Fact 2.4. If T is $\omega $ -categorical and $\mathcal {M} \models T$ then, for every $n < \omega $ , $Th(\mathcal {M}_n)$ is $\omega $ -categorical.

Fact 2.5. (i) If T is $\omega $ -categorical then, for all $0 < n < \omega $ , there are only finitely many complete types over $\emptyset $ in the free variables $x_1, \ldots , x_n$ and each one of them is implied (modulo T) by a single formula (the formula that isolates the type). Hence, there are only finitely many nonequivalent (modulo T) formulas in the same free variables.

(ii) If T is $\omega $ -categorical, $\mathcal {M} \models T$ , $n < \omega $ , and $B \subseteq M^{\mathrm {eq}}$ is finite, then $\mathrm {acl}^n(B)$ is finite.

Proof. Let $n < \omega $ , let $A \subseteq M_n$ be finite, with cardinality $s < \omega $ , say, and let $B \subseteq M^{\mathrm {eq}}$ . Suppose that $a_1, \ldots , a_m \in \mathrm {acl}^n(A)$ and, for all $k = 1, \ldots , m$ , $a_k \notin \mathrm {acl}^{\mathrm {eq}}(a_1, \ldots , a_{k-1}, B)$ . Then $a_i \neq a_j$ if $i \neq j$ so $|\mathrm {acl}^n(A)| \geq m$ . Since $Th(\mathcal {M}_n)$ is $\omega $ -categorical (by Fact 2.4) it follows that there is $r < \omega $ such that whenever $A' \subseteq M^n$ and $|A'| \leq s$ , then $|\mathrm {acl}^n(A')| \leq r$ . Hence $m \leq |\mathrm {acl}^n(A)| \leq r$ .

Proof. We prove by induction that if $\mathrm {rk}^n(A / B) \geq \alpha $ then $\mathrm {rk}^{n+1}(A / B) \geq \alpha $ . This is clear for $\alpha = 0$ . So suppose that $\mathrm {rk}^n(A / B) \geq \alpha + 1$ . Then there is $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ such that $\mathrm {rk}^n(A / aB) \geq \alpha $ . By the induction hypothesis we get $\mathrm {rk}^{n+1}(A / aB) \geq \alpha $ . We have $a \in \mathrm {acl}^n(A) \subseteq \mathrm {acl}^{n+1}(A)$ and $a \notin \mathrm {acl}^n(B)$ so (by Lemma 2.3) $a \notin \mathrm {acl}^{n+1}(B)$ . Hence $\mathrm {rk}^{n+1}(A / B) \geq \alpha + 1$ . If $\alpha $ is a limit ordinal and $\mathrm {rk}^n(A / B) \geq \alpha $ then $\mathrm {rk}^n(A / B) \geq \beta $ for all $\beta < \alpha $ , so by the induction hypothesis $\mathrm {rk}^{n+1}(A / B) \geq \beta $ for all $\beta < \alpha $ , hence $\mathrm {rk}^{n+1}(A / B) \geq \alpha $ .

Proof. Let $\alpha < \omega $ and $A, B \subseteq M^{\mathrm {eq}}$ . We prove (i) by induction on $\alpha $ . For $\alpha = 0$ the statement is vacuous. Suppose that $\mathrm {rk}^n(A / B) \geq \alpha + 1$ . Then there is $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ such that $\mathrm {rk}^n(A / aB) \geq \alpha $ . By the induction hypothesis there are $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ such that, for all $k = 1, \ldots , \alpha $ , $a_k \notin \mathrm {acl}^n(\{a, a_1, \ldots , a_{k-1}\} \cup B)$ . If we rename $a_i$ by $a_{i+1}$ for $i = 1, \ldots , \alpha $ and then rename a by $a_1$ we get $a_i \notin \mathrm {acl}^n(\{a_1, \ldots , a_{i-1}\} \cup B)$ for all $i = 1, \ldots , \alpha + 1$ .

Now suppose that $a_1, \ldots , a_{\alpha +1} \in \mathrm {acl}^n(A)$ and, for all $k = 1, \ldots , \alpha + 1$ ,

$$\begin{align*}a_k \notin \mathrm{acl}^n(\{a_1, \ldots, a_{k-1}\} \cup B). \end{align*}$$

By the induction hypothesis we have $\mathrm {rk}^n(A / a_1 B) \geq \alpha $ . By the choice of $a_1, \ldots , a_{\alpha + 1}$ we have $a_1 \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ , so $\mathrm {rk}^n(A / B) \geq \alpha + 1$ . Now we have proved part (i). Part (ii) follows directly from part (i).

(iii) Suppose that $\mathrm {rk}^n(A / B) = \alpha $ and that $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ are such that, for all $k = 1, \ldots , \alpha $ , $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup B)$ . Then $\mathrm {rk}^n(a_k / \{a_1, \ldots , a_{k-1}\} \cup B) \geq 1$ for all $k = 1, \ldots , \alpha $ . Suppose, for a contradiction, that for some k, $\mathrm {rk}^n(a_k / \{a_1, \ldots , a_{k-1}\} \cup B) \geq 2$ . Choose the least such k. Then there is

$$\begin{align*}a \in \mathrm{acl}^n(a_k) \setminus \mathrm{acl}^n(\{a_1, \ldots, a_{k-1}\} \cup B) \end{align*}$$

such that $\mathrm {rk}^n(a_k / \{a, a_1, \ldots , a_{k-1}\} \cup B) \geq 1$ which in particular means that

$$\begin{align*}a_k \notin \mathrm{acl}^n(\{a, a_1, \ldots, a_{k-1}\} \cup B). \end{align*}$$

Since $a \in \mathrm {acl}^n(a_k)$ it follows from the choice of $a_1, \ldots , a_\alpha $ that, for all $i = k+1, \ldots , \alpha $ , $a_i \notin \mathrm {acl}^n(\{a, a_1, \ldots , a_{i-1}\} \cup B)$ . Since $a \in \mathrm {acl}^n(a_k) \subseteq \mathrm {acl}^n(A)$ it now follows from part (i) that $\mathrm {rk}^n(A / B) \geq \alpha + 1$ , contradicting the assumption.

By assumption and part (ii), for every $k = 1, \ldots , \alpha $ , the number $\alpha - k$ is maximal such that there are $a^{\prime }_{k+1}, \ldots , a^{\prime }_\alpha \in \mathrm {acl}^n(A)$ such that, for all $i = k+1, \ldots , \alpha $ ,

$$\begin{align*}a^{\prime}_i \notin \mathrm{acl}^{\mathrm{eq}}(\{a_1, \ldots, a_k, a^{\prime}_{k+1}, \ldots, a^{\prime}_{i-1}\} \cup B). \end{align*}$$

Hence $\mathrm {rk}(A / \{a_1, \ldots , a_k\} \cup B) = \alpha - k$ for all $k = 1, \ldots , \alpha $ .

Let $k \in \{1, \ldots , \alpha \}$ . By part (i), the sequence $a_1, \ldots , a_k$ witnesses that $\mathrm {rk}(a_1, \ldots , a_k / B)\, \geq k$ . Suppose, for a contradiction, that $\mathrm {rk}(a_1, \ldots , a_k / B) \geq k+1$ . Then (by part (i)) we find

$$\begin{align*}a^{\prime}_0, \ldots, a^{\prime}_k \in \mathrm{acl}^n(a_1, \ldots, a_k) \subseteq \mathrm{acl}^n(A) \end{align*}$$

such that, for all $i = 0, \ldots , k$ , $a^{\prime }_i \notin \mathrm {acl}^n(\{a^{\prime }_1, \ldots , a^{\prime }_{i-1}\} \cup B)$ . Then $\mathrm {acl}^n(\{a^{\prime }_0, \ldots , a^{\prime }_k\} \cup B) \subseteq \mathrm {acl}^n(\{a_1, \ldots , a_k\} \cup B)$ so, for all $i = k+1, \ldots , \alpha $ , $a_i \notin \mathrm {acl}^n(\{a^{\prime }_0, \ldots , a^{\prime }_k, a_{k+1}, \ldots , a_{i-1}\} \cup B)$ By part (i), the sequence $a^{\prime }_0, \ldots , a^{\prime }_k, a_{k+1}, \ldots , a_\alpha $ witnesses that $\mathrm {rk}(A / B) \geq \alpha +1$ which contradicts the assumption.

Proof. Let $A, B \subseteq M^{\mathrm {eq}}$ where A is finite. Then $|\mathrm {acl}^n(A)| = \alpha $ for some $\alpha < \omega $ . If $\mathrm {rk}^n(A / B) \geq \alpha + 1$ then, by Lemma 3.5(i), there are $a_1, \ldots , a_{\alpha + 1} \in \mathrm {acl}^n(A)$ such that, for all $k = 1, \ldots , \alpha + 1$ , $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup B)$ . Then $a_i \neq a_j$ if $i \neq j$ so $|\mathrm {acl}^n(A)| \geq \alpha + 1$ , a contradiction.

