Abstract
AbstractLet S and T be countable complete theories. We assume that T is superstable without the dimensional order property, and S is interpretable in T in such a way that every model of S is coded in a model of T. We show that S does not have the dimensional order property, and we discuss the question of whether Depth(S) ≤ Depth(T). For Mekler's uniform interpretation of arbitrary theories S of finite similarity type into suitable theories TS of groups we show that Depth(S) ≤ Depth(TS) ≤ 1 + Depth(S).
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