Proof (1) Clear from definitions using Proposition 2.2.

(2) Suppose $\varphi \left (x,y\right )\land \psi \left (x,y\right )$ is not a semi-equation. By Proposition 2.2, there are b and an indiscernible sequence $\left (a_{i}\right )_{i\in \mathbb {Z}}$ such that $\models \varphi \left (a_{i},b\right )\land \psi \left (a_{i},b\right )\iff i\neq 0$ . Either $\not \models \varphi \left (a_{0},b\right )$ , in which case $\varphi \left (x,y\right )$ is not a semi-equation, or $\not \models \psi \left (a_{0},b\right )$ , in which case $\psi \left (x,y\right )$ is not a semi-equation. And $\varphi (x,y)$ is a semi-equation if and only if $\varphi ^*(y,x) := \varphi (x,y)$ is a semi-equation by the symmetry of the definition.

(3) For conjunctions, the same as the proof of (2), but with the stipulation that $\left (a_{i}\right )_{i\neq 0}$ is b-indiscernible added. Now suppose $\varphi \left (x,y\right )\lor \psi \left (x,y\right )$ is not a weak semi-equation. Then there are b and an indiscernible sequence $\left (a_{i}\right )_{i\in \mathbb {Z}}$ such that $\left (a_{i}\right )_{i\neq 0}$ is b-indiscernible, and $\models \varphi \left (a_{i},b\right )\lor \psi \left (a_{i},b\right )\iff i\neq 0$ . Either $\models \varphi \left (a_{1},b\right )$ or $\models \psi \left (a_{1},b\right )$ , and then, by b-indiscernibility, either $\models \varphi \left (a_{i},b\right )$ for all $i\neq 0$ or $\models \psi \left (a_{i},b\right )$ for all $i\neq 0$ . In the first case, $\varphi \left (x,y\right )$ is not a weak semi-equation, and in the second case, $\psi \left (x,y\right )$ is not a weak semi-equation.

Proof (1) Left to right is immediate from the definition (Proposition 2.2). For the converse, assume that in some model of T we can find an infinite sequence $\left (a_{i},b_{i}\right )_{i\in I}$ such that for all $i,j \in I$ , $\models \varphi \left (a_{i},b_{j}\right )\iff i\neq j$ . By completeness of T, we can find arbitrarily long finite sequences with the same property in every model of T, in particular in some model of $T'$ . By compactness we can thus find an infinite sequence with the same property in a model of $T'$ , demonstrating that $\varphi (x,y)$ is not a semi-equation in $T'$ .

(2) If $\varphi \left (x,y\right ) \in \mathcal {L}$ is not a weak semi-equation in $T'$ , then (in a monster model of $T'$ , and hence of T) there are b and an $\mathcal { L}'$ -indiscernible $\left (a_{i}\right )_{i\in I_{L}+\left (0\right )+I_{R}}$ such that $\left (a_{i}\right )_{i\in I_{L}+I_{R}}$ is $\mathcal { L}'$ -indiscernible over b and $\models \varphi \left (a_{i},b\right )$ for $i\in I_{L}+I_{R}$ , but $\not \models \varphi \left (a_{0},b\right )$ , for infinite linear orders $I_{L},I_{R}$ . Then, in particular, $\left (a_{i}\right )_{i\in I_{L}+(0) + I_{R}}$ is $\mathcal { L}$ -indiscernible, and $\left (a_{i}\right )_{i\in I_{L}+I_{R}}$ is $\mathcal { L}$ -indiscernible over b, so $\varphi \left (x,y\right )$ is a not a weak semi-equation in T.

Proof (1) If $\varphi \left (x,y\right )$ is not NIP, then there are an indiscernible sequence $\left (a_{i}\right )_{i\in \mathbb {N}}$ and b such that $\models \varphi \left (a_{i},b\right ) \iff i$ is even. For any finite set of formulas $\Delta \left (x_{1}, \ldots , x_{n},y\right )$ , by Ramsey’s theorem, there is an infinite $I\subseteq 2\mathbb {N}$ on which the truth value of all formulas in $\Delta \left (a_{i_{1}}, \ldots , a_{i_{n}},b\right )$ is constant for all $i_{1}<\cdots < i_{n}\in I$ . Thus, by letting $a_{0}':=a_{i}$ for some sufficiently large odd i, we can find an indiscernible sequence $\left (a_{i}'\right )_{i\in I_{L}+\left (0\right )+I_{R}}$ (using $I_{L}\sqcup I_{R}=I$ , and $a_{i}'=a_{i}$ for $i\in I$ ) for some infinite $I_{R}$ and arbitrarily large finite $I_{L}$ , such that $\left (a_{i}'\right )_{i\in I_{L}+I_{R}}$ is $\Delta $ -indiscernible over b. By compactness, it follows that $\varphi \left (x,y\right )$ is not a weak semi-equation.

(2) If $\varphi \left (x,y\right )$ is not a semi-equation, then there is a sequence $\left (a_{i},b_{i}\right )_{i\in \mathbb {N}}$ such that $\models \varphi \left (a_{i},b_{j}\right )\iff i\neq j$ . In particular, $\models \varphi \left (a_{i},b_{j}\right )$ for all $j<i$ , and $\not \models \varphi \left (a_{i},b_{i}\right )$ , so this is a counterexample to the descending chain condition, and $\varphi \left (x,y\right )$ is not an equation.

(3) If $\varphi \left (x,y\right )$ is not an equation, then by Ramsey and compactness there is an indiscernible sequence $\left (a_{i},b_{i}\right )_{i\in \mathbb {N}}$ such that $\models \varphi \left (a_{i},b_{j}\right )$ for all $j<i$ , and $\not \models \varphi \left (a_{i},b_{i}\right )$ . If $\varphi \left (a_{i},b_{j}\right )$ holds for $i<j$ then $\varphi \left (x,y\right )$ is not a semi-equation. Otherwise, $\varphi \left (x,y\right )$ is not stable.