Proof. (i) Suppose that $\mathrm {rk}^n(A / B) \geq \alpha $ . By Lemma 3.5, there are $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ such that, for all $k = 1, \ldots , \alpha $ , $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup B)$ . By induction on $\alpha $ we prove that there are $a^{\prime }_1, \ldots , a^{\prime }_\alpha \in \mathrm {acl}^0(A)$ such that $a^{\prime }_k \notin \mathrm {acl}^0(\{a^{\prime }_1, \ldots , a^{\prime }_{k-1}\} \cup B)$ and $a^{\prime }_k \in \mathrm {acl}^0(a_k)$ for all $k = 1, \ldots , \alpha $ , and it follows from Lemma 3.5 that $\mathrm {rk}^0(A / B) \geq \alpha $ . If $\alpha = 0$ there is nothing to prove, so suppose that $\alpha> 0$ . By the induction hypothesis there are $a^{\prime }_1, \ldots , a^{\prime }_{\alpha -1} \in \mathrm {acl}^0(A)$ such that $a^{\prime }_k \notin \mathrm {acl}^0(\{a^{\prime }_1, \ldots , a^{\prime }_{k-1}\} \cup B)$ and $a^{\prime }_k \in \mathrm {acl}^0(a_k)$ for all $k = 1, \ldots , \alpha -1$ . Since T is assumed to have geometric elimination of imaginaries, there is a finite sequence $\bar {c}$ of elements from $M_0 = M$ such that $\mathrm {acl}^{\mathrm {eq}}(a_\alpha ) = \mathrm {acl}^{\mathrm {eq}}(\bar {c})$ . Since $a_\alpha \notin \mathrm {acl}^n(\{a_1, \ldots , a_{\alpha -1}\} \cup B)$ we have $\mathrm {rng}(\bar {c}) \not \subseteq \mathrm {acl}^n(\{a_1, \ldots , a_{\alpha -1}\} \cup B)$ . As $a^{\prime }_k \in \mathrm {acl}^0(a_k)$ for all $k = 1, \ldots , \alpha - 1$ it follows that $\mathrm {rng}(\bar {c}) \not \subseteq \mathrm {acl}^0(\{a^{\prime }_1, \ldots , a^{\prime }_{\alpha - 1}\} \cup B)$ . Thus we can choose some element from $\bar {c}$ , which we call $a^{\prime }_\alpha $ , such that $a^{\prime }_\alpha \notin \mathrm {acl}^0(\{a^{\prime }_1, \ldots , a^{\prime }_{\alpha -1}\} \cup B)$ . Now Lemma 3.5 implies that $\mathrm {rk}^0(A / B) \geq \alpha $ .

(ii) Suppose that A is finite. By Lemma 3.7, $\mathrm {rk}^n(A / B) = \alpha $ for some $\alpha < \omega $ . By Lemma 3.4, $\mathrm {rk}^n(A / B) \geq \mathrm {rk}^0(A / B)$ , and by part (i) $\mathrm {rk}^n(A / B) \leq \mathrm {rk}^0(A / B)$ .

Proof. Suppose that $A, B, C, D \subseteq M^{\mathrm {eq}}$ where $A \supseteq B$ , $C \subseteq D,$ and A is finite, so all mentioned n-ranks in (a)–(d) are finite. Part (a) is obvious from the definition of n-rank.

For part (b), suppose that $\mathrm {rk}^n(B / C) \geq \alpha $ (where $\alpha < \omega $ ). By Lemma 3.5, there are $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(B)$ such that, for all $k = 1, \ldots , \alpha $ , $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup C)$ . Since $A \supseteq B$ it follows that $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ , so, by Lemma 3.5 again, $\mathrm {rk}^n(A / C) \geq \alpha $ .

Now consider part (c). Suppose that $\mathrm {rk}^n(A / D) \geq \alpha $ . Then there are $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ such that, for all $k = 1, \ldots , \alpha $ , $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup D)$ . Since $C \subseteq D$ we get $a_k \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k-1}\} \cup C)$ for all $k = 1, \ldots , \alpha $ , so $\mathrm {rk}^n(A / C) \geq \alpha $ .

Finally we consider part (d). Let $\beta = \mathrm {rk}^n(B / C)$ and $\alpha = \mathrm {rk}^n(A / BC)$ , so $\alpha , \beta < \omega $ . By Lemma 3.5, there are $b_1, \ldots b_\beta \in \mathrm {acl}^n(B)$ and $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ such that, for all $k = 1, \ldots , \beta $ , $b_k \notin \mathrm {acl}^n(\{b_1, \ldots , b_{k-1}\} \cup C)$ , and for all $l = 1, \ldots , \alpha $ , $a_l \notin \mathrm {acl}^n(\{a_1, \ldots , a_{l-1}\} \cup BC)$ . Moreover, $\beta $ and $\alpha $ are maximal such that such sequences exist. For a contradiction, suppose that, for some $l \in \{1, \ldots , \alpha \}$ , $a_l \in \mathrm {acl}^n(\{b_1, \ldots , b_\beta , a_1, \ldots , a_{l-1}\} \cup C)$ . Since $b_1, \ldots , b_\beta \in \mathrm {acl}^n(B)$ we get $a_l \in \mathrm {acl}^n(\{a_1, \ldots , a_{l-1}\} \cup BC)$ which contradicts the choice of $a_1, \ldots , a_\alpha $ . Hence we conclude that $a_l \notin \mathrm {acl}^n(\{b_1, \ldots , b_\beta , a_1, \ldots , a_{l-1}\} \cup C)$ for all $l = 1, \ldots , \alpha $ . Since $B \subseteq A$ we have $\mathrm {acl}^n(B) \subseteq \mathrm {acl}^n(A)$ and therefore $b_1, \ldots , b_\beta , a_1, \ldots , a_\alpha \in \mathrm {acl}^n(A)$ . By Lemma 3.5, we get $\mathrm {rk}(A / C) \geq \beta + \alpha $ .

Proof. Note that (by Lemma 3.7), if $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ then $\mathrm {rk}^n(a / B) < \omega $ . We first prove the following claim.

Claim. If $a' \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ and $\mathrm {rk}^n(a' / B) = \alpha + 1$ where $\alpha \geq 1$ , then there is $a \in \mathrm {acl}^n(a') \subseteq \mathrm {acl}^n(A)$ such that $1 \leq \mathrm {rk}^n(a / B) \leq \alpha $ .

Proof of the claim.

Suppose that $a' \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ and ${\mathrm {rk}^n(a' / B) = \alpha + 1}$ where $\alpha \geq 1$ . Suppose that $a' \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ and $\mathrm {rk}^n(a' / B) = \alpha + 1$ where $\alpha \geq 1$ . By the definition of $\mathrm {rk}^n$ there is $a \in \mathrm {acl}^n(a') \setminus \mathrm {acl}^n(B)$ such that ${\mathrm {rk}^n(a' / aB) = \alpha }$ . If $a' \in \mathrm {acl}^n(aB)$ then $\mathrm {rk}^n(a' / aB) = 0$ , so $\alpha = 0$ , which contradicts that $\alpha \geq 1$ . Hence $a' \notin \mathrm {acl}^n(aB)$ . As $a \notin \mathrm {acl}^n(B)$ we have $\mathrm {rk}^n(a / B) \geq 1$ .

We now show that $\mathrm {rk}^n(a / B) \leq \alpha $ . For a contradiction, suppose that $\mathrm {rk}^n(a / B) \geq \alpha + 1$ . By Lemma 3.5 there are $a_1, \ldots , a_{\alpha + 1} {\kern-1pt}\in{\kern-1pt} \mathrm {acl}^n(a)$ such that for all ${i {\kern-1pt}={\kern-1pt} 1, \ldots , \alpha {\kern-1pt}+{\kern-1pt} 1}$ , $a_i \notin \mathrm {acl}^n(\{a_1, \ldots , a_{i-1}\} \cup B)$ . If $a' \in \mathrm {acl}^n(\{a_1, \ldots , a_{\alpha + 1}\} \cup B)$ , then $a' \in \mathrm {acl}^n(aB)$ contradicting what we concluded above. Hence $a' \notin \mathrm {acl}^n(aB)$ . Therefore the sequence $a_1, \ldots , a_{\alpha + 1}, a' \in \mathrm {acl}^n(a')$ and Lemma 3.5 witness that $\mathrm {rk}^n(a' / B) \geq \alpha + 2$ which contradicts the assumption about $a'$ . This concludes the proof of the claim.

Now suppose that $a' \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ . Let $\alpha = \mathrm {rk}^n(a' / B)$ . Since $a' \notin \mathrm {acl}^n(B)$ we have $\alpha \geq 1$ . If $\alpha = 1$ then we are done (by letting $a = a'$ ). So suppose that $\alpha \geq 2$ . By repeatedly using the claim we eventually find $a \in \mathrm {acl}^n(a') \subseteq \mathrm {acl}^n(A)$ such that $\mathrm {rk}^n(a / B) = 1$ . This ends the proof of the lemma.

Proof. (i) Let $A \subseteq M^{\mathrm {eq}}$ be finite and let $B \subseteq C \subseteq M^{\mathrm {eq}}$ . We prove the claim by induction on $\mathrm {rk}^n(A / B)$ . If $\mathrm {rk}^n(A / B) = 0$ then there is nothing to prove (as we cannot have $\mathrm {rk}^n(A / C) < \mathrm {rk}^n(A / B)$ ). So suppose that $\mathrm {rk}^n(A / B) = \alpha + 1$ . Then there is $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ such that $\mathrm {rk}^n(A / \{a\} \cup B) \geq \alpha $ . In fact we must have $\mathrm {rk}^n(A / \{a\} \cup B) = \alpha $ because if $\mathrm {rk}^n(A / \{a\} \cup B) \geq \alpha +1$ then $ \mathrm {rk}^n(A / B) \geq \alpha +2$ which contradicts the assumption. Since A is finite it follows that $\mathrm {acl}^n(A)$ is finite and therefore there is finite $C_1 \subseteq C$ such that

$$\begin{align*}\mathrm{acl}^n(A) \cap \mathrm{acl}^n(C) = \mathrm{acl}^n(A) \cap \mathrm{acl}^n(C_1). \end{align*}$$

By the induction hypothesis, if $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ is such that $\mathrm {rk}^n(A / \{a\} \cup C) < \mathrm {rk}^n(A / \{a\} \cup B) = \alpha $ , then there is finite $C_{a} \subseteq C$ such that $\mathrm {rk}^n(A / \{a\} \cup B \cup C_{a}) < \mathrm {rk}^n(A / \{a\} \cup B)$ . Let $C_2$ be the union of all $C_{a} \subseteq C$ where a ranges over the members of the finite set $\mathrm {acl}^n(A)$ such that $\mathrm {rk}^n(A / \{a\} \cup C) < \mathrm {rk}^n(A / \{a\} \cup B)$ . Let $C' = C_1 \cup C_2$ so $C'$ is finite.

