, the value of f at U cannot lie in UFor a symmetric function f defined on P, there exists a finite list L of pairs f...

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    Related propositions within the same area of thought.
    For any pair U in P but not in L, a permutation π can be found that fixes all el...
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    Because π does not fix the members of U, the value of f at U cannot li...86%Because π fixes the support of f, π also fixes f (i.e., πf = f).86%For any pair U in P but not in L, a permutation π can be found that fi...82%π(U) = U because π only interchanges c and d, which are both in U, so ...79%

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    SEP: axiom-choice
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    The point here is that for a symmetric function \(f\) defined on \(P\) there is a finite list \(L\) of pairs from \(P\) the fixing of all of whose elements suffices to fix \(f\), and hence also all the values of \(f\). Now, for any pair \(U\) in \(P\) but not in \(L\) , a permutation \(\pi\) can always be found which fixes all the elements of the pairs in \(L\), but does not fix the members of \(U\). Since \(\pi\) must fix the value of \(f\) at \(U\), that value cannot lie in \(U\). Therefore \(

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