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Proof: For each 0139-001.gif, let Lf denote the single-valued total language: 0139-002.gif. For each 0139-003.gif, let 0139-004.gif denote the collection of single-valued total languages 0139-005.gif.
Let p1 be a recursive function that translates an index for a (partial) recursive function  y  into an index for the corresponding single-valued language L y . Formally, for all i, 0139-006.gif,
0139-007.gif
Similarly, let p2 be a recursive function that translates an index for a language L into an index for a partial recursive function represented by L. Formally, for all i, 0139-008.gif,
0139-009.gif
In the first condition of the above definition of  j p2(i)(x), we assume that the z used is the first appropriate one produced by some uniform enumeration of Wi.
Let 0139-010.gif and let 0139-011.gif. It is easy to verify that if i is an index for an a-variant of f such that i does not make any convergent errors (with respect to f), then pl (i) is an index for an a-variant of the single-valued total language Lf. Similarly, if j is an index for an a-variant for the single-valued total language L, then p2(j) is an index for an a-variant of the recursive function represented by L.
Choose an 0139-012.gif. (By Proposition 6.5, such an S exists.) We argue that 0139-013.gif.
Claim: 0139-014.gif. Proof: Let M be a scientist that searches a text for an element of the form 0139-015.gif and keeps on emitting p1(z'), where z' is a program obtained by patching the convergent errors of z with respect to the function represented by the input language. Clearly, M TxtExm+1-identified 0139-016.gif. Hence, the claim follows.
Claim: 0139-017.gif. Proof: Suppose by way of contradiction that a scientist M TxtExm-identifies 0139-018.gif. Let 0139-019.gif be such that t((x, y)) = <x, y> and t(#) = #. Extend t to 0139-020.gif by taking t( s ) to be the result of applying t to each element of  s . Now, let M' be a scientist such that, for each segment  s , M'( s ) = p2(M(t( s ))). Since M TxtExm-identifies 0139-021.gif, it follows that M' must Exm-identify S, a contradiction. Hence, the claim and part (a) follows.
A similar proof works for part (b) using ASD* of Corollary 6.6.
6.20 Corollary 0139-022.gif.

 
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