. Clearly, we have N 1 (z) ? b?N k?2 (x) N 1 (b) By definition, N 1 (z) = K(O k?3 (z) and b?N k?2 (x) N 1 (b) = K(O k?2 (x)). We then obtain K(O k?3 (z)) ? K(O k?2 (x)), which gives, pp.3-5

?. Since-z, Again, since z ? W k?2 (x), we have O k?3 (x) ? O k?3 (z) ? O k?2 (x) Since we already proved that N k?1 (x) = W k?1 (x), O k?4 (x)), and then, pp.3-3

N. Since, |N k?2 (x)| > 1, necessarily O k?3 (x) O k?2 (x) Similarly, the fact that N k?1 (x) = W k?1 (x) and |N k?1 (x)| > 1 implies that O k?2 (x) O k?1 (x) At last, from Remark 3, we have K(O k?1 (x)) = N 1 (x), and since |N 1 (x)| ? 2, it follows that |K(O k?1 (x))| ? 2. Combined with the fact that |O 1 (x)| ? 2

H. Proof-of and . .. Sur, From recursion hypothesis H sur (k ? 1), for any P ? O such that O k?3 P , there exists y P ? V k?1 such that S(y P ) = (O 1 , We denote by Y the set = (O 1 , . . . , O k?3 ) and O k?2 ? V * k and is therefore an element of V k . In order to conclude the proof of H sur (k), let us now show that the element x of V k has the desired characterising sequence (O 1 , . . . , O k?1 ). First, from H N (k), which we already proved, we know that O k?3 ), from the definition of x = Y y?Y N (y). Second, from Remark 3, we have K(O k?1 (x)) = N 1 (x) and we already know that |N 1 (x)| ? 2 (see beginning of the proof of H sur (k)), k?1 ) ? O k?1 such that O 1 . . . O k?1 O k?3 (y)) for any y ? N k?1 (x), which gives (O 1 (x), . . . , O k?3 (x)) = (O 1 This gives |K(O k?1 (x))| ? 2, and as we have |O 1 (x)| ? 2 and O 1 (x) ? . . . ? O k?2 (x) ? O k?1 (x), we obtain that O k?2 (x), O k?1 (x) ? O. Now, from H N (k), we know that the couple (O k?2 (x), O k?1 (x)) is such that N k?1 (x) = {y ? V k?1 | O k?2

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