(4) If $\varphi \left (x,y\right )$ is not an equation, by Ramsey and compactness we can choose an indiscernible sequence $\left (a_{i},b_{i}\right )_{i\in \mathbb {Z}}$ such that $\models \varphi \left (a_{i},b_{j}\right )$ for all $j<i$ , and $\not \models \varphi \left (a_{i},b_{i}\right )$ . The indiscernible sequence $(a_i,b_i)_{i \in \mathbb {Z}}$ is totally indiscernible by stability of T; hence, we have $\models \varphi (a_i, b_0) \iff i \neq 0$ , and also $(a_i : i \neq 0)$ is indiscernible over $b_0$ . This shows that $\varphi \left (x,y\right )$ is not a weak semi-equation.

Fact 2.12. A stable theory T is 1-based if and only if in T, every formula $\varphi (x,y) \in \mathcal {L}$ , with $x,y$ arbitrary finite tuples of variables, is equivalent to a Boolean combination of finitely many weakly normal formulas $\psi _1(x,y), \ldots , \psi _n(x,y) \in \mathcal {L}$ .

Proof (1) Clear from the definitions.

(2) If $\varphi \left (x,y\right )$ is not a semi-equation, let $\left (a_{i},b_{i}\right )_{i\in \mathbb {N}}$ be such that $\models \varphi \left (a_{i},b_{j}\right )\iff i\neq j$ . Then for any $\left (k,n\right )$ we have $\models \varphi \left (a_{0},b_{1}\right )\land \cdots \land \varphi \left (a_{0},b_{k}\right )$ , but for any pairwise distinct $i_{1}, \ldots , i_{n},j\in \left [k\right ]$ , $\models \varphi \left (a_{j},b_{i_{1}}\right )\land \cdots \land \varphi \left (a_{j},b_{i_{n}}\right )\land \neg \varphi \left (a_{j},b_{j}\right )$ ; hence, $\varphi (x,y)$ is not a $(k,n)$ -semi-equation. Conversely, for any $k \in \mathbb {N}$ , if $\varphi \left (x,y\right )$ is not a $\left (k,k-1\right )$ -semi-equation, then there exist $b_{1}, \ldots , b_{k}$ such that for each $j\in \left [k\right ]$ , there is $a_{j}$ such that $\models \varphi \left (a_{j},b_{i}\right )$ for $i\neq j$ , but $\not \models \varphi \left (a_{j},b_{j}\right )$ . Hence if $\varphi \left (x,y\right )$ is not a $\left (k,k-1\right )$ -semi-equation for any k, then by compactness $\varphi \left (x,y\right )$ is not a semi-equation. And if $\varphi \left (x,y\right )$ is not an n-semi-equation, then it is not an $\left (n+1,n\right )$ -semi-equation by definition, so a formula that is not an n-semi-equation for any n is also not a $\left (k,k-1\right )$ -semi-equation for any k.

Proof The family of sets $\left \{ \varphi \left (\mathbb {M},b\right )\mid b\in \mathbb { M}^{y}\right \} $ has breadth at most k if and only if every finite consistent conjunction of instances of $\varphi $ is implied by the conjunction of at most k of those instances. In particular this applies to consistent conjunctions of $(k+1)$ instances of $\varphi $ , showing that if the breadth of $\mathcal {F}_{\varphi }$ is $\leq k$ , then it is a $(k+1,k)$ -semi-equation. Conversely, assume $\varphi (x,y)$ is a $(k+1,k)$ -semi-equation. Given any consistent conjunction of $n>k$ instances of $\varphi $ , any $(k+1)$ of them contain an instance implied by the other k instances. Removing this implied instance, we reduce to the case of a consistent conjunction of $(n-1)$ instances, and after $(n-k)$ steps to a conjunction of k instances of $\varphi $ implying all the other ones. The “in particular” part is Proposition 2.14(2).

Proof Clearly every k-weakly normal formula is a $\left (k,1\right )$ -semi-equation and is also an equation, hence stable. Conversely, suppose that $\varphi \left (x,y\right )$ is a $\left (k,1\right )$ -semi-equation and is stable, or just NSOP: there is some $\ell \in \omega $ such that there is no strictly increasing chain of sets of the form $\varphi \left (\mathbb {M},b_{0}\right ) \subsetneq \cdots \subsetneq \varphi \left (\mathbb {M},b_{\ell }\right )$ . We will show that then $\varphi (x,y)$ is $k^{\ell }$ -weakly normal. Let $\left (b_{\eta }\right )_{\eta \in \left [k\right ]^{\ell }}$ be such that $\models \exists x\,\bigwedge _{\eta \in \left [k\right ]^{\ell }}\varphi \left (x,b_{\eta }\right )$ . For $\sigma \in \left [k\right ]^{\leq \ell }$ , we will show by induction on $m:=\ell -\left |\sigma \right |$ that there are pairwise distinct $\eta _{0}, \ldots , \eta _{m}\in \left [k\right ]^{\ell }$ extending $\sigma $ (as sequences) such that $\varphi \left (\mathbb { M},b_{\eta _{0}}\right )\subseteq \varphi \left (\mathbb { M},b_{\eta _{1}}\right )\subseteq \cdots \subseteq \varphi \left (\mathbb { M},b_{\eta _{m}}\right )$ . With $m=\ell $ , so that $\sigma = \langle \rangle $ is the empty sequence, this implies by the choice of $\ell $ that there are $\eta \neq \eta '\in \left [k\right ]^{\ell }$ such that $\varphi \left (\mathbb { M},b_{\eta }\right )=\varphi \left (\mathbb { M},b_{\eta '}\right )$ , as desired. The base case ( $m=0$ ) is trivial, with $\eta _{0}=\sigma $ .

Assume the claim holds for m, and let $\sigma \in \left [k\right ]^{\ell -\left (m+1\right )}$ . For each $i\in \left [k\right ]$ , there exist pairwise distinct $\eta _{i,0}, \ldots , \eta _{i,m}\in \left [k\right ]^{\ell }$ extending $\sigma ^{\frown } i$ such that $\varphi \left (\mathbb { M},b_{\eta _{i,0}}\right )\subseteq \cdots \subseteq \varphi \left (\mathbb { M},b_{\eta _{i,m}}\right )$ . Among the sets $\left \{ \varphi \left (\mathbb { M},b_{\eta _{i,0}}\right )\mid i\in \left [k\right ]\right \} $ , one must be contained in another by $\left (k,1\right )$ -semi-equationality. Say $\varphi \left (\mathbb { M},b_{\eta _{j,0}}\right )\subseteq \varphi \left (\mathbb { M},b_{\eta _{i,0}}\right )$ for some $i\neq j$ . Then $\varphi \left (\mathbb { M},b_{\eta _{j,0}}\right )\subseteq \varphi \left (\mathbb { M},b_{\eta _{i,0}}\right )\subseteq \varphi \left (\mathbb { M},b_{\eta _{i,1}}\right ) \subseteq \cdots \subseteq \varphi \left (\mathbb { M},b_{\eta _{i,m}}\right )$ , and $\eta _{j,0},\eta _{i,0}, \eta _{i,1}, \ldots ,\eta _{i,m}$ are pairwise distinct and extend $\sigma $ , as desired. The “in particular” part follows by Fact 2.12.

