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		§4.4.3 Performability on [·]f

		If n is an index for N, then Theorem 4.28 shows that no superset of {n} is performable on [·]f. However, nonperformability also infects much simpler sets of indexes, as the following proposition reveals.

		4.41 Proposition (Jantke [98]) Let be the set of all indexes x with the following property: there is exactly one such that  j j is total. Then I is not perform able on [·]f.

		So, no parameterized scientist can take an arbitrary and adapt itself to identify the sole with an index in Wi.

		Proof:4 Let total recursive h be the function guaranteed by Proposition 4.31. Then:

		4.42 For all , .

		Given , let gi be the sole member of with an index in Wh(i). It follows immediately from 4.31 that:

		4.43 is not identifiable.

		Now for a contradiction, suppose that parameterized scientist  G  performs I on [·]f. Then:

		4.44 For all , identifies gi.

		Define (unparameterized) scientist M as follows. For all , if content( s ) does not contain exactly one occurrence of a pair of the form (0, i), then M is undefined. Otherwise, if then M( s ) =  G  (h(i),  s ). It follows from 4.44 that M identifies , contradicting 4.43.

		4.45 Corollary is not performable on [·]f.

		It is the presence of indexes for nontotal functions in the descriptions delivered by [·]f that causes nonperformability. To see this, we first define the set of descriptions free of this problem.

		4.46 Definition .

		The symbol "" may be read as "all total." We leave the following proposition as an exercise.

		4 A different proof is given in [98].