Now suppose that $\mathrm {rk}^n(A / C) < \mathrm {rk}^n(A / B)$ . Then for every $a \in \mathrm {acl}^n(A) \setminus \mathrm {acl}^n(B)$ such that $\mathrm {rk}^n(A / \{a\} \cup B) = \alpha $ we have either $a \in \mathrm {acl}^{\mathrm {eq}}(C)$ or $\mathrm {rk}^n(A / \{a\} \cup C) < \alpha $ which in turn implies that either $a \in \mathrm {acl}^n(C')$ or $\mathrm {rk}^n(A / \{a\} \cup B \cup C') < \mathrm {rk}^n(A / \{a\} \cup B)$ . It follows that $\mathrm {rk}^n(A / B \cup C') < \alpha +1 = \mathrm {rk}^n(A / B)$ .

(ii) Suppose that $\mathrm {rk}^n(A / B) = \alpha $ (where A is finite) and $\mathrm {rk}^n(A / \emptyset ) = \beta $ so $\alpha \leq \beta $ (by Lemma 3.9). If $\alpha = \beta $ we are done. So suppose that $\alpha < \beta $ . Let $B_0 = \emptyset $ . By part (i) there is finite $B_1 \subseteq B$ such that $\mathrm {rk}^n(A / B_1) < \mathrm {rk}^n(A / B_0)$ . If $\mathrm {rk}^n(A / B) < \mathrm {rk}^n(A / B_1)$ then we use part (i) again and get a finite $B_2 \subseteq B$ such that $\mathrm {rk}^n(A / B_0 \cup B_1 \cup B_2) < \mathrm {rk}^n(A / B_0 \cup B_1)$ . We can repeat this procedure as long as $\mathrm {rk}^n(A / B) < \mathrm {rk}^n(A / B_0 \cup \cdots \cup B_m)$ . But as there is no infinite decreasing sequence of natural numbers we eventually find $m < \omega $ and finite $B_k \subseteq B$ for $k \leq m$ such that $\mathrm {rk}^n(A / B_0 \cup \cdots \cup B_m) = \mathrm {rk}^n(A / B)$ .

Proof. Suppose that $A \subseteq M^{\mathrm {eq}}$ is finite and $B \subseteq M^{\mathrm {eq}}$ . Lemma 3.11 says that for every $n < \omega $ there is a finite $B_n \subseteq B$ such that $\mathrm {rk}^n(A / B_n) = \mathrm {rk}^n(A / B)$ . Let $C = \bigcup _{n < \omega } B_n$ , so $C \subseteq B$ is countable. By Lemma 3.9, for all $n < \omega $ , $\mathrm {rk}^n(A / B_n) \geq \mathrm {rk}^n(A / C) \geq \mathrm {rk}^n(A / B) = \mathrm {rk}^n(A / B_n)$ , and hence $\mathrm {rk}^n(A / C) = \mathrm {rk}^n(A / B_n) = \mathrm {rk}^n(A / B)$ .

Proof. (i) Suppose that $n \leq m < \omega $ . Since T is $\omega $ -categorical it follows that $Th(\mathcal {M}_m)$ is $\omega $ -categorical and from this it follows that, for all $k < \omega $ , the $(k+1)$ -ary relation on $M_m$ which holds for $(b, a_1, \ldots , a_k) \in (M_m)^{k+1}$ if and only if $b \in \mathrm {acl}^n(a_1, \ldots , a_k)$ is $\emptyset $ -definable in $\mathcal {M}_m$ . It follows that for all $r, k, l < \omega $ , the $(k+l)$ -ary relation on $M_m$ which holds for $(a_1, \ldots , a_k, b_1, \ldots , b_l) \in (M_m)^{k+l}$ if and only if r is maximal such that

(3.1) $$ \begin{align} &\text{there are } \ a^{\prime}_1, \ldots, a^{\prime}_r \in \mathrm{acl}^n(a_1, \ldots, a_k) \ \text{ such that,}\\ &\text{for all } i = 1, \ldots, r, \ a^{\prime}_i \notin \mathrm{acl}^n(a^{\prime}_1, \ldots, a^{\prime}_{i-1}, b_1, \ldots, b_l) \nonumber \end{align} $$

is $\emptyset $ -definable in $\mathcal {M}_m$ , by $\varphi _r^m(x_1, \ldots , x_k, y_1, \ldots , y_l)$ say. Then the formula $\psi _r^m(x_1, \ldots , x_k,\, y_1, \ldots , y_l)$ which expresses that “all $x_1, \ldots , x_k, y_1, \ldots , y_l$ belong to $M_m$ and $\varphi _r^m(x_1, \ldots , x_k,\, y_1, \ldots , y_l)$ holds” defines the same relation in $\mathcal {M}^{\mathrm {eq}}$ .

By Lemma 3.5, $\mathrm {rk}^n(a_1, \ldots , a_k / b_1, \ldots , b_l) = r$ if and only if r is maximal such that (3.1) holds. It follows that, for every $r < \omega $ , the $(k+l)$ -ary relation on $(M_m)^{k+l}$ which holds for $(a_1, \ldots , a_k, b_1, \ldots , b_l) \in (M_m)^{k+l}$ if and only if $\mathrm {rk}^n(a_1, \ldots , a_k / b_1, \ldots , b_l) = r$ is $\emptyset $ -definable in $\mathcal {M}_m$ by $\varphi _r^m(x_1, \ldots , x_k, y_1, \ldots , y_l)$ . The same relation is definable in $\mathcal {M}^{\mathrm {eq}}$ by $\psi _r^m(x_1, \ldots , x_k, y_1, \ldots , y_l)$ .

(ii) Suppose that $\bar {a}, \bar {a}' \in (M^{\mathrm {eq}})^k$ , $\bar {b}, \bar {b}' \in (M^{\mathrm {eq}})^l$ and that $\mathrm {rk}^n(\bar {a} / \bar {b}) = r$ . Then there is $m < \omega $ such that $n \leq m$ , $\bar {a}, \bar {a}' \in (M_m)^k$ , and $\bar {b}, \bar {b}' \in (M_m)^l$ . Then $\mathcal {M}^{\mathrm {eq}} \models \psi _r^m(\bar {a}, \bar {b})$ , and if $\mathrm {tp}(\bar {a}, \bar {b}) = \mathrm {tp}(\bar {a}', \bar {b}')$ then also $\mathcal {M}^{\mathrm {eq}} \models \psi _r^m(\bar {a}', \bar {b}')$ , so $\mathrm {rk}^n (\bar {a}' / \bar {b}') = r$ .

Proof. Let S be the sort of a. Since $a \notin \mathrm {acl}^{\mathrm {eq}}(B)$ it follows that $\mathrm {tp}(a / B)$ has infinitely many realizations. As C is finite it follows (from $\omega $ -categoricity) that $\mathrm {acl}^{\mathrm {eq}}(C) \cap S$ is finite and hence there is $a' \notin \mathrm {acl}^{\mathrm {eq}}(C)$ such that $\mathrm {tp}(a' / B) = \mathrm {tp}(a / B)$ .

Proof. Let $\bar {d}$ be a finite sequence of elements from $M_n$ . Suppose that ${\alpha = \mathrm {rk}^n(\bar {d} / B)}$ . By Lemma 3.5 there are $a_1, \ldots , a_\alpha \in \mathrm {acl}^n(\bar {d})$ such that,

(3.2) $$ \begin{align} \text{ for all } k = 1, \ldots, \alpha, \ &\mathrm{rk}^n(a_k / \{a_1, \ldots, a_{k-1}\} \cup B) = 1, \\ &\mathrm{rk}^n(a_1, \ldots, a_k / B) = k, \text{ and} \nonumber \\ &\mathrm{rk}^n(\bar{d} / \{a_1, \ldots, a_k\} \cup B) = \alpha-k. \nonumber \end{align} $$

In particular, $\mathrm {rk}^n(a_1, \ldots , a_\alpha / B) = \alpha $ and $\mathrm {rk}^n(\bar {d} / \{a_1, \ldots , a_\alpha \} \cup B) = 0$ , so (as $\mathrm {rng}(\bar {d}) \subseteq M_n$ ) $\mathrm {rng}(\bar {d}) \subseteq \mathrm {acl}^n(\bar {d}) \subseteq \mathrm {acl}^n(\{a_1, \ldots , a_\alpha \} \cup B)$ and hence $\mathrm {acl}^n(\bar {d}B) = \mathrm {acl}^n(\{a_1, \ldots , a_\alpha \} \cup B)$ .

