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		Proof (Lemma 5.22): Suppose first that TxtEx = [TxtEx]prudent. Fix an . We argue that is r.e. extendible. Let M be a prudent scientist that identifies . Then is an r.e. set. Since M is prudent, M identifies . Clearly, is r.e. indexable and a superset of .

		Now suppose that every is r.e extendible. Fix an and let be a superset of that is r.e. indexable, say by the r.e. set S. Without loss of generality we take S to be infinite. By Proposition 5.29, there is a total order-independent scientist, M, that identifies . We show how to construct a prudent scientist M' that identifies (and hence ). Let S0, Sl, . . . be a one-one recursive enumeration of S. Let Ti be a text for Wsi that can be obtained recursively from i. Now, for each  s , define

		.

		Since M is order-independent and identifies , M' will converge on any text T for to an si such that Ws i = L. Also, every index in S is an index for a language in , and M' outputs only indexes from S. Thus M' is prudent and identifies .

		Proof (Lemma 5.23): Our goal is to show that every is r.e extendible. Fix an . By Proposition 5.29, is identifiable by some scientist M such that M identifies L if and only if there is a locking sequence for M on L. Without loss of generality we assume that . We exhibit an r.e. indexable collection such that and . There are two cases:

		Case 1: M identifies N. Define a recursive function  ƒ  by

		To see that  ƒ  defined in this way is recursive, we argue informally. Given  s , to enumerate W ƒ ( s ), compute M( s ) and enumerate nothing in W ƒ ( s ) until, if ever, a stage s is reached such that WM( s ),s contains all of content( s ). At this point begin enumerating all of WM( s ) into W ƒ ( s ) until, if ever, there is discovered a sequence  g  such that , and . If such a  g  exists, then begin enumerating all of N into W ƒ ( s .

		Since  ƒ  is recursive, is r.e. On the other hand, since and N are identified by M and since M identifies every L such that there is a locking sequence for M on L, M identifies every language with an index in S.