We show that a hyperarithmetic set can be truth table reduced to a HIl-path through 0 iff it is truth table reducible to some r.e. set. It is known from results of Feferman and Spector [1] that while there are El-paths through 0, there do exist HIl such paths. Here by a path is meant a linearly ordered subset P of 0, closed under <0 and having order type co,. In this note we prove THEOREM 1. If P is a RI-path through 0, A is hyperarithmetical, and A is truth table reducible to P, then the Turing degree of A is at most 0'. Thus in a certain sense such paths contain very little information. It is not known if a hyperarithmetic set of Turing degree greater than O' can be Turing reduced to such a path. PROOF. We shall actually prove a somewhat stronger fact. If P is a I-path through 0 and A is truth table reducible to P, then either A is truth table reducible to a proper segment of P (which is necessarily r.e.) or else all of P is used in an essential way, and then one can go backwards and arithmetically decide P from A. So suppose the Turing degree of A is not <0' and A is truth table reducible to such a path P. Then there exists a z such that, for all n, CA(n) = U(ayTP(z, n,y)) and moreover, for all XcN, U(atyTX(z,.n,y)) is a total function of n. (See [4, p. 143, Theorem XIX].) Now by (1) there exists a d E 0* and an r.e. ordering < such that Pc2Pl={y1y?d}, < is a linear ordering on P1 and P is the maximal well ordered segment of P1. (In fact, < is that r.e. linear ordering which restricts on 0 to <O.) Consider pairs (a, b), a, b e P1, a<Kb, and consider a z-computation of CA(n) where: whenever the machine asks, does m e P? the answer given is yes if m?a, no if m q P1 or b?m, and answer is given if neither condition holds. We will say that (a, b) is adequate for n if the correct value of CA(n) is computed in this way. Received by the editors April 11, 1972. AMS (MOS) subject classifications (1970). Primary 02F27.