Suppose that $k < \alpha $ and that there are $a^{\prime }_1, \ldots , a^{\prime }_k \in M_n$ such that

(3.3) $$ \begin{align} &\mathrm{tp}(a^{\prime}_1, \ldots, a^{\prime}_k / B) = \mathrm{tp}(a_1, \ldots, a_k / B), \ \text{ and} \end{align} $$

(3.4) $$ \begin{align} &\text{for all } i = 1, \ldots, k, \ a^{\prime}_i \notin \mathrm{acl}^n(\{a^{\prime}_1, \ldots, a^{\prime}_{i-1}\} \cup C). \end{align} $$

We will find $a^{\prime }_{k+1}$ so that the above also holds with k replaced by $k+1$ . Let

$$\begin{align*}p(x, y_1, \ldots, y_k) = \mathrm{tp}(a_{k+1}, a_1, \ldots, a_k / B). \end{align*}$$

By (3.2), $p(x, a_1, \ldots , a_k)$ is non-algebraic. By (3.3), also $p(x, a^{\prime }_1, \ldots , a^{\prime }_k)$ is non-algebraic, so by Lemma 3.14 there is $a^{\prime }_{k+1} \in M_n \setminus \mathrm {acl}^n(\{a^{\prime }_1, \ldots , a^{\prime }_k\} \cup C)$ which realizes $p(x, a^{\prime }_1, \ldots , a^{\prime }_k)$ , that is, $\mathrm {tp}(a^{\prime }_1, \ldots , a^{\prime }_k, a^{\prime }_{k+1} / B) = \mathrm {tp}(a_1, \ldots , a_k, a_{k+1} / B)$ . Now (3.3) and (3.4) hold if k is replaced by $k+1$ . By induction it follows that there are $a^{\prime }_1, \ldots , a^{\prime }_\alpha \in M_n$ such that

(3.5) $$ \begin{align} &\mathrm{tp}(a^{\prime}_1, \ldots, a^{\prime}_\alpha / B) = \mathrm{tp}(a_1, \ldots, a_\alpha / B), \ \text{ and} \end{align} $$

(3.6) $$ \begin{align} &\text{for all } i = 1, \ldots, \alpha, \ a^{\prime}_i \notin \mathrm{acl}^n(\{a^{\prime}_1, \ldots, a^{\prime}_{i-1}\} \cup C). \end{align} $$

From (3.5) it follows that there is $\bar {d}'$ such that

(3.7) $$ \begin{align} \mathrm{tp}(\bar{d}', a^{\prime}_1, \ldots, a^{\prime}_\alpha / B) = \mathrm{tp}(\bar{d}, a_1, \ldots, a_\alpha / B). \end{align} $$

Then $a^{\prime }_1, \ldots , a^{\prime }_\alpha \in \mathrm {acl}^n(\bar {d}')$ . This together with (3.6) implies that $\mathrm {rk}^n(\bar {d}' / C) \geq \alpha $ . From (3.7) and Lemma 3.13 we get $\mathrm {rk}^n(\bar {d}' / B) = \mathrm {rk}^n(\bar {d} / B) = \alpha $ . Since we must (by Lemma 3.9) have $\mathrm {rk}^n(\bar {d}' / C) \leq \mathrm {rk}^n(\bar {d}' /B)$ we get $\mathrm {rk}^n(\bar {d}' / C) = \alpha $ . From (3.7) we get $\mathrm {tp}(\bar {d}' / B) = \mathrm {tp}(\bar {d} / B)$ so the proof is completed.

Proof. Let $\bar {d}$ be a finite sequence of elements from $M_n$ and let $\alpha = \mathrm {rk}^n(\bar {d} / B)$ . Hence $\mathrm {rk}^n(\bar {d} / B') \geq \alpha $ for every finite $B' \subseteq B$ . By Lemma 3.13, for all finite $B' \subseteq B$ and finite $C' \subseteq C$ there is a formula $\varphi _{B'C'}(\bar {x})$ with parameters from $B' \cup C'$ that expresses that $\mathrm {tp}(\bar {x} / B') = \mathrm {tp}(\bar {d} / B')$ and $\mathrm {rk}^n(\bar {x} / B'C') \geq \alpha $ . Let $\Phi (\bar {x})$ be the set of all such formulas $\varphi _{B'C'}(\bar {x})$ as $B'$ and $C'$ varies over finite subsets of B and C, respectively. From Lemma 3.15 it follows that every finite subset of $\Phi (\bar {x})$ is consistent. From compactness it follows that $\Phi (\bar {x})$ is consistent. Let $\bar {d}'$ be a realization of the type $\Phi (\bar {x})$ . It follows that $\mathrm {tp}(\bar {d}' / B) = \mathrm {tp}(\bar {d} / B)$ and, by Lemma 3.11, that $\mathrm {rk}^n(\bar {d}' / C) \geq \alpha $ . Since we must (by Lemmas 3.9 and 3.13) have $\mathrm {rk}^n(\bar {d}' / C) \leq \mathrm {rk}^n(\bar {d}' / B) = \mathrm {rk}^n(\bar {d} / B) = \alpha $ it follows that $\mathrm {rk}^n(\bar {d}' / C) = \alpha $ .

Proof. Suppose that $\sigma $ is an elementary function and that $A, B$ , and C are subsets of the domain of $\sigma $ . Suppose that $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } B$ , so by Definition 4.1 of ${\smile} \hspace {-6.8pt}{|^n} $ we have $\mathrm {rk}^n(A' / BC) = \mathrm {rk}^n(A' / C)$ for all finite $A' \subseteq A$ . It suffices to show that for every finite $A' \subseteq A$ , $\mathrm {rk}^n(\sigma (A') / \sigma (B)\sigma (C)) = \mathrm {rk}^n(\sigma (A') / \sigma (C))$ . This follows if we can show that for all $A, B \subseteq M^{\mathrm {eq}}$ where A is finite, $\mathrm {rk}^n(A / B) = \mathrm {rk}^n(\sigma (A) / \sigma (B))$ . Since also the inverse of $\sigma $ is an elementary function, it actually suffices to show that $\mathrm {rk}^n(A / B) \geq \mathrm {rk}^n(\sigma (A) / \sigma (B))$ .

So let $A = \{a_1, \ldots , a_k\} \subseteq A$ be finite. Lemma 3.11 tells that there is finite ${B' = \{b_1, \ldots , b_l\} \subseteq B}$ such that $\mathrm {rk}^n(A / B') = \mathrm {rk}^n(A / B)$ . Since $\mathrm {tp}(a_1, \ldots , a_k, b_1, \ldots , b_l) = \mathrm {tp}(\sigma (a_1), \ldots , \sigma (a_k), \sigma (b_1), \ldots , \sigma (b_l))$ it follows from Lemmas 3.9 and 3.13 that

$$\begin{align*}\mathrm{rk}^n(\sigma(A) / \sigma(B)) \leq \mathrm{rk}^n(\sigma(a_1), \ldots, \sigma(a_k)/ \sigma(b_1), \ldots, \sigma(b_l)) = \mathrm{rk}^n(A / B') = \mathrm{rk}^n(A / B). \end{align*}$$

Proof. Let $A, B, C, D \subseteq M^{\mathrm {eq}}$ where $B \subseteq C \subseteq D$ . Suppose that $A \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } D$ and let $A' \subseteq A$ be finite. Then $\mathrm {rk}^n(A' / D) = \mathrm {rk}^n(A' / B)$ . By Lemma 3.9, $\mathrm {rk}^n(A' / B) = \mathrm {rk}^n(A' / D) \leq \mathrm {rk}^n(A' / C) \leq \mathrm {rk}^n(A' / B)$ . Hence $\mathrm {rk}^n(A' / D) = \mathrm {rk}^n(A' / C) = \mathrm {rk}^n(A' / B)$ and thus $A \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } C$ and $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } D$ .

Proof. Suppose that $A, B, C, D \subseteq M^{\mathrm {eq}}$ and $B \subseteq C \subseteq D$ . By the definition of n-independence it suffices to prove that for all finite $A' \subseteq A$ , if $A' \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } C$ and $A' \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } D$ then $A' \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } D$ . Therefore we can assume that A is finite. So suppose that $A \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } C$ and $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } D$ . Then $\mathrm {rk}^n(A / B) = \mathrm {rk}^n(A / C) = \mathrm {rk}^n(A / D)$ , so $A \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } D$ .

Proof. If $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n}\hspace {-12pt} {\diagup} } B$ then (by definition of ${\smile} \hspace {-6.8pt}{|^n}\hspace {-12pt} {\diagup} $ ) $A' \underset {C}{{\smile} \hspace {-6.8pt}{|^n}\hspace {-12pt} {\diagup} } B$ for some finite $A' \subseteq A$ . Then $\mathrm {rk}^n(A' / BC) < \mathrm {rk}^n(A' / C)$ . By Lemma 3.11(i) there is finite $B' \subseteq B$ such that $\mathrm {rk}^n(A' / B'C)\, <\, \mathrm {rk}^n(A' / C)$ . Hence $A' \underset {C}{{\smile} \hspace {-6.8pt}{|^n}\hspace {-12pt} {\diagup} } B'$ .

Proof. Let $A, B \subseteq M^{\mathrm {eq}}$ and suppose that A is finite. From Lemma 3.11(ii) it follows that there is a finite $C \subseteq B$ such that $\mathrm {rk}^n(A / C) = \mathrm {rk}^n(A / B)$ . Hence $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } B$ .

Proof. Under the given assumptions, Lemma 3.16 implies that there is a finite sequence $\bar {a}'$ of elements from $M_n$ such that $\mathrm {tp}(\bar {a}' / B) = \mathrm {tp}(\bar {a} / B)$ and $\mathrm {rk}^n(\bar {a}' / C) = \mathrm {rk}^n(\bar {a} / B)$ . Hence $\mathrm {rk}^n(\bar {a}' / B) = \mathrm {rk}^n(\bar {a} / B)$ , so $\bar {a}' \underset {B}{{\smile} \hspace {-6.8pt}{|^n} } C$ .

