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By R. L. Epstein

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Od(Tn+l(~)) ~ ~n(Od(~]) is IS-O, # rn(i* such of length £ ~n+l" i = 0,I such that is now fixed. that * 0)(y) z for all these Od(T0(T)) First rn(0) (I) is none, for all T0(T) are IS-0 we say choose { ~n+l" Since of n+l: Now we ensure Tn+ I. of that l)(y)), ~ ~(n+l) Tn(i lies = Tn(i for any :~ 0)(i y on , i) , 0). and lying are S-O. is even. Tn, ~ See Let if there is an ~n(Od(~))(x)+. If there is one, on let If ~ be least. III-7 ~ ~ T~ If ~n(Od(~))(x) ~(n + i) = w. w = Tn(i rn(i , i) / 0) ~ Tn(i Tn(i) Tn, S-O # ~(x) If not, * 0) * a * i) * a = w as Od(w) rn first note for some a.

H. See the Notes A Minimal We're = Cn(rn(i))(x). for some r e f i n e m e n t s of this Lemma. Degree n o w going to show that the o r d e r i n g A degree b < on ~ is not dense. of u n s o l v a b i l i t y minimal if b ~ 0 no _0 < _c < _b. b and there is is I-5 in here c, is m i n i m a l That is, in the o r d e r i n g it of degrees. Theorem: There is a m i n i m a l Proof: by stages. Tn+l c~ Tn degree. We c o n s t r u c t At stage n+l we w i l l and Tn(~) ~ Tn+l(~). B, a set w h i c h construct We w i l l a full have has m i n i m a l degree, tree such that Tn+ 1 B = Un Tn (~)" 21 Stage 0: T O = identity Stage n+l: See if there no pair of strings Case i: Case 2: tree.

That is, we require l. Od(B) is not recursive 2. Ev(B) is not recursive 3. B ~T Od(B) 4. S ~r Ev(B) 5. For every n, i f On(B) is t o t a l , IV-2 B Ev(B) ~ Od(B) then On(B ) ~T B, or 0n(B ) ~T Ev(B), or 0n(B ) ~T Od(B), or On(B ) is r e c u r s i v e . This will be enough as we then cannot have would imply Ev(B) ~T B. l'. Ev(B) and for that conditions and 2'. is d i c t a t e d by what the form for the general s o l u t i o n will be. How can we satisfy these conditions? , and 4. by just two: Od(B) ~T Ev(B).

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