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If A B =B A , then (i) A am +Ban =Ben + A . m, (ii) ( A + B ) . m = A . m + B - r n . PROOF. (ii). 5 THEOREM. If B . n+D, D+E=B and E + B - W = C . We consider only the non-trivial case when A , B, C are all non-zero. 1) there exist A E A and C E Csuch that A)(C and B . (vi). By assumption C90+4, B some n'20. hence there is a C E C Cwhere c = j ( b , n')for some ~ E Cand Therefore F = {2(a):a~CA} Ch. 61 67 MULTIPLICATION is a set of natural numbers bounded by n'. F therefore has a maximum, let it be n.

6. We shall now write A < B for A s B & A+B when B is a quord and A< B for AI B and A+ B. 5 LEMMA. If B is a quord, A S B E B and A= COT(A), then A< B. Left to the reader. 9 for quords. Ch. 6 49 THE ORDERINGS THEOREM. If A or B is a quord, then A S B &B<*A*A= B. PROOF. Assume the hypotheses and that A I B , then there exist A, A E A , B E B and recursive isomorphisms p , q, r such that p : A 2: B' IB , q : B N A" < * A and r : A N A', where both of the inequalities (l), (2) are strict since A + B.

C , = B + A = C , + 1 + . . + c, + c1 + - . a + c,. , r ) , where 0 is the permutation53 1 m+l on r letters. n n + 1 ... r ) so that z(i)=i+l =1 if if i