Proof. By definition of ${\smile} \hspace {-6.8pt}{|^n} $ , $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } B$ if and only if $\mathrm {rk}^n(A' / BC) = \mathrm {rk}^n(A' / C)$ for all finite $A' \subseteq A$ . Under the assumption that T has geometric elimination of imaginaries, Lemma 3.8 implies that $\mathrm {rk}^n(A' / BC) = \mathrm {rk}^0(A' / BC)$ and $\mathrm {rk}^n(A' / C) = \mathrm {rk}^0(A' / C)$ for all finite $A' \subseteq A$ . Hence $A \underset {C}{{\smile} \hspace {-6.8pt}{|^n} } B$ if and only if $A \underset {C}{{\smile} \hspace {-6.8pt}{|^0} } B$ .

Assumption 5.1. (Exchange property with respect to n)

In this section we fix an arbitrary $n < \omega $ . We assume that if $C \subseteq M^{\mathrm {eq}}$ , $2 \leq k < \omega $ , $a_1, \ldots , a_k \in M_n$ , $\mathrm {rk}^n(a_i / C) = 1$ for all $i = 1, \ldots , k$ , and $a_k \in \mathrm {acl}^{\mathrm {eq}}(\{a_1, \ldots , a_{k-1}\} \cup C) \setminus \mathrm {acl}^{\mathrm {eq}}(\{a_2, \ldots , a_{k-1}\} \cup C)$ , then $a_1 \in \mathrm {acl}^{\mathrm {eq}}(\{a_2, \ldots , a_k\} \cup C)$ .

Proof. It is well-known that for all $A \subseteq M^{\mathrm {eq}}$ , $A \subseteq \mathrm {acl}^{\mathrm {eq}}(A)$ , $\mathrm {acl}^{\mathrm {eq}}(\mathrm {acl}^{\mathrm {eq}}(A)) = \mathrm {acl}^{\mathrm {eq}}(A)$ , and if $a \in \mathrm {acl}^{\mathrm {eq}}(A)$ then $a \in \mathrm {acl}^{\mathrm {eq}}(A')$ for some finite $A' \subseteq A$ . It follows straightforwardly that $\mathrm {cl}$ has properties (1)–(3) in Definition 2.10, even without Assumption 5.1. Suppose that $A \subseteq X$ , $b, c \in X$ , and $b \in \mathrm {cl}(A \cup \{c\}) \setminus \mathrm {cl}(A)$ . Then there is finite $A' \subseteq A$ such that $b \in \mathrm {cl}(A' \cup \{c\}) \setminus \mathrm {cl}(A)$ . By Assumption 5.1, we get $c \in \mathrm {cl}(A \cup \{b\})$ . Hence $(X, \mathrm {cl})$ is a pregeometry.

Proof. We prove part (i) and point out during the argument why the uniqueness properties of part (ii) follow. Suppose that $\mathrm {acl}^n(A) \not \subseteq \mathrm {acl}^n(B)$ for otherwise the empty sequence has the required properties. By convention let $k_0 = 0$ . By Lemma 3.10 the following set is nonempty:

$$\begin{align*}X_1 = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / B) = 1 \}. \end{align*}$$

By Lemma 5.2 $(\{d \in M_n : \mathrm {rk}^n(d / B) = 1 \}, \mathrm {cl})$ with $\mathrm {cl}(Y) = \mathrm {acl}^n(YB) \cap \{d \in M_n : \mathrm {rk}^n(d / B) = 1 \}$ for all $Y \subseteq \{d \in M_n : \mathrm {rk}^n(d / B) = 1 \}$ is a pregeometry. Let $k_1$ be the dimension of $X_1$ and let $\{a_1, \ldots , a_{k_1}\} \subseteq X_1$ be a basis of $X_1$ . Note that $k_1$ is determined only by A, $B,$ and n. Also observe that for any other basis $\{a^{\prime }_1, \ldots , a^{\prime }_{k_1}\} \subseteq X_1$ we have $\mathrm {acl}^{\mathrm {eq}}(\{a^{\prime }_1, \ldots , a^{\prime }_{k_1}\} \cup B) = \mathrm {acl}^{\mathrm {eq}}(\{a_1, \ldots , a_{k_1}\} \cup B)$ .

Now suppose that, for some m, $k_1 < \cdots < k_m$ and $a_1, \ldots , a_{k_m} \in \mathrm {acl}^n(A)$ have been defined in such a way that, for all $j = 0, \ldots , m-1$ , $\{a_{k_j + 1}, \ldots , a_{k_{j+1}}\}$ is a basis of

$$\begin{align*}X_j = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_j}\} \cup B) = 1 \}. \end{align*}$$

Also suppose that the uniqueness properties of part (ii) hold for $k_1, \ldots , k_m$ and $\mathrm {acl}^{\mathrm {eq}}(\, \{a_1, \ldots , a_{k_i}\}\, \cup B)$ for $i \leq m$ . If $\mathrm {acl}^n(A) \subseteq \mathrm {acl}^n(\{a_1, \ldots , a_{k_m}\} \cup B)$ then we let $\alpha = k_m$ and then $a_1, \ldots , a_\alpha $ is an n-ccs for $A/B$ .

Suppose that $\mathrm {acl}^n(A) \not \subseteq \mathrm {acl}^n(\{a_1, \ldots , a_{k_m}\} \cup B)$ . By Lemma 3.10 the set

$$\begin{align*}X_{m+1} = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_m}\} \cup B) = 1 \} \end{align*}$$

is nonempty. Let d be the dimension of $X_{m+1}$ , let $k_{m+1} = k_m + d$ , and let

$$\begin{align*}\{a_{k_m + 1}, \ldots, a_{k_{m+1}}\} \subseteq X_{m+1} \end{align*}$$

be a basis of $X_{m+1}$ . Note that $k_{m+1}$ depends only on $k_m$ and the dimension of $X_{m+1}$ where the latter depends only on A, B, $a_1, \ldots , a_{k_m}$ , and n. The inductive assumption that for any choice of $a^{\prime }_1, \ldots , a^{\prime }_{k_m}$ with the same properties as $a_1, \ldots , a_{k_m}$ we have $\mathrm {acl}^{\mathrm {eq}}(\{a^{\prime }_1, \ldots , a^{\prime }_{k_m}\} \cup B) = \mathrm {acl}^{\mathrm {eq}}(\{a_1, \ldots , a_{k_m}\} \cup B)$ implies that $\mathrm {acl}^{\mathrm {eq}}(\{a^{\prime }_1, \ldots , a^{\prime }_{k_{m+1}}\} \cup B) = \mathrm {acl}^{\mathrm {eq}}(\{a_1, \ldots , a_{k_{m+1}}\} \cup B)$ for any basis $\{a^{\prime }_{k_m + 1}, \ldots , a^{\prime }_{k_{m+1}}\} \subseteq X_{m+1}$ . It also implies that, ultimately, $k_{m+1}$ depends only on A, $B,$ and n.

Observe that for all $i = 1, \ldots , k_{m+1}$ , $a_i \notin \mathrm {acl}^n(\{a_1, \ldots , a_{i-1}\} \cup B)$ so by Lemma 3.5 we have $\mathrm {rk}^n(A / B) \geq k_{m+1}$ . Since $\mathrm {rk}^n(A / B)$ is finite (as A is finite) it follows that the process of extending the current sequence $a_1, \ldots , a_{k_{m+1}}$ in the described way will terminate after finitely many steps and then we have an n-ccs for $A/B$ .

Proof. Suppose that $a_1, \ldots , a_\alpha $ is an n-ccs for $A/B$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_m = \alpha $ . From Definition 5.3 of an n-ccs it follows that, for all $i = 1, \ldots , \alpha $ , $a_i \notin \mathrm {acl}^n(\{a_1, \ldots , a_{i-1}\} \cup B)$ . Since $\alpha = \mathrm {rk}^n(A / B)$ it follows from Definition 3.6 of an n-cs that $a_1, \ldots , a_\alpha $ is an n-cs for $A/B$ .

Proof. Suppose that $\mathrm {rk}^n(A/B) = \alpha $ . By Lemma 5.4, it suffices to prove that there is at least one n-ccs for $A/B$ that has length $\alpha $ . By Lemma 3.5 and Definition 3.6 of n-cs, there is an n-cs $a_1, \ldots , a_\alpha $ for $A/B$ . By the same definition and lemma we have

(5.1) $$ \begin{align} &a_i \notin \mathrm{acl}^n(\{a_1, \ldots, a_{i-1}\} \cup B) \ \text{ and } \\ &\mathrm{rk}^n(a_i / \{a_1, \ldots, a_{i-1}\} \cup B) = 1 \ \ \text{ for all } i = 1, \ldots ,\alpha. \nonumber \end{align} $$

The rest of the proof will show how to transform $a_1, \ldots , a_\alpha $ into an n-ccs for $A/B$ with length $\alpha $ . This will be done step by step via a sequence of claims.

Claim 5.7. Suppose that $0 \leq k_1 < k_2 < \alpha $ and that $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is an independent subset of

$$\begin{align*}X = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_1}\} \cup B) = 1 \}. \end{align*}$$

If $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is not a basis of X then there are $d \in X$ and $k_2 < l \leq \alpha $ such that $\{a_{k_1 + 1}, \ldots , a_{k_2}, d\}$ is an independent subset of X and $a_1, \ldots , a_{l-1}, d, a_{l+1}, \ldots , a_\alpha $ is an n-cs for $A/B$ .

