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		If the first argument, i, of a parameterized scientist  G  is fixed, the resulting function is just a scientist in the usual sense. This scientist can be denoted by . The signifies that  G  is to be considered as a function only of its second argument, the first argument having been fixed at i. Thus, if  G , is a parameterized scientist for languages, then for every , is a computable function from SEQ to N.

		In general, a parameterized scientist can identify some of the collections described to it but not others; and it may succeed if the description is described in one way but not another. To characterize the competence of a parameterized scientist we rely on the following definition.

		4.35 Definition Let , parameterized scientist  G , and descriptor DL (either for languages or functions) be given.  G  performs X on DL just in case for all , identifies DL(i). X is performable on DL just in case some parameterized scientist performs X on DL.

		Thus, to perform on DL,  G  must convert any description into a scientist that identifies DL(i). In this sense a parameterized scientist may be conceived as a device that synthesizes ordinary scientists on the basis of input problem descriptions. The synthesizer provides trustworthy responses only to the set of descriptions that it performs.

		§4.4.2 Performability on [·]

		Suppose that is such that contains an index for N and also an index for each finite set. Then by Corollary 3.28, no for which is performable on [·]. Is unidentifiability the only reason for nonperformability? That is, if for all , does it follow that X is performable on [·]? We shall now see that the answer is negative inasmuch as every finite collection of languages is identifiable (Exercise 4-1), yet is not performable on [·]. Indeed, the next proposition makes an even stronger claim.

		4.36 Proposition is not performable on [·].

		Proof: Fix a parameterized scientist  G . Let p be an index for N and let h be a recursive function such that for all j, Wh(j) = {p, j }. By an application of the recursion theorem we shall construct an e such that fails to identify . The existence of such an e implies the proposition.