Fact 2.22 [Reference Guingona and Laskowski16, Claim 1 in the proof of Proposition 2.5].

Let X be a set and $\mathcal {F} \subseteq \mathcal {P}(X)$ a family of subsets of X such that there are no $A,B \in \mathcal {F}$ satisfying $A \cap B \neq \emptyset , B \setminus A \neq \emptyset $ and $B \setminus A \neq \emptyset $ simultaneously. Then there exists a linear order $<$ on X so that every $A \in \mathcal {F}$ is a $<$ -convex subset of X.

Proof (1) Let $(S,<), f,g$ be as in Definition 2.21 for $R_{\varphi }$ . Given any $b_1, b_2 \in \mathbb {M}^{y}$ , the sets $\left \{x \in S : x < g(b_i) \right \}$ for $i \in \{1,2\}$ are initial segments of S. Say $g(b_1) \leq g(b_2)$ . Then for any $a \in \mathbb {M}^x$ , $f(a) < g(b_1) \Rightarrow f(a) < g(b_2)$ , so $\varphi (\mathbb {M},b_1) \subseteq \varphi (\mathbb {M},b_2)$ , and the other case is symmetric.

(2) If $\varphi (x,y)$ is a $(2,1)$ -semi-equation, then the family $\mathcal {F}_{\varphi }$ of subsets of $\mathbb {M}^{x}$ satisfies the assumption in Fact 2.22. Hence there exists a (not necessarily definable) linear ordering $<'$ of $\mathbb {M}^{x}$ so that for every $b \in \mathbb {M}^y$ , $\varphi (\mathbb {M},b)$ is $<'$ -convex. Let $(S,<)$ be the Dedekind completion of $\left ( \mathbb {M}^x, <' \right )$ . Consider the functions $g_1, g_2: \mathbb {M}^{y} \to S$ so that $g_1(b)$ is the infimum of $\varphi (\mathbb {M},b)$ in S, and $g_2(b)$ is the supremum of $\varphi (\mathbb {M},b)$ in S. Then $R_{\varphi } = \left \{ (a,b) \in \mathbb {M}^x \times \mathbb {M}^y : g_1(b)\leq a \right \} \cap \left \{ (a,b) \in \mathbb {M}^x \times \mathbb {M}^y: a \leq g_2(b) \right \}$ , and both of these relations are basic (see [Reference Basit, Chernikov, Starchenko, Tao and Tran5, Remark 2.7]).

Fact 3.1. Let $T = {\operatorname {Th}}(\mathcal {M})$ , with $\mathcal {M} = \left (M; <, +, \ldots \right )$ a linear o-minimal expansion of a group. Let $\mathcal {L} = (<,+,\ldots )$ be the language of T. A partial endomorphism of $\mathcal {M}$ is a map $f: (-c,c) \to M$ , for c an element of M or $\infty $ , such that if $a,b,a+b$ are all in the domain, then $f(a+b) = f(a) + f(b)$ . Let $\mathcal {M}'$ be the reduct of $\mathcal {M}$ to the language $\mathcal {L}'$ consisting of $+,<$ , constant symbols naming $\operatorname {acl}_{\mathcal {L}}(\emptyset )$ , and for each $\mathcal {L}(\emptyset )$ -definable partial endomorphism $f: (-c,c) \to M$ with $c \in \operatorname {acl}_{\mathcal {L}}(\emptyset )$ or $c = \infty $ , a unary function symbol interpreted as f on $(-c,c)$ and as $0$ outside of the domain of f. Let $T' := {\operatorname {Th}}_{\mathcal {L}'}(\mathcal {M}')$ .

1. (1) [Reference Loveys and Peterzil23, Proposition 4.2] A subset of $M^n$ is $\emptyset $ -definable in $\mathcal {M}$ if and only if it is $\emptyset $ -definable in $\mathcal {M}'$ .

2. (2) [Reference Loveys and Peterzil23, Corollary 6.3] $T'$ admits quantifier elimination in the language $\mathcal {L}'$ .

Proof (1) Let $\mathcal {L} = (<,+,\ldots )$ , $\mathcal {M}'$ and $\mathcal {L}'$ be as in Fact 3.1. By Fact 3.1(1) it suffices to show that $T' := {\operatorname {Th}}_{\mathcal {L}'} \left (\mathcal {M}'\right )$ is $(2,1)$ -semi-equational. By Fact 3.1(2), it then suffices to show that every atomic $\mathcal {L}'$ -formula $\varphi (x,y)$ , with $x,y$ arbitrary finite tuples of variables, is equivalent in $T'$ to a Boolean combination of $(2,1)$ -semi-equations. By the proof of Theorem 4.3 in [Reference Anderson1], every atomic $\mathcal {L}'$ -formula $\varphi (x,y)$ is equivalent in $T'$ to a Boolean combination of atomic formulas of the form $f(x) \square g(y) + c$ , where $\square \in \left \{<, = ,> \right \}$ , $f: M^{|x|} \to M, g: M^{|y|} \to M$ are total multivariate $\mathcal {L}'(\emptyset )$ -definable homomorphisms and $c \in \operatorname {dcl}_{\mathcal {L}'}(\emptyset )$ . Every formula of this form clearly defines a basic relation on $M^{|x|} \times M^{|y|}$ , hence is a $(2,1)$ -semi-equation by Proposition 2.23(1).

(2) By the o-minimal trichotomy theorem (see [Reference Peterzil and Starchenko30] and Remark 2 after the statement of Theorem 1.7 there), if $\mathcal {M}$ is not linear, then it defines an infinite field. But then Remark 2.24(1) implies that T is not $(2,1)$ -semi-equational.

