By John N. Crossley

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It is a rooted tree ordering if it is a tree ordering and there is an element 0 such that 0 I A for every A . 1). The following theorem sums up all the important properties of the orderings that we have so far obtained and adds some more. 8 THEOREM. (i) Iis a rooted tree partial well-ordering of %, (ii) I *is a tree ordering of 9 (and hence of %), (iii) Iis a rooted tree ordering but not a partial well-ordering of 9. PROOF. 3 (iv). 3 (ii) that Iis a rooted tree ordering. We leave the details of the proofs of (ii) and (iii) to the reader except for the proof that I is not a partial well-ordering of 9.

Hence A N Df p ( A )$ p ( C ) and therefore P(A)= D T P(A) $ P(C)Y since 6 p 2 A. s. A -0=O, A . 10 DEFINITION. a n +A. 11 THEOREM. (i) If m>O and A . m l B . m , then A I B , (ii) If m>O and A . m l * B - m , then A 5 * B . PROOF. By duality it suffices to prove (i). If m= 1 there is nothing to prove. m + A 5 B - m + B = B . rn2B-m and A f C < * B. In the former case the result follows by the induction hypothesis. In the latter the induction hypothesis yields B S A . Hence there exist D and A (E A) such that B f D =A.

Case 2. If D nA =8, then B = 0 or B nC 8. In the former case put E= 0 (since A , B, C, D are then all 0) and in the latter case the proof of the existence of an E such that A E= B and C = E+ D is merely a literal variant of that for case 1. This completes the proof. 3 COROLLARY. If E + A , + . - + A , , + F = B o + . . + B , , then there exist C, D such that for some I, m, 0 < 1, m 5 n B, = C + A, + D . PROOF. The assertion is trivial if n = 0. Suppose it holds for all numbers less than n, the hypothesis is true but (3C, D ) ( B , = C + A , + D ) is false for all I , m l n .