T. , ?. N}-?-{nar-(-t-)-t-?-t-n, and }. , Then T is defined to be the union of all the trees in T n for some n. For two trees T 1 , T 2 in T , we say that T 2 extends T 1 or also T 1 T 2 if T 2 ? T 1, addition if T 1 ? T 2 we write T 1 T 2 . We illustrate the tree of trees by the following picture: 7.3. NON-UNIVERSALITY IN CONTINUOUS RELATIVIZATION

T. Therefore, [[i]] for some i, j ? N. Also as B ? i Nar i (T n ) and as ? B we have that ? ? Nar(Nar j (T n )) In particular by Fact 7.3.2 we have that ? i (Nar j (T n )) ? and as ? k (T ) ? i (Nar j (T n )) we also have that ? k (T ) ? which is a contradiction Construction claims

. Also, the search for ? is done in priority over e-strategies. If we find no such prefix, then we set ? s = ? s?1 Otherwise let n = ? and do the following: In case o < ? we set ? s = ? s?1 n ?(o + 1)?0 ? . In case o = ?, let ? be the smallest prefix of A s?1 such that [X e,s d ?1] ? U ? e,s . Using Lemma 7.3.2 we find k such that ? ? k (T ? ) cancels noise around A s?1 above Nar(T ? ) = T ??o . Using Construction claim 2 and claim 3, let t be the last stage smaller than s such that ? t n+1 = ??k. Note that we have A, case at stage s ? 1, some value ? s?1 (n) is equal to ? + 1 whereas ? s?2

N. In, Then either k ? i in which case A s cannot not extend stem(T ? ) and (*) is true at stage s, or k = i, in which case, if stem(T ? ) A s , the string stem(T ? ) is comparable with stem Then by Fact 7.3.3 we have that stem(T ? ) A s iff stem(T ? ) ? j (T ?, iff stem(T ? ) A s ? . Therefore, by induction hypothesis, as (*) is true at stage s ? it is true at stage s

. Actually, 1) is not in S, therefore we can apply Lemma 7.3.5: For every stage t ? s + 1, we have A t stem(T ?sm?k?sm?k ) But as stem(T ?sm?k?sm?k ) stem(T ? ) we then have for any t ? s + 1 that A t stem

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