1967
DOI: 10.1002/malq.19670131302
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Arithmetical Sets and Retracing Functions

Abstract: in Manchester (England) Retraceable sets and retracing functions were introduced by DEKKER and MYHILL [3] and have been further investigated in [l], [2], [lo], [ll], [12], and [18].Two questions were left open in [3], the second of which we answered in [HI. The first question was concerned with characterising retracing functions amongst the partial recursive functions, and one of our purposes in the present note is to show that in a sense no interesting characterisation can exist. For) it is easy to prove (The… Show more

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Cited by 7 publications
(6 citation statements)
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“…We should like to remark that a result similar to Corollary 5.1 was obtained by Yates [16,Theorem 7]. An analogue of Corollary 5.2 was obtained by Yates as Theorem 6 of the same paper.…”
Section: ) Is a Path Through U Such That En A = Asupporting
confidence: 59%
“…We should like to remark that a result similar to Corollary 5.1 was obtained by Yates [16,Theorem 7]. An analogue of Corollary 5.2 was obtained by Yates as Theorem 6 of the same paper.…”
Section: ) Is a Path Through U Such That En A = Asupporting
confidence: 59%
“…If P is chosen so that H(P) has no member recursive in O' (Theorem 3.1), then T easily yields a finite-one retracing function which retraces no infinite set recursive in O'. Such a retracing function was shown to exist by Yates [13,Theorem 6]; another proof is given in [5,Theorem 4.22]. §5.…”
Section: If P Is a Recursive Basic Partition Then H(p) Has Members Amentioning
confidence: 95%
“…We The construction of I from k is very easy: Comment. It is known ( [5,Theorem 5.2]; [4,Lemma 3]) that if J is the unique infinite set retraced by some partial recursive, finiteto-one function, then J has degree of unsolvability ^ 0'. So, in particular, the set R of Theorem 5 has degree ^ 0'.…”
Section: Partial Function /: K-> N K S N Is Downward <=> Df Range (mentioning
confidence: 99%
“…Hence J = {λ n \ n e N} is the only infinite set retraced by g. It remains only to prove that there exists a finite-to-one partial recursive restriction g' of g which retains the property of retracing /. The existence of such a function g', however, follows from [5,Theorem 8], since it is easily verified that {X n \neN} has degree of unsolvability ^ 0' (Simply note that λ« is a recursive function of n and s if we set X s n = 0 whenever Λ n is unattached at the end of Stage s.) That completes the proof of Theorem 1*.…”
Section: ) But It Is Trivial To Show By Induction On S That Ifmentioning
confidence: 99%
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