Proof. Suppose that $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is not a basis of X. There is $d \in X$ such that $d \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k_2}\} \cup B)$ . If $d \notin \mathrm {acl}^n(\{a_1, \ldots , a_\alpha \} \cup B)$ then it follows (by (5.1) and Lemma 3.5) that $\mathrm {rk}^n(A / B) \geq \alpha + 1$ , which contradicts the assumption. Hence there is a minimal $1 \leq l \leq \alpha $ such that $d \in \mathrm {acl}^n(\{a_1, \ldots , a_l\} \cup B)$ . Since $d \notin \mathrm {acl}^n(\{a_1, \ldots , a_{k_2}\} \cup B)$ we must have $l> k_2$ . Then $d \notin \mathrm {acl}^n(\{a_1, \ldots , a_{l-1}\} \cup B)$ . As also $d \in X$ it follows (using Lemma 3.9) that $\mathrm {rk}^n(d / \{a_1, \ldots , a_{l-1}\} \cup B) = 1$ . By (5.1), we have $\mathrm {rk}^n(a_l / \{a_1, \ldots , a_{l-1}\} \cup B) = 1$ . Now Assumption 5.1 implies that $a_l \in \mathrm {acl}^n( \{a_1, \ldots , a_{l-1}, d\} \cup B)$ .

Suppose, for a contradiction, that there is $i> l$ such that

$$\begin{align*}a_i \in \mathrm{acl}^n(\{a_1, \ldots, a_{l-1}, d, a_{l+1}, \ldots, a_{i-1}\} \cup B). \end{align*}$$

Since $d \in \mathrm {acl}^n(\{a_1, \ldots , a_l\} \cup B)$ it follows that $a_i \in \mathrm {acl}^n(\{a_1, \ldots , a_{i-1}\} \cup B)$ which contradicts (5.1). Hence, no such $i> l$ exists and it follows that $a_1, \ldots , a_{l-1}, d, a_{l+1}, \ldots , a_\alpha $ is an n-cs for $A/B$ .

Claim 5.8. Suppose that, for some $0 \leq k < \alpha $ and $1 \leq s \leq \alpha - k$ , we have

$$\begin{align*}\mathrm{rk}^n(a_{k+i} / \{a_1, \ldots, a_k\} \cup B) = 1 \ \text{ for all } i = 1, \ldots, s. \end{align*}$$

If $k+s < l \leq \alpha $ and $\mathrm {rk}^n(a_l / \{a_1, \ldots , a_k\} \cup B) = 1$ then

$$\begin{align*}a_1, \ldots, a_{k+s}, a_l, a_{k+s+1}, \ldots, a_{l-1}, a_{l+1}, \ldots, a_\alpha \ \ \text{ is an } n-\text{cs for } A/B. \end{align*}$$

Proof. Let k and s be as assumed. Suppose that $k+s < l \leq \alpha $ and $\mathrm {rk}^n(a_l / \{a_1, \ldots , a_k\} \cup B) = 1$ . Suppose, for a contradiction, that there is $k+s < j < l$ such that $a_j \in \mathrm {acl}^n(\{a_1, \ldots , a_{k+s}, a_l, a_{k+s+1}, \ldots , a_{j-1}\} \cup B)$ . By (5.1) and Lemma 3.9,

$$ \begin{align*} 1 &= \mathrm{rk}^n(a_l / \{a_1, \ldots, a_k\} \cup B) \\ &\geq \mathrm{rk}^n(a_l / \{a_1, \ldots, a_{j-1}\} \cup B) \\ &\geq \mathrm{rk}^n(a_l / \{a_1, \ldots, a_{l-1}\} \cup B) = 1. \end{align*} $$

Hence $\mathrm {rk}^n(a_l / \{a_1, \ldots , a_{j-1}\} \cup B) = 1$ . By (5.1) we also have $\mathrm {rk}^n(a_j / \{a_1, \ldots , a_{j-1}\} \cup B) = 1$ . Now Assumption 5.1 implies that $a_l \in \mathrm {acl}^n(\{a_1, \ldots , a_j\} \cup B)$ and since $j < l$ this contradicts (5.1).

For a contradiction, suppose that there is $l < j \leq \alpha $ such that

$$\begin{align*}a_j \in \mathrm{acl}^n(\{a_1, \ldots, a_{k+s}, a_l, a_{k+s+1}, \ldots, a_{l-1}, a_{l+1}, \ldots, a_{j-1}\} \cup B). \end{align*}$$

Since $\{a_1, \ldots , a_{k+s}, a_l, a_{k+s+1}, \ldots , a_{l-1}, a_{l+1}, \ldots , a_{j-1}\} = \{a_1, \ldots a_{j-1}\}$ we have a contradiction to (5.1). It follows that $a_1, \ldots , a_{k+s}, a_l, a_{k+s+1}, \ldots , a_{l-1}, a_{l+1}, \ldots , a_\alpha $ is an n-cs for $A/B$ .

Claim 5.9. Suppose that $0 \leq k_1 < k_2 < \alpha $ and that $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is an independent subset of

$$\begin{align*}X = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_1}\} \cup B) = 1\}. \end{align*}$$

If $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is not a basis of X then there is $d \in X$ such that $\{a_{k_1 + 1}, \ldots , a_{k_2}, d\}$ is an independent subset of X and, for some $k_2 < i \leq \alpha $ ,

$$\begin{align*}a_1, \ldots, a_{k_2}, d, a_{k_2 + 1}, \ldots, a_{i-1}, a_{i+1}, \ldots, a_\alpha \ \ \text{ is an } n-\text{cs for } A/B. \end{align*}$$

Proof. Let us make the assumptions of the claim. Moreover, suppose that $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is not a basis of X. Now we first apply Claim 5.7 to find $d \in X$ and $k_2 < i \leq \alpha $ such that $\{a_{k_1 + 1}, \ldots , a_{k_2}, d\}$ is an independent subset of X and $a_1, \ldots , a_{i-1}, d, a_{i+1}, \ldots , a_\alpha $ is an n-cs for $A/B$ . Then Claim 5.8 implies that also $a_1, \ldots , a_{k_2}, d, a_{k_2 + 1}, \ldots , a_{i-1}, a_{i+1}, \ldots , a_\alpha $ is an n-cs for $A/B$ .

Claim 5.10. Suppose that $0 \leq k_1 < k_2 < \alpha $ and that $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is an independent subset of

$$\begin{align*}X = \mathrm{acl}^n(A) \cap \{d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_1}\} \cup B) = 1\}. \end{align*}$$

If $\{a_{k_1 + 1}, \ldots , a_{k_2}\}$ is not a basis of X then there are $1 \leq s \leq \alpha - k_2$ , $d_1, \ldots , d_s \in X$ , and $a^{\prime }_{\alpha - k_2 - s}, \ldots , a^{\prime }_\alpha \in M_n$ such that $\{a_{k_1 + 1}, \ldots , a_{k_2}, d_1, \ldots , d_s\}$ is a basis of X and $a_1, \ldots , a_{k_2}, d_1, \ldots , d_s, a^{\prime }_{\alpha - k_2 - s}, \ldots , a^{\prime }_\alpha $ is an n-cs for $A/B$ .

Proof. The claim follows from repeated uses of Claim 5.9 and the fact that all bases of X have the same finite cardinality.

Claim 5.11. Suppose that there are $m < \alpha $ and $0 = k_0 < k_1 < \cdots < k_m < \alpha $ such that, for all $j = 0, \ldots , m-1$ , $\{a_{k_j + 1}, \ldots , a_{k_{j+1}}\}$ is a basis of

$$\begin{align*}X_j = \mathrm{acl}^n(A) \cap \{ d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_j}\} \cup B) = 1\}. \end{align*}$$

Then there are $k_m < k_{m+1} \leq \alpha $ , a basis $\{a^{\prime }_{k_m + 1}, \ldots , a^{\prime }_{k_{m+1}}\}$ of

$$\begin{align*}X_{m+1} = \mathrm{acl}^n(A) \cap \{ d \in M_n : \mathrm{rk}^n(d / \{a_1, \ldots, a_{k_m}\} \cup B) = 1\}, \end{align*}$$

and $a^{\prime }_{k_{m+1} + 1}, \ldots , a^{\prime }_\alpha \in M_n$ such that $a_1, \ldots , a_{k_m}, a^{\prime }_{k_m + 1}, \ldots , a^{\prime }_\alpha $ is an n-cs for $A/B$ .

Proof. From (5.1) we get $\mathrm {rk}^n(a_{k_m + 1} / \{a_1, \ldots , a_{k_m}\} \cup B) = 1$ . If $\{a_{k_m + 1}\}$ is a basis of $X_{m+1}$ then let $a^{\prime }_i = a_i$ for all $i = k_m + 1, \ldots , \alpha $ and we are done. Otherwise we get the conclusion of the claim by Claim 5.10.

Proof. Suppose that B is finite and $\mathrm {rk}^n(B / AC) = \mathrm {rk}^n(B / C) = \beta $ .

