Page 176

		Proof: For each define:

		Consider the collection of functions . We first show that . Suppose, by way of contradiction, that some scientist M InlEx*-identifies . We describe a scientist M' which Ex*-identifies —yielding a contradiction. For each f, let Gf denote a text such that for all k, x, Gf(<k, x>) = (1 + <k, x>, f(x)). Note that Gf is a 1-incomplete text for f'. Also, note that Gf[n] can be effectively computed from f[n]. Let g be a recursive function such that for all p, x,  j g(p)(x) =  j p(<1 + <0, x>>). Define M' such that M'(f[n])). Note that M' can easily be constructed from a description of M. Clearly, for each , if  j p =* f', then  j g(p) =* f. Since M InlEx*-identifies , it follows that M' Ex*-identifies . However, (Corollary 6.11); thus .

		We now show that . Let F be a mapping from SEGI to SEGI such that for each ,

		Also, for each text G, define the text G' such that for all n,

		Now let G be a *-noisy text for . It is easy to see that for all but finitely many n, G'[n] = F(G[n]). Also, for all i, x, and y, we have . Moreover, from the finitely many candidates for f'(0) (at least one