Fact 3.5. Let $\mathcal {M} = \left ( M, <, \left (C_i \right )_{i \in I}, \left (R_j \right )_{j \in J} \right )$ be a linear order expanded by arbitrary unary ( $C_i$ ) and monotone binary ( $R_j$ ) relations. Then ${\operatorname {Th}} \left ( \mathcal {M} \right )$ is $(2,1)$ -semi-equational.

Proof Let $\mathcal {M}'$ be an expansion of $\mathcal {M}$ obtained by naming all $\mathcal {L}_{\mathcal {M}}(\emptyset )$ -definable unary and monotone binary relations, then a subset of $M^n$ is $\emptyset $ -definable in $\mathcal {M}$ if and only if it is $\emptyset $ -definable in $\mathcal {M}'$ , so it suffices to show that $T' := {\operatorname {Th}} \left (\mathcal {M}' \right )$ is $(2,1)$ -semi-equational. By [Reference Simon35, Proposition 4.1], $T'$ eliminates quantifiers. Note that if $R(x,y)$ is monotone, then it is a $(2,1)$ -semi-equation (given any $b_1 \leq b_2 \in M$ , for any $a \in M$ we have $\models R(a,b_1) \rightarrow R(a,b_2)$ by monotonicity; hence, $R(M,b_1) \subseteq R(M,b_2)$ ). And any unary relation $C_i(x)$ is trivially a $(2,1)$ -semi-equation; hence, $T'$ is $(2,1)$ -semi-equational.

Proof (1) We show that $ \psi (x_1,x_2; y) := {\circlearrowright }\left (x_{1},x_{2};y\right )$ is not a Boolean combination of semi-equations. By quantifier elimination, the formulas ${\circlearrowright }\left (x_{1},x_{2};y\right )$ and ${\circlearrowright }\left (x_{2},x_{1};y\right )$ each isolate a complete $3$ -type (over $\emptyset $ ). Any Boolean combination of formulas that is equivalent to ${\circlearrowright }\left (x_{1},x_{2};y\right )$ must contain some formula $\varphi \left (x_{1},x_{2};y\right )$ that is implied by ${\circlearrowright }\left (x_{1},x_{2};y\right )$ and is inconsistent with ${\circlearrowright }\left (x_{2},x_{1};y\right )$ , or vice versa. Assume the former. Let $\left (c_{i}\right )_{i\in \mathbb {Z}}$ be such that $\models {\circlearrowright }\left (c_{k},c_{i},c_{j}\right )$ for $i<j<k$ . Let $a_{1,i}=c_{2i}$ , $a_{2,i}=c_{2i+2}$ , and $b_{i}=c_{2i+1}$ . Then $\models {\circlearrowright }\left (a_{2,i},a_{1,i};b_{j}\right )\Leftrightarrow i=j$ and $\models {\circlearrowright }\left (a_{1,i},a_{2,i};b_{j}\right ) \Leftrightarrow i\neq j$ , so $\models \varphi \left (a_{1,i},a_{2,i};b_{j}\right ) \Leftrightarrow i\neq j$ , so $\varphi \left (x_{1},x_{2};y\right )$ is not a semi-equation. If instead, $\varphi \left (x_{1},x_{2};y\right )$ is implied by ${\circlearrowright }\left (x_{2},x_{1};y\right )$ and inconsistent with ${\circlearrowright }\left (x_{1},x_{2};y\right )$ , we switch the roles of $x_{1}$ and $x_{2}$ to get the same result.

(2) Let $<$ be defined by $x<y\Leftrightarrow {\circlearrowright }\left (x,y,c\right )$ . Then $<$ is a dense linear order on the complement of $\left \{ c\right \} $ , so $x < y$ is a $\left (2,1\right )$ -semi-equation. We have that ${\circlearrowright }\left (x,y,z\right )$ is equivalent to $x<y<z\lor y<z<x\lor z<x<y\lor \left (z=c\land x<y\right )\lor \left (y=c\land z<x\right )\lor \left (x=c\land y<z\right )$ . Hence ${\circlearrowright }\left (x,y,z\right )$ is a Boolean combination of $\left (2,1\right )$ -semi-equations (with c as a parameter), under any partition of the variables. By quantifier elimination, it follows that every formula is a Boolean combination of $\left (2,1\right )$ -semi-equations (using c as a parameter).

Proof Since every auxiliary sort is finite and linearly ordered by a (definable) relation in $\mathcal {L}_{\operatorname {CH}}$ , all auxiliary sorts are contained in $\operatorname {dcl}(\emptyset )$ . Hence we only need to verify that every formula $\varphi (x,y)$ with $x,y$ tuples of the main sort $\mathcal {G}$ is a Boolean combination of $1$ -semi-equations in the expansion with every element of every auxiliary sort named by a new constant symbol (countably many in total). As explained in [Reference Aschenbrenner, Chernikov, Gehret and Ziegler3, Proposition 3.14], it then follows from the relative quantifier elimination that $\varphi (x,y)$ is equivalent to a Boolean combination of atomic formulas of the form $\pi _{\alpha }(f(x)) \diamond _{\alpha } \pi _{\alpha }(g(y)) + k_{\alpha }$ , where $\diamond \in \left \{ =, <, \equiv _{m} \right \}$ , $k \in \mathbb {Z}$ , $\alpha $ is an element of an auxiliary sort, $f,g$ are $\mathbb {Z}$ -linear functions on $\mathcal {G}$ , $G_{\alpha }$ is a corresponding convex subgroup of G, $\pi _{\alpha }: G \to G/G_{\alpha }$ is the quotient map, $1_{\alpha }$ is the minimal positive element of $G/G_{\alpha }$ if it is discrete or $0 \in G/G_{\alpha }$ otherwise, and $k_{\alpha } = k \cdot 1_{\alpha }$ in $G/G_{\alpha }$ , and for $g,h \in G/G_{\alpha }$ we have $g \equiv _m h$ if $g - h \in m \left ( G/G_{\alpha } \right )$ (note that these relations on G are expressible in the pure language of ordered abelian groups).