(i) Let $b_1, \ldots , b_\beta $ be an n-ccs for $B/AC$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_t = \beta $ . Then, for all $m = 0, \ldots , t-1$ ,

(5.2) $$ \begin{align} &\{b_{k_m + 1}, \ldots, b_{k_{m+1}}\} \ \text{ is a basis of } \\ &X_m^{AC} = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b_1, \ldots, b_{k_m}\} \cup AC) = 1 \} \ \text{ and,} \nonumber \\ &\text{for all } i = k_m + 1, \ldots, k_{m+1}, \mathrm{rk}^n(b_i / \{b_1, \ldots, b_{k_m}\} \cup AC) = 1. \nonumber \end{align} $$

This implies that

(5.3) $$ \begin{align} &\text{for all } \ i = 1, \ldots, \beta, b_i \notin \mathrm{acl}^n(\{b_1, \ldots, b_{i-1}\} \cup AC) \ \text{ and hence} \\ &b_i \notin \mathrm{acl}^n(\{b_1, \ldots, b_{i-1}\} \cup C). \nonumber \end{align} $$

It also follows (using Lemma 3.9) that, for all $m = 0, \ldots , t-1$ and all $ i = k_m + 1, \ldots , k_{m+1}$ ,

$$\begin{align*}\mathrm{rk}^n(b_i / \{b_1, \ldots, b_{k_m}\} \cup C) \geq 1. \end{align*}$$

Towards a contradiction, suppose that there are $m \in \{0, \ldots , t-1\}$ and $i \in \{k_m +1, \ldots , k_{m+1}\}$ such that $\mathrm {rk}^n(b_i / \{b_1, \ldots , b_{k_m}\} \cup C) \geq 2$ . Then there is $d \in \mathrm {acl}^n(b_i)$ such that $d \notin \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}\} \cup C)$ and $\mathrm {rk}^n(b_i / \{d, b_1, \ldots , b_{k_m}\} \cup C) \geq 1$ . Since $d \in \mathrm {acl}^n(b_i)$ it follows from (5.3) that for all $k_m < j \leq \alpha $ ,

$$\begin{align*}b_j \notin \mathrm{acl}^n(\{b_1, \ldots, b_{k_m}, d, b_{k_m + 1}, \ldots, b_{j-1}\} \cup C). \end{align*}$$

This implies (via Lemma 3.5) that $\mathrm {rk}^n(B / C) \geq \beta + 1$ , contradicting the assumption. Thus we conclude that

(5.4) $$ \begin{align} &\text{for all } m = 0, \ldots, t-1 \text{ and all } i = k_m + 1, \ldots, k_{m+1}, \\ &\mathrm{rk}^n(b_i / \{b_1, \ldots, b_{k_m}\} \cup C) = 1. \nonumber \end{align} $$

By (5.3) and (5.4), for all $m = 0, \ldots , t-1$ , $\{b_{k_m + 1}, \ldots , b_{k_{m+1}}\}$ is an independent subset of

$$\begin{align*}X^C_m = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b_1, \ldots, b_{k_m}\} \cup C) = 1\}. \end{align*}$$

Towards a contradiction, suppose that for some $m \in \{0, \ldots , t-1 \}$ there is $d \in X^C_m$ such that $d \notin \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}\} \cup C)$ . Since $\{b_1, \ldots , b_{k_m}\}$ is a basis of $X_m^{AC}$ it follows that $d \in \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}\} \cup AC)$ . If, for some $i> k_m$ , we would have that $b_i \in \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}, d, b_{k_m + 1}, \ldots , b_{i-1}\} \cup C)$ , then we would get

$$\begin{align*}b_i \in \mathrm{acl}^n(\{b_1, \ldots, b_{k_m}, b_{k_m + 1}, \ldots, b_{i-1}\} \cup AC) \end{align*}$$

which would contradict (5.3). Hence we conclude that for all $i> k_m$ ,

$$\begin{align*}b_i \notin \mathrm{acl}^n(\{b_1, \ldots, b_{k_m}, d, b_{k_m + 1}, \ldots, b_{i-1}\} \cup C). \end{align*}$$

But this implies that $\mathrm {rk}^n(B / C) \geq \beta + 1$ which contradicts the assumption. Hence there cannot be any $m \in \{0, \ldots , t-1 \}$ and $d \in X^C_m$ such that $d \notin \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}\} \cup C)$ , and therefore $\{b_1, \ldots , b_{k_m}\}$ is a basis of $X^C_m$ . It follows that $b_1, \ldots , b_\beta $ is an n-ccs for $B/C$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_t = \beta $ .

(ii) Now suppose that $b_1, \ldots , b_\beta $ is an n-ccs for $B/C$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\delta = \beta $ . It follows that for all $m = 0, \ldots , \delta - 1$ and all $i = k_m +1, \ldots , k_{m+1}$ ,

(5.5) $$ \begin{align} &\mathrm{rk}^n(b_i / \{b_1, \ldots, b_{i-1}\} \cup C) = \mathrm{rk}^n(b_i / \{b_1, \ldots, b_{k_m}\} \cup C) = 1, \ \text{ and } \\ &b_i \notin \mathrm{acl}^n(\{b_1, \ldots, b_{i-1}\} \cup C). \nonumber \end{align} $$

By Lemmas 5.4 and 5.6, there is an n-ccs $b^{\prime }_1, \ldots , b^{\prime }_\beta $ for $B/AC$ with core sequence of indices $0 = l_0 < l_1 < \cdots < l_\gamma = \beta $ . By part (i), $b^{\prime }_1, \ldots , b^{\prime }_\beta $ is also an n-ccs for $B/C$ with the same core sequence of indices $0 = l_0 < l_1 < \cdots < l_\gamma = \beta $ . By Lemma 5.4, we then have $\gamma = \delta $ and $l_m = k_m$ for all $m = 0, \ldots , \delta $ . The same lemma also tells that, for all $m = 1, \ldots , \delta $ ,

(5.6) $$ \begin{align} \mathrm{acl}^{\mathrm{eq}}(\{b_1, \ldots, b_{k_m}\} \cup C) = \mathrm{acl}^{\mathrm{eq}}(\{b^{\prime}_1, \ldots, b^{\prime}_{k_m}\} \cup C). \end{align} $$

Now we prove that

(5.7) $$ \begin{align} \mathrm{acl}^{\mathrm{eq}}(\{b_1, \ldots, b_{k_m}\} \cup AC) = \mathrm{acl}^{\mathrm{eq}}(\{b^{\prime}_1, \ldots, b^{\prime}_{k_m}\} \cup AC). \end{align} $$

Suppose that $d \in \mathrm {acl}^{\mathrm {eq}}(\{b_1, \ldots , b_{k_m}\} \cup AC)$ . Then there are a formula $\varphi (u, \bar {x}, \bar {y}, \bar {z})$ , a tuple $\bar {b}$ of elements from $\{b_1, \ldots , b_{k_m}\}$ , a tuple $\bar {a}$ of elements from A, and a tuple $\bar {c}$ of elements from C, such that $\mathcal {M}^{\mathrm {eq}} \models \varphi (d, \bar {b}, \bar {a}, \bar {c})$ and only finitely many elements satisfy $\varphi (u, \bar {b}, \bar {a}, \bar {c})$ . Let m be such that d and all elements that belong to any of $\bar {b}$ , $\bar {a}$ , or $\bar {c}$ , are members of $M_m$ . As $\mathcal {M}_m$ is $\omega $ -categorical there is a formula ${\unicode{x00FE}} eta (u, \bar {x}, \bar {y}, \bar {z})$ which isolates $\mathrm {tp}_{\mathcal {M}_m}(d, \bar {b}, \bar {a}, \bar {c})$ , where ‘ $\mathrm {tp}_{\mathcal {M}_m}$ ’ denotes the type computed in $\mathcal {M}_m$ . Then only finitely many elements satisfy ${\unicode{x00FE}} eta (u, \bar {b}, \bar {a}, \bar {c})$ , all of them belong to $M_m$ , and d is one among them. From (5.6) we have $\mathrm {rng}(\bar {b}) \subseteq \mathrm {acl}^m(\{b^{\prime }_1, \ldots , b^{\prime }_{k_m}\} \cup C)$ , so there are a formula $\psi (\bar {x}, \bar {y}, \bar {z})$ , a tuple $\bar {b}'$ of elements from $\{b^{\prime }_1, \ldots , b^{\prime }_{k_m}\}$ , and a tuple $\bar {c}'$ of elements from C such that $\mathcal {M}^{\mathrm {eq}} \models \psi (\bar {b}, \bar {b}', \bar {c}')$ and $\psi (\bar {x}, \bar {b}', \bar {c}')$ is satisfied by only finitely many tuples. It is now clear that d satisfies the formula

$$\begin{align*}\chi(u, \bar{a}, \bar{b}', \bar{c}, \bar{c}') := \exists \bar{x} \big[ {\unicode{x00FE}} eta(u, \bar{x}, \bar{a}, \bar{c}) \wedge \psi(\bar{x}, \bar{b}', \bar{c}')\big]. \end{align*}$$

The formula $\psi (\bar {x}, \bar {b}', \bar {c}')$ is satisfied by only finitely many tuples, and if $\bar {b}^*$ is one of them and $\mathcal {M}^{\mathrm {eq}} \models {\unicode{x00FE}} eta (d', \bar {b}^*, \bar {a}, \bar {c})$ for some $d'$ , then $d'$ and all members of $\bar {b}^*$ belong to $M_m$ , and $\mathrm {tp}_{\mathcal {M}_m}(d', \bar {b}^*, \bar {a}, \bar {c}) = \mathrm {tp}_{\mathcal {M}_m}(d, \bar {b}, \bar {a}, \bar {c})$ , so only finitely many elements satisfy ${\unicode{x00FE}} eta (u, \bar {b}^*, \bar {a}, \bar {c})$ . It now follows that $\chi (u, \bar {a}, \bar {b}', \bar {c}, \bar {c}')$ is satisfied by only finitely many elements. Hence

$$\begin{align*}d \in \mathrm{acl}^{\mathrm{eq}}(\bar{a}, \bar{b}', \bar{c}, \bar{c}') \subseteq \mathrm{acl}^{\mathrm{eq}}(\{b^{\prime}_1, \ldots, b^{\prime}_{k_m}\} \cup AC). \end{align*}$$

This shows the inclusion from left to right in (5.7) and by a similar argument (letting the $b_i$ and $b^{\prime }_i$ switch roles) the converse inclusion follows. This proves (5.7).

