  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 \(f\) cannot “choose” an element of \(U\), so a fortiori \(f\) cannot be a choice function on \(P\).