It is straightforward from Definition 2.13 that if $\varphi (x,y)$ is a $(k,n)$ -semi-equation and $f(x), g(y)$ are $\emptyset $ -definable functions, then the formula $\psi (x,y) := \varphi (f(x), g(y))$ is also a $(k,n)$ -semi-equation. Using this (in an expansion of $\bar {G}$ naming $\pi _{\alpha }$ , and the ordered group structure on $G/G_{\alpha }$ together with the constants for $k_{\alpha }$ ), we only have to show that the relations $x = y, x < y, x \in y + m\left (G/G_{\alpha } \right )$ on $G/G_{\alpha }$ are $(2,1)$ -semi-equations, which is straightforward.

Fact 3.10 (see, e.g., [Reference Chernikov and Mennen7, Lemma 3.14] or [Reference Mennuni27, Section 1]).

The theory of infinitely branching dense trees is complete and eliminates quantifiers in the language $\left \{ \land \right \} $ .

Proof By [Reference Simon35, Corollary 4.6] (using that $x \leq y \iff x \land y = x$ ), in any tree $\mathcal {M}=(M, \land )$ we have: two tuples $\bar {a} = (a_i : i \in [n]),\bar {b} = \left ( b_j: j \in [n] \right ) \in M^{n}$ have the same type if and only if $\left (a_{i},a_{j},a_{k}\right )$ and $\left (b_{i},b_{j},b_{k}\right )$ have the same type for every $i,j,k\in \left [n\right ]$ . Hence for any $\bar {a},\bar {b}$ , $\operatorname {tp}\left (\bar {a}\bar {b}\right )$ is implied by the set of formulas satisfied by three-element subtuples of $\bar {a}\bar {b}$ . So if every partitioned formula with three total free variables is a Boolean combination of semi-equations, then $\operatorname {tp}\left (\bar {a}\bar {b}\right )$ is implied by a Boolean combination of semi-equations. In view of this, it is sufficient to assume that every formula of the form $\varphi \left (x;y_{1},y_{2}\right )$ is a Boolean combination of semi-equations, because then by symmetry, every formula of the form $\varphi \left (x_{1},x_{2};y\right )$ is as well, and every partitioned formula with one of the parts empty (i.e., $\varphi \left (;y_{1},y_{2},y_{3}\right )$ or $\varphi \left (x_{1},x_{2},x_{3};\right )$ ) is automatically a semi-equation.

Proof Let $\mathcal { M} = (M, \land )$ be an infinitely branching dense tree. By Lemma 3.11, it is enough to check that every formula $\varphi \left (x;y_{1},y_{2}\right )$ is a Boolean combination of semi-equations, and, by Fact 3.10, it is enough to check this for positive atomic formulas $\varphi \left (x;y_{1},y_{2}\right )$ . Using the fact that $\land $ is associative, commutative, and idempotent, each such formula is equivalent to a formula of the form $\bigwedge A=\bigwedge B$ for non-empty $A,B\subseteq \left \{ x,y_{1},y_{2}\right \} $ . By a direct case analysis (see [Reference Chernikov and Mennen7, Theorem 3.16] for the details) every such formula is either a tautology, or does not mention x, or an equality between two variables, or is equivalent to a Boolean combination of the following formulas (possibly replacing $y_2$ by $y_1$ ):

(1) $x=x\land y_{1}$ , i.e., $x\leq y_{1}$ —a semi-equation: given $\left (a_{i},b_{i}\right )_{i\in \mathbb {Z}}$ such that $\models a_{i}\leq b_{j}\iff i\neq j$ , $a_{i}\leq b_{0}$ for $i\neq 0$ , so $\left (a_{i}\right )_{i\neq 0}$ forms a chain. This is not consistent with $a_{1}\leq b_{2}$ , $a_{2}\leq b_{1}$ , $a_{1}\not \leq b_{1}$ , $a_{2}\not \leq b_{2}$ .

(2) $x \land y_1 \land y_2 = y_1 \land y_2$ , i.e., $x \geq y_1 \land y_2$ —a semi-equation for the same reason.

(3) $x\land y_{1}=x\land y_{2}$ —a negated semi-equation: given $\left (a_{i},b_{i},b_{i}'\right )_{i\in \mathbb {Z}}$ such that $\models a_{i}\land b_{j}=a_{i}\land b_{j}'\iff i=j$ , for every $i\neq 0$ we have: either $a_{i}\land b_{0}>a_{0}\land b_{0}$ or $a_{i}\land b_{0}'>a_{0}\land b_{0}$ . By pigeonhole, there are $i_{1}\neq i_{2}$ such that the same case holds for both. Without loss of generality, $a_{1}\land b_{0}>a_{0}\land b_{0}$ and $a_{2}\land b_{0}>a_{0}\land b_{0}$ . But then $a_{1}\land a_{2}>a_{0}\land b_{0}=a_0\land a_1$ , so $a_{1}$ and $a_{2}$ meet strictly closer to each other than to $a_{0}$ . But, since $a_1\land b_1\leq a_0\land b_1$ and $a_1\land b_1\leq a_2\wedge b_1$ , it must also be true that $a_{0}\land a_{2}\geq a_1\land b_1=a_1\land a_0$ , so $a_{0}$ and $a_{2}$ meet at least as closely to each other as to $a_{1}$ . These are inconsistent.

(4) $x\land y_{1}=x\land y_{1}\land y_{2}$ (i.e., $x\land y_{1}\leq y_{2}$ )—a negated semi-equation: given $\left (a_{i},b_{i},b_{i}'\right )_{i\in \mathbb {Z}}$ such that $\models a_{i}\land b_{j}\leq b_{j}'\iff i=j$ , in particular $a_{0}\land b_{0}\leq b_{0}'$ and $a_i\land b_0\not \leq b_{0}'$ for $i\neq 0$ . Since the initial segment below $b_0$ is totally ordered, it follows that $a_0\land b_0 < a_i \land b_0$ for $i\neq 0$ . $a_1\land a_2\geq \left (a_1\land b_0\right )\land \left (a_2\land b_0\right )>a_0\land b_0=a_0\land a_1$ . That is, $a_1$ and $a_2$ meet strictly closer together with each other than with $a_0$ . But, by switching the roles of the indices $0$ and $2$ in that argument, $a_0$ and $a_1$ must meet strictly closer together with each other than with $a_2$ as well, a contradiction.