For all $m = 0, \ldots , \delta - 1$ , let

$$\begin{align*}X_m^{AC} = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b_1, \ldots, b_{k_m}\} \cup AC) = 1\} \end{align*}$$

and note that by (5.7) we also have

(5.8) $$ \begin{align} X_m^{AC} = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b^{\prime}_1, \ldots, b^{\prime}_{k_m}\} \cup AC) = 1\}. \end{align} $$

Since $b^{\prime }_1, \ldots , b^{\prime }_\beta $ is an n-ccs for $B/AC$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\delta = \beta $ it follows that, for all $m = 0, \ldots , \delta - 1$ , $\{b^{\prime }_{k_m + 1}, \ldots , b^{\prime }_{k_{m+1}}\}$ is a basis of $X_m^{AC}$ , so $X_m^{AC}$ has dimension $k_{m+1} - k_m$ . From (5.7) it follows that

$$\begin{align*}X_m^{AC} \subseteq \mathrm{acl}^n(\{b_1, \ldots, b_{k_{m+1}}\} \cup AC). \end{align*}$$

Fix any $m \in \{0, \ldots , \delta - 1\}$ . From (5.5), we get

$$\begin{align*}\mathrm{rk}^n(b_i / \{b_1, \ldots, b_{k_m}\} \cup AC) \leq 1 \ \text{ for all } i = k_m + 1, \ldots, k_{m+1}. \end{align*}$$

Suppose, for a contradiction, that there is $i \in \{ k_m + 1, \ldots , k_{m+1}\}$ such that

• • $\mathrm {rk}^n(b_i / \{b_1, \ldots , b_{k_m}\} \cup AC) = 0$ (equivalently, $b_i \in \mathrm {acl}^n(\{b_1, \ldots , b_{k_m}\} \cup AC)$ ),

• • or $b_i \in \mathrm {acl}^n(B_iAC)$ where $B_i = \{b_{k_m + 1}, \ldots , b_{k_{m+1}}\} \setminus \{b_i\}$ .

Then some proper subset of $\{b_{k_m + 1}, \ldots , b_{k_{m+1}}\}$ is a basis of $X_m^{AC}$ which contradicts our previous conclusion that $X_m^{AC}$ has dimension $k_{m+1} - k_m$ . Hence we conclude that $\mathrm {rk}^n(b_i / \{b_1, \ldots , b_{k_m}\} \cup AC) = 1$ for all $i = k_m + 1, \ldots , k_{m+1}$ and that $\{b_{k_m + 1}, \ldots , b_{k_{m+1}}\}$ is independent (in the pregeometry $\{d \in M_n : \mathrm {rk}^n(d / \{b_1, \ldots , b_{k_m}\} \cup AC) = 1\}$ ) and hence a basis of $X_m^{AC}$ . Since the argument holds for all $m = 0, \ldots , \delta - 1$ it follows that $b_1, \ldots , b_\beta $ is an n-ccs for $B/AC$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\delta = \beta $ .

Proof. Let $A, C \subseteq M^{\mathrm {eq}}$ and $B \subseteq M_n$ where A and B are finite. We prove the result by induction on $\mathrm {rk}^n(B / C)$ . If $\mathrm {rk}^n(B / C) = 0$ then, since $B \subseteq M_n$ and by the definition of $\mathrm {rk}^n$ , $B \subseteq \mathrm {acl}^n(B) \subseteq \mathrm {acl}^n(C)$ so $\mathrm {acl}^n(BC) = \mathrm {acl}^n(C)$ . Hence (by the definition of $\mathrm {rk}^n$ ) $\mathrm {rk}^n(A / BC) = \mathrm {rk}^n(A / C)$ and the statement is vacuously satisfied.

The induction step remains. The induction hypothesis will be:

1. (IH) For every finite $B' \subseteq M_n$ , if $\mathrm {rk}^n(B' / C) < \mathrm {rk}^n(B / C)$ and $\mathrm {rk}^n(A / B'C) < \mathrm {rk}^n(A / C)$ , then $\mathrm {rk}^n(B' / AC) < \mathrm {rk}^n(B' / C)$ .

Suppose that $\mathrm {rk}^n(A / BC) < \mathrm {rk}^n(A / C)$ . Also let $\beta = \mathrm {rk}^n(B / C)$ . (We will show that $\mathrm {rk}^n(B / AC) < \beta $ .) Then there is an n-ccs $b_1, \ldots , b_\beta $ for $B/C$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\delta = \beta $ . In particular, $b_1, \ldots , b_\beta \in \mathrm {acl}^n(B)$ . We divide the argument into two main cases.

Case 1. Suppose that there is $s < \beta $ such that $\mathrm {rk}^n(A / \{b_1, \ldots , b_s\} \cup C) < \mathrm {rk}^n(A / C)$ .

We may assume that s is minimal such that the above holds. As $b_1, \ldots , b_\beta $ is an n-ccs for $B/C$ , hence (by Lemma 5.5) an n-cs for $B/C$ , it follows from Lemma 3.5 that

$$\begin{align*}\mathrm{rk}^n(b_1, \ldots, b_s / C) = s < \beta = \mathrm{rk}^n(B / C). \end{align*}$$

Now the induction hypothesis (IH) implies that

$$\begin{align*}\mathrm{rk}^n(b_1, \ldots, b_s / AC) < \mathrm{rk}^n(b_1, \ldots, b_s / C) = s. \end{align*}$$

By Lemma 3.5 there is $t \leq s$ such that

(5.9) $$ \begin{align} b_t \in \mathrm{acl}^n(\{b_1, \ldots, b_{t-1}\} \cup AC). \end{align} $$

Let t be minimal such that the above holds. Towards a contradiction, suppose that $\mathrm {rk}^n(B / AC) = \mathrm {rk}^n(B / C) = \beta $ . Then there is an n-ccs $b^{\prime }_1, \ldots , b^{\prime }_\beta $ for $B/AC$ with core sequence of indices $0 = l_0 < l_1 < \cdots < l_\gamma = \beta $ . By Lemma 5.12, $b_1, \ldots , b_\beta $ is also an n-ccs for $B/C$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\gamma = \beta $ . By Lemma 5.4 we have $\gamma = \delta $ and $l_i = k_i$ for all $i = 0, \ldots , \delta $ . From Lemma 5.4 it follows that if, for any $m \in \{0, \ldots , \delta - 1\}$ , we define

$$\begin{align*}X_m^{AC} = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b_1, \ldots, b_{k_m}\} \cup AC) = 1 \} \end{align*}$$

then

$$\begin{align*}X_m^{AC} = \mathrm{acl}^n(B) \cap \{d \in M_n : \mathrm{rk}^n(d / \{b^{\prime}_1, \ldots, b^{\prime}_{k_m}\} \cup AC) = 1 \}. \end{align*}$$

Also, $b_{k_m + 1}, \ldots , b_{k_{m+1}}, b^{\prime }_{k_m + 1}, \ldots , b^{\prime }_{k_{m+1}} \in X_m^{AC}$ . Let m be such that $k_m < t \leq k_{m+1}$ . Now (5.9) implies that some proper subset of $\{b_{k_m + 1}, \ldots , b_{k_{m+1}}\}$ is a basis of $X_m^{AC}$ . But since $b^{\prime }_1, \ldots , b^{\prime }_\beta $ is an n-ccs for $B/AC$ with core sequence of indices $0 = k_0 < k_1 < \cdots < k_\delta = \beta $ it follows that $\{b^{\prime }_{k_m + 1}, \ldots , b^{\prime }_{k_{m+1}}\}$ is a basis of $X_m^{AC}$ . So there are two bases of $X_m^{AC}$ with different cardinalities, which is impossible. Thus we conclude that $\mathrm {rk}^n(B / AC) < \beta $ .

Case 2. Suppose that Case 1 does not hold, that is, suppose that for all $s < \beta $ , $\mathrm {rk}^n(A / \{b_1, \ldots , b_s\} \cup C) = \mathrm {rk}^n(A / C)$ .

Let $\alpha = \mathrm {rk}^n(A / C)$ , so by assumption,

$$\begin{align*}rk^n(A / \{b_1, \ldots, b_{\beta - 1}\} \cup C) = \mathrm{rk}^n(A / C) = \alpha. \end{align*}$$

Since $b_1, \ldots , b_\beta $ is an n-ccs for $B/C$ , hence an n-cs for $B/C$ , it follows (by Lemma 3.5 and the assumption that $B\subseteq M_n$ ) that ${B {\kern-1pt}\subseteq{\kern-1pt} \mathrm {acl}^n(B) {\kern-1pt}\subseteq{\kern-1pt} \mathrm {acl}^n(\{b_1, \ldots , b_\beta \} {\kern-1pt}\cup{\kern-1pt} C)}$ and hence $\mathrm {rk}^n(A / BC) = \mathrm {rk}^n(A / \{b_1, \ldots , b_\beta \} \cup C)$ . From the assumption that $\mathrm {rk}^n(A / BC) < \mathrm {rk}^n(A /C)$ we now get

(5.10) $$ \begin{align} \mathrm{rk}^n(A / \{b_1, \ldots, b_\beta\} \cup C) < \mathrm{rk}^n(A / \{b_1, \ldots, b_{\beta-1}\} \cup C) = \alpha. \end{align} $$

Let $a_1, \ldots , a_\alpha $ be an n-ccs for $A/C$ with core sequence of indices $0 = l_0 < l_1 < \cdots < l_\gamma = \alpha $ . By Lemma 5.12, $a_1, \ldots , a_\alpha $ is also an n-ccs for $A/\{b_1, \ldots , b_{\beta -1}\}\cup C$ with the same core sequence of indices $0 = l_0 < l_1 < \cdots < l_\gamma = \alpha $ . This implies that,