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**Example text**

Groundwork The next result follows immediately from the previous exercise. Both results are easily proved by induction on the complexity of formulas. See [Chang & Keisler 1973]. 24 Lemma. 25 Definition. Let E C M be linearly ordered by <, which need not be a definable relation in M. The ordered sequence (E, <) is a sequence of order indiscernibles if for any φ(x) and any pair e, e' of finite sequences from E, which are both in increasing order, (= φ(e) <-* φ(e'). The following result is proved using the compactness theorem and Ramsey's theorem.

7 it is rather easy to see that if φ(x) is strongly minimal in T then the dependence relation defined by algebraic closure allows us to assign a dimension to the strongly minimal set. If T is not No-categorical, it is easy to conclude that infinitely many distinct finite dimensions are possible. Thus when φ contains no parameters one concludes that T has NO countable models. 34, some finite inessential extension of every NI but not N0-categorical theory has infinitely many countable models. To prove that T itself has infinitely many models is much more difficult.

Formula of the form 0(0; 6 — y) or is inconsistent. 10 Exercise. p. formula without parameters then φ(M) is a submodule of Mn. p. p. definable subgroups where the meet of two such subgroups is their intersection and the join is defined by (φ + ψ)(v) = (3w)φ(w) Λ ψ(v — w). The following exercise is crucial to our development. 11 Exercise. p. formula φ and any α, 6, either 0(z;α) and 0(x;6) are equivalent or they are contradictory. Here is the main theorem of this section. 12 Theorem. Fix an R- module M.