Proof By quantifier elimination, there are four complete $2$ -types over $\emptyset $ axiomatized by $\left \{ x=y,\,x>y,\,x<y,\,x\perp y\right \} $ , where $\perp $ denotes incomparable elements. Thus, up to equivalence, there are only $16$ formulas $\varphi \left (x,y\right )$ with $x,y$ singletons without parameters. By a direct case analysis (see [Reference Chernikov and Mennen7, Theorem 3.17] for the details) the only $1$ -semi-equations among them are $x\neq x$ , $x=x$ , $x=y$ , $x>y$ , $x\geq y$ . None of them separate $x<y$ from $x\perp y$ , so any Boolean combination of $1$ -semi-equations implied by $x<y$ must also be implied by $x\perp y$ , so $x<y$ is not equivalent to a Boolean combination of $1$ -semi-equations.

Proof Suppose $x<y$ is equivalent to a Boolean combination of $1$ -semi-equations with parameters $c=\left (c_1, \ldots , c_n\right )$ . Say $x<y\iff \Phi (\varphi _1 (x,y,c ), \ldots , \varphi _k (x,y,c ))$ , where $\Phi $ is a Boolean formula in k variables, and $\varphi _1\left (x,y,c\right ), \ldots , \varphi _k\left (x,y,c\right )$ are $1$ -semi-equations. Let d be an element such that $d\perp \bigwedge _{i\leq n}c_i$ . For each i, let $\psi _i\left (x,y\right )$ be the formula $\exists z \left ( \operatorname {tp}(z) = \operatorname {tp}\left (c\right ) \& \left (x\wedge y\perp \bigwedge _{i\leq n}z_i\right ) \& \, \varphi _i\left (x,y,z\right ) \right )$ . As $\operatorname {tp}\left (c\right )$ is isolated by quantifier elimination, this is indeed a first-order formula. For $a,b>d$ , if $\models \varphi _i\left (a,b,c\right )$ , then $\models \psi _i\left (a,b\right )$ . By quantifier elimination and [Reference Simon35, Lemma 4.4], the converse also holds. Thus, for $a,b>d$ , $ \models a<b\iff \models \Phi \left (\varphi _1\left (a,b,c\right ), \ldots , \varphi _k\left (a,b,c\right )\right )\iff \models \Phi \left (\psi _1\left (a,b\right ), \ldots ,\psi _k\left (a,b\right )\right )$ . Since all singletons have the same type, it follows that this holds for all $a,b$ . It thus remains to show that each $\psi _i\left (x,y\right )$ is a $1$ -semi-equation, contradicting Theorem 3.13. If this were not the case for some $i\leq k$ , then there would be $\left (b_j\right )_{j\in \mathbb N}$ and a such that $\models \psi _i\left (a,b_j\right )$ for all $j\in \mathbb N$ , but such that for every $j\neq \ell \in \mathbb N$ , there is $a_{j,\ell }$ such that $\models \psi _i\left (a_{j,\ell },b_j\right )$ but $\not \models \psi _i\left (a_{j,\ell },b_{\ell }\right )$ . But, again because all singletons have the same type, and every finite set of elements has a lower bound, it is consistent that furthermore all of these elements are above d. But then this would also provide a counterexample to $\varphi _i\left (x,y\right )$ being a $1$ -semi-equation.

Proof Assume there exist an array $\left ( a_{i,j} : i,j \in \mathbb {Z} \right )$ and b with these properties. For any fixed $i\neq 0$ , we have $\models \varphi \left (a_{i,j},b\right )$ for all $j \neq 0$ , $\left (a_{i,j}\right )_{j \in \mathbb {Z}}$ is indiscernible, and $\left (a_{i,j}\right )_{j\neq 0}$ is b-indiscernible, so, by weak semi-equationality of $\varphi $ , $\models \varphi \left (a_{i,0},b\right )$ . But $\not \models \varphi \left (a_{i,0},b\right )\land \neg \psi \left (a_{i,0},b\right )$ , so $\models \psi \left (a_{i,0},b\right )$ . Now the sequence $\left (a_{i,0}\right )_{i \in \mathbb {Z}}$ is indiscernible, $\left (a_{i,0}\right )_{i\neq 0}$ is b-indiscernible, and $\models \psi (a_{i,0},b)$ for all $i \neq 0$ , so, by weak semi-equationality of $\psi $ , $\models \psi \left (a_{0,0},b\right )$ —contradicting $\models \varphi \left (a_{0,0},b\right )\land \neg \psi \left (a_{0,0},b\right )$ .

Proof Any conjunction of finitely many weak semi-equations and negations of weak semi-equations is of the form $\psi \left (x,y\right )\land \neg \theta \left (x,y\right )$ for some weak semi-equations $\psi \left (x,y\right )$ and $\theta \left (x,y\right )$ , because weak semi-equations are closed under conjunction and under disjunction (Proposition 2.3(3)), so negations of weak semi-equations are also closed under conjunction. Thus any Boolean combination of weak semi-equations is equivalent, via its disjunctive normal form, to $\bigvee _{k\in I} \left ( \psi _{k}\left (x,y\right )\land \neg \theta _{k}\left (x,y\right ) \right )$ for some finite index set I and weak semi-equations $\psi _{k}\left (x,y\right )$ and $\theta _{k}\left (x,y\right )$ for $k\in I$ . Given b and $\left (a_{i,j}\right )_{i.j\in \mathbb {Z}}$ as above, since $i=0\lor j\neq 0\iff \models \varphi \left (a_{i,j},b\right )\iff \models \bigvee _{k\in I} \left ( \psi _{k}\left (a_{i,j},b\right )\land \neg \theta _{k}\left (a_{i,j},b\right ) \right )$ , and all $a_{i,j}$ with $i=0$ or $j\neq 0$ have the same type over b, there is some k such that $\models \psi _{k}\left (a_{i,j},b\right )\land \neg \theta _{k}\left (a_{i,j},b\right )\iff i=0\lor j\neq 0$ , contradicting Lemma 5.2.

Claim 5.6. The elements $\gamma _{0},\gamma _{1},\gamma _{2},\gamma _{3},\gamma _{4},\gamma _{5}$ are $\mathbb {Z}$ -linearly independent (in $\Gamma $ viewed as a $\mathbb {Z}$ -module). If K has mixed characteristic $\left (0,p\right )$ , then $\nu \left (p\right ),\gamma _{0},\gamma _{1},\gamma _{2},\gamma _{3},\gamma _{4},\gamma _{5}$ are $\mathbb {Z}$ -linearly independent.

