We describe the important role that the conjectures and questions posed at the end of the two editions of Gerald Sacks's Degrees of Unsolvability have had in the development of recursion theory over the past thirty years.

The biinterpretability conjecture for the r.e. degrees asks whether, for each sufficiently large k, the [Formula: see text] relations on the r.e. degrees are uniformly definable from parameters. We solve a weaker version: for each k ≥ 7, the [Formula: see text] relations bounded from below by a nonzero degree are uniformly definable. As applications, we show that Low 1 is parameter definable, and we provide methods that lead to a new example of a ∅-definable ideal. Moreover, we prove that (...) automorphisms restricted to intervals [d, 1], d ≠ 0, are [Formula: see text]. We also show that, for each c ≠ 0, can be interpreted in [0, c] without parameters. (shrink)

§1. Introduction. Natural sets that can be enumerated by a computable function always seem to be either actually computable or of the same complexity as the Halting Problem, the complete r.e. set K. The obvious question, first posed in Post [1944] and since then called Post's Problem is then just whether there are r.e. sets which are neither computable nor complete, i.e., neither recursive nor of the same Turing degree as K?Let be the r.e. degrees, i.e., the r.e. sets modulo (...) the equivalence relation of equicomputable with the partial order induced by Turing computability. This structure is a partial order with least element 0, the degree of the computable sets, and greatest element 1 or 0′, the degree of K. Post's problem then asks if there are any other elements of.The solution of Post's problem by Friedberg [1957] and Muchnik [1956] was followed by various algebraic or order theoretic results that were interpreted as saying that the structure was in some way well behaved:Theorem 1.1. Every countable partial ordering or even uppersemilattice can be embedded into.Theorem 1.2. For every nonrecursive r.e. degreeathere are r.e. degreesb, c < asuch thatb ∨ c = a.Theorem 1.3. For every pair of nonrecursive r.e. degreesa < bthere is an r.e. degreecsuch thata < c < b. (shrink)