. Then, ) are equivalent and imply (1) If T is NIP, then they are all equivalent

. Proof, This is enough Assume A = a i : i < ?, B = b i : i < ? and c are a witness to ¬(3) Then there are two tuples (i 1 < ... < i n ), (j 1 < ... < j n ) and a formula ?(x; y 1 , ..., y n ) such that | = ?(c; a i 1 , ..., a in ) ? ¬?(c; a j 1 , ..., a jn ) Take an ? < ? greater than all the i k and the j k . Then, exchanging the i k and j k if necessary, we may assume that | = ?(c; a i 1 , ..., a in ) ? ¬?(c; a n.? , ..., a n.?+n?1 ). Define §6.4 Dp-minimal groups We study inp-minimal groups. Note that by an example of Simonetta, ([64]), not all such groups are abelian-by-finite. It is proven in [39] that C-minimal groups are abelian-by-torsion, We generalize the statement here to all inp-minimal theories

. Proof and A. Let, To see this A for all n > 0. Then, for n = m, the cosets a m B and a n B are distinct, as are A.b m and A.b n . Now we obtain an independence pattern of length two by considering the sequences of formulas ? k (x) = ''x ? a k B " and ? k (x) = ''x ? A.b k " . For x ? G, let C(x) be the centralizer of x. By compactness, there is k such that for x, y ? G, for some k ? k, either x k ? C(y) or y k ? C(x) In particular, letting n = k!, x n and y n commute n )), the bicommutant of the nth powers of G. It is an abelian definable subgroup of G and for all x ? G, x n ? H. Finally, if H contains all n powers then it is also the case of all conjugates of H, so replacing H by the intersection of its conjugates, we obtain what we want Now we work with ordered groups, Note that in such a group, the convex hull of a subgroup is again a subgroup

. Proof, We may assume that H and C are ?-definable. So without loss, assume C = G. If H is not of finite index, there is a coset of H that forks over ?

?. One-color, C. Such, and M. , ? For each k, monotone relations R k and S k such that M | = x 0

U. Define, {t : tR k x k } and V k (x) = {t : xS k t}. The U k (x) are initial segments of M and the V k (x) final segments. For each k, k , either U k (x k ) ? U k (x k ) or U k

. Proof, By the previous result, we may assume that T = Th(M) eliminates quantifiers

. Proof, We can define such a structure in a linear order with monotone relations : see [52] More precisely, there exists a structure P = (P, R j ) in which is a linear order and the R j are monotone relations, and there is a definable relation O(x, y) such that the structure (P, O) is isomorphic to (M, ?) The result therefore follows from the previous one

. Proof, A straightforward back-and-forth, noticing that tp(¯ x/a)?tp(¯ y/a) tp qf (¯ x?¯ y/a) (quantifier-free type)

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