Proof If $n_{0}\gamma _{0}+ \cdots + n_{5}\gamma _{5}=0$ with $n_{0}, \ldots , n_{5}\in \mathbb {Z}$ not all $0$ , let $i\leq 5$ be maximal such that $n_{i}\neq 0$ . Now $n_{0}\gamma _{0}+ \cdots + n_{i}\gamma _{i}=0$ , and $n_{i}\neq 0$ . By indiscernibility of the sequence $\left (\gamma _1, \ldots , \gamma _6 \right )$ , $n_{0}\gamma _{0}+ \cdots + n_{i-1}\gamma _{i-1}+n_{i}\gamma _{6}=0$ , but then $n_{i}\left (\gamma _{i}-\gamma _{6}\right )=0$ , contradicting that $n_{i}\neq 0$ , $\gamma _{i}\neq \gamma _{6}$ , and $\Gamma $ is ordered and thus torsion-free. In mixed characteristic, the same argument can be repeated starting from $n_{0}\gamma _{0}+ \cdots +n_{5}\gamma _{5}=m\nu \left (p\right )$ with $n_{0}, \ldots , n_{5},m\in \mathbb {Z}$ .

Claim 5.7. (1) Let $\varphi \left (x;z;w;b',a_{J}'\right )$ be a formula with parameters $b'$ and $a_{J}'$ for some $J\subseteq \mathbb {Z}$ , tuples of variables x of sort K, z of sort k, and w of sort $\Gamma _{\infty }$ . Let $I_{1},I_{2}$ be tuples of distinct indices from $\mathbb {Z}$ , with $\left |I_{1}\right |=\left |I_{2}\right |=\left |x\right |$ . Let $\sigma \in \operatorname {Aut}\left (k\right )$ be such that $\sigma \left (\tilde {a}_{I_{1}}\right )=\tilde {a}_{I_{2}}$ (preserving the ordering of the tuples), $\sigma \left (\tilde {a}_{J}'\right )=\tilde {a}_{J}'$ , and $\sigma \left (\tilde {b}'\right )=\tilde {b}'$ . Then for any tuples $c \in k^z, d \in \Gamma _{\infty }^w$ we have $\models \varphi \left (a_{I_{1}};c;d;b';a_{J}'\right )\iff \models \varphi \left (a_{I_{2}};\sigma \left (c\right );d;b';a_{J}'\right )$ .

(2) Likewise, let $\varphi \left (y;z;w;b,a_{I}\right )$ be a formula with parameters b and $a_I$ for some $I\subseteq \mathbb {Z}$ , tuples of variables y of sort K, z of sort k, and w of sort $\Gamma _{\infty }$ . Let $J_1,J_2$ be tuples of distinct indices from $\mathbb {Z}$ , with $\left |J_1\right |=\left |J_2\right |=\left |y\right |$ , and let $\sigma \in \operatorname {Aut}\left (k\right )$ be such that $\sigma \left (\tilde {a}_{J_1}'\right )=\tilde {a}^{\prime }_{J_2}$ , $\sigma \left (\tilde {a}_I\right )=\tilde {a}_I$ , and $\sigma \left (\tilde {b}\right )= \tilde {b}$ . Then for any tuples $c \in k^z, d \in \Gamma _{\infty }^w$ we have

$$ \begin{align*}\models\varphi\left(a_{J_{1}}';c;d;b;a_{I}\right)\iff\models\varphi\left(a_{J_{2}}';\sigma\left(c\right);d;b;a_{I}\right).\end{align*} $$

Claim 5.8. Let $\varphi \left (x;y;z;w;b;b'\right )$ be a formula with parameters $b,b'$ , where x and y are single variables of sort K, and z and w are tuples of variables of sort k and $\Gamma _{\infty }$ , respectively. Let $\sigma _{i}\in \operatorname {Aut} \left (k\right )$ be such that $\sigma _{i}\left (\tilde {a}_{i}\right )=\tilde {b}$ , $\sigma _{i}\left (\tilde {a}_{0}\right )=\tilde {a}_{0}$ , $\sigma _{i}\left (\tilde {a}_{0}'\right )=\tilde {a}_{0}'$ , and $\sigma _{i}\left (\tilde {b}'\right )=\tilde {b}'$ . Let $\sigma _{j}'\in \operatorname {Aut} \left (k\right )$ be such that $\sigma _{j}'\left (\tilde {b}'\right )=\tilde {a}^{\prime }_j$ , $\sigma _{j}'\left (\tilde {a}_{0}'\right )=\tilde {a}_{0}'$ , $\sigma _{j}'\left (\tilde {a}_{i}\right )=\tilde {a}_{i}$ , and $\sigma _{j}'\left (\tilde {b}\right )=\tilde {b}$ . Let $\pi \in \operatorname {Aut} \left (\Gamma _{\infty }\right )$ be such that $\pi \left (\gamma _{2}\right )=\gamma _{4}$ , $\pi \left (\gamma _{0}\right )=\gamma _{0}$ , $\pi \left (\gamma _{1}\right )=\gamma _{1}$ , and $\pi \left (\gamma _{5}\right )=\gamma _{5}$ , and let $\tau \in \operatorname {Aut}\left (\Gamma _{\infty }\right )$ be such that $\tau \left (\gamma _{5}\right )=\gamma _{3}$ , $\tau \left (\gamma _{0}\right )=\gamma _{0}$ , $\tau \left (\gamma _{1}\right )=\gamma _{1}$ , and $\tau \left (\gamma _{2}\right )=\gamma _{2}$ . Then, for $i,j\neq 0$ , $c\in k^{z}$ , and $d\in \Gamma _{\infty }^{w}$ , $\models \varphi \left (a_{0};a_{0}';\sigma _{i}\left (c\right );\pi \left (d\right );b;b'\right )\iff \models \varphi \left (a_{i};a_{0}';c;d;b;b'\right )\iff \models \varphi \left (a_{i};a_{j}';\sigma _{j}'\left (c\right );\tau \left (d\right );b;b'\right )$ .

Claim 5.9. The elements $a_{\infty },a_{\infty }',\left (\alpha \operatorname {lift} \left ( \tilde {a}_{i} \right )\right )_{i\in \mathbb {Z}}, \left (\beta \operatorname {lift} \left ( \tilde {a}_{j}' \right )\right )_{j\in \mathbb {Z}}, b-a_0, b'-a_{0}'$ are valuationally independent.

Proof Define a valuation $\nu ^{*}:\mathbb {Z}\left [u,v,x,y,z,w\right ]\rightarrow \Gamma _{\infty }$ (with $\left |u\right |=\left |v\right |=\left |z\right |=\left |w\right |=1$ , $\left |x\right |,\left |y\right |$ arbitrary), by, for monomials (which in case of mixed characteristic is taken to include its coefficient),

$$ \begin{gather*} \nu^{*}\left(n \cdot u^{r_{\infty}}x_{1}^{r_{1}} \ldots x_{\left|x\right|}^{r_{\left|x\right|}}v^{s_{\infty}}y_{1}^{s_{1}} \ldots y_{\left|y\right|}^{s_{\left|y\right|}}z^{t_1}w^{t_2}\right) \\ :=\nu\left(n\right)+r_{\infty}\gamma_{0}+s_{\infty}\gamma_{1}+\left(r_{1}+ \cdots +r_{\left|x\right|}\right)\gamma_{2}+\left(s_{1}+ \cdots +s_{\left|y\right|}\right)\gamma_{3} + t_1\gamma_{4} + t_2\gamma_{5}, \end{gather*} $$

and the valuation of a polynomial is the minimum of the valuations of its monomials. That way, for any $I,J\subseteq \mathbb {Z}$ with $\left |I\right |=\left |x\right |$ and $\left |J\right |=\left |y\right |$ we have

$$ \begin{gather*} \nu^{*}\left(f\left(u,v,x,y,z,w\right)\right)=\nu\left(f\left(a_{\infty},a_{\infty}',\alpha\cdot\operatorname{lift}\left(\tilde{a}_{I}\right),\beta\cdot\operatorname{lift}\left(\tilde{a}_{J}'\right),b-a_0,b'-a_{0}'\right)\right) \end{gather*} $$

when f is a monomial (where $\alpha \cdot \operatorname {lift}(\tilde {a}_I) := \left (\alpha \operatorname {lift}(\tilde {a}_i)\right )_{i\in I}$ ), and we need to prove that this holds for all polynomials f. Given a polynomial $f\left (u,v,x,y,z,w\right )$ ,

$$ \begin{align*}\nu^{*}\left(f\right)=\nu\left(n\right)+m_{0}\gamma_{0}+m_{1}\gamma_{1}+m_{2}\gamma_{2}+m_{3}\gamma_{3}+m_4\gamma_4+m_5\gamma_5\end{align*} $$

for some $n,m_{0},m_{1},m_{2},m_{3},m_{4},m_{5}\in \mathbb {N}$ (with $\nu \left (n\right ),m_{0},m_{1},m_{2},m_{3},m_{4},m_{5}$ unique by Claim 5.6). Let $\tilde {f}\left (u,v,x,y,z,w\right )$ be the sum of monomials in f of the same valuation as f, so that every monomial appearing in $\tilde {f}\left (u,v,x,y,z,w\right )$ has degree $m_{0}$ in u, degree $m_{1}$ in v, total degree $m_{2}$ in x, total degree $m_{3}$ in y, degree $m_{4}$ in z, degree $m_{5}$ in w, and has leading coefficient with valuation $\nu \left (n\right )$ , and $\nu ^{*}\left (f-\tilde {f}\right )>\nu ^{*}\left (f\right )$ . Thus

$$ \begin{align*}\frac{\tilde{f}\left(u,v,x,y,z,w\right)}{n \cdot u^{m_{0}}v^{m_{1}}z^{m_4}w^{m_5}}\end{align*} $$

is a non-zero polynomial in $x,y$ , all coefficients having valuation $0$ , so it reduces under the residue map to a nonzero polynomial in $x,y$ . Since the set of elements in the tuples $\tilde {a}_{I},\tilde {a}_{J}'$ is algebraically independent (they come from an infinite indiscernible sequence), it follows that

$$ \begin{align*}\frac{\tilde{f}\left(u,v,\tilde{a}_{I},\tilde{a}_{J}',z,w\right)}{n \cdot u^{m_{0}}v^{m_{1}}z^{m_4}w^{m_5}}\neq0,\end{align*} $$

and thus a lift of it,

$$ \begin{align*}\frac{\tilde{f}\left(u,v,\operatorname{lift}\left(\tilde{a}_{I}\right),\operatorname{lift}\left(\tilde{a}_{J}'\right),z,w\right)}{n \cdot u^{m_{0}}v^{m_{1}}z^{m_4}w^{m_5}},\end{align*} $$

has valuation $0$ . Thus

$$ \begin{align*}\nu\left(\tilde{f}\left(a_{\infty},a_{\infty}',\operatorname{lift}\left(\tilde{a}_{I}\right),\operatorname{lift}\left(\tilde{a}_{J}'\right),b-a_0,b'-a_{0}'\right)\right)=\nu\left(n\right)+m_{0}\gamma_{0}+m_{1}\gamma_{1}+m_4\gamma_4+m_5\gamma_5\end{align*} $$

and, by homogeneity of $\tilde {f}$ ,

$$ \begin{gather*}\nu\left(\tilde{f}\left(a_{\infty},a_{\infty}',\alpha\cdot\operatorname{lift}\left(\tilde{a}_{I}\right),\beta\cdot\operatorname{lift}\left(\tilde{a}_{J}'\right),b-a_0,b'-a_{0}'\right)\right) \\=\nu\left(n\right)+m_{0}\gamma_{0}+m_{1}\gamma_{1}+m_{2}\gamma_{2}+m_{3}\gamma_{3}+m_4\gamma_4+m_5\gamma_5=\nu^{*}\left(f\right).\end{gather*} $$

We have

$$ \begin{align*}\nu\left(\left(f-\tilde{f}\right)\left(a_{\infty},a_{\infty}',\alpha\cdot\operatorname{lift}\left(\tilde{a}_{I}\right),\beta\cdot\operatorname{lift}\left(\tilde{a}_{J}'\right),b-a_0,b'-a_{0}'\right)\right)\geq\nu^{*}\left(f-\tilde{f}\right)>\nu^{*}\left(f\right)\end{align*} $$

(the first inequality holds by the ultrametric property, combined with the fact that it holds for monomials), so it follows that