Tangle-tree duality in abstract separation systems

TL;DR

This paper establishes a broad duality theorem linking local cohesion structures called tangles to global tree-like decompositions across various combinatorial and non-traditional settings.

Contribution

It introduces a unified tangle-tree duality framework applicable to diverse structures, generalizing classical graph width dualities and enabling new applications.

Findings

Proves a general width duality theorem for abstract separation systems.

Unifies the concept of tangles across different structures.

Extends duality theorems to new contexts like image analysis and social sciences.

Abstract

We prove a general width duality theorem for combinatorial structures with well-defined notions of cohesion and separation. These might be graphs and matroids, but can be much more general or quite different. The theorem asserts a duality between the existence of high cohesiveness somewhere local and a global overall tree structure. We describe cohesive substructures in a unified way in the format of tangles: as orientations of low-order separations satisfying certain consistency axioms. These axioms can be expressed without reference to the underlying structure, such as a graph or matroid, but just in terms of the poset of the separations themselves. This makes it possible to identify tangles, and apply our tangle-tree duality theorem, in very diverse settings. Our result implies all the classical duality theorems for width parameters in graph minor theory, such as path-width, tree-width, branch-width or rank-width. It yields new, tangle-type, duality theorems for tree-width and path-width. It implies the existence of width parameters dual to cohesive substructures such as $k$-blocks, edge-tangles, or given subsets of tangles, for which no width duality theorems were previously known. Abstract separation systems can be found also in structures quite unlike graphs and matroids. For example, our theorem can be applied to image analysis by capturing the regions of an image as tangles of separations defined as natural partitions of its set of pixels. It can be applied in big data contexts by capturing clusters as tangles. It can be applied in the social sciences, e.g. by capturing as tangles the few typical mindsets of individuals found by a survey. It could also be applied in pure mathematics, e.g. to separations of compact manifolds.

Equations

$$(A,B)\le (C,D):\iff A\subseteq C\text{ and }B\supseteq D,$$

$$\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{r}\le \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\iff \stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\leftarrow \text{vss}}{r}\ge \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}.$$

$$(\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{r}\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}{)}^{\ast }=\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\leftarrow \text{vss}}{r}\wedge \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}.$$

$$\vec{E}(T):=\{(x,y):\{x,y\}\in E(T)\}$$

$${\vec{F}}_t:=\{(x,t):xt\in E(T)\}$$

$$O^{-}:=\{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\mid \{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}\}\in \mathcal{F}\}.$$

$${\mathcal{F}}_1:=\mathcal{F}\cup \{\{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}\}\}\text{and}{\mathcal{F}}_2:=\mathcal{F}\cup \{\{\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s_0}\}\},$$

$${\alpha }^{\prime }(\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{e}):=\alpha (\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{e})\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}\text{if }\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{e_x}\le \stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{e}$$

$$f\downarrow {}_{\text{raise}0.03012pt\text{vbox}to0.0pt\text{vss}\stackrel{\text{raise}0.1507pt\text{vbox}to0.0pt\to \text{vss}}{s_0}}^{\stackrel{\text{raise}0.43054pt\text{vbox}to0.0pt\to \text{vss}}{r}}(\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}):=\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}\text{and}f\downarrow {}_{\text{raise}0.03012pt\text{vbox}to0.0pt\text{vss}\stackrel{\text{raise}0.1507pt\text{vbox}to0.0pt\to \text{vss}}{s_0}}^{\stackrel{\text{raise}0.43054pt\text{vbox}to0.0pt\to \text{vss}}{r}}(\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}):=(\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}{)}^{\ast }$$

$${\alpha }^{\prime }=f\downarrow {}_{\text{raise}0.03012pt\text{vbox}to0.0pt\text{vss}\stackrel{\text{raise}0.1507pt\text{vbox}to0.0pt\to \text{vss}}{s_0}}^{\stackrel{\text{raise}0.43054pt\text{vbox}to0.0pt\to \text{vss}}{r}}\circ \alpha ,$$

$$\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{r}<\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}={\alpha }^{\prime }(y,t)=(\alpha (t,y)\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}{)}^{\ast }=(\stackrel{\text{raise}0.06029pt\text{vbox}to0.0pt\leftarrow \text{vss}}{r\text{lower}0.5pt{}^{\prime }}\vee \stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s_0}{)}^{\ast }\underset{\text{eqref}deMorgan}{=}\stackrel{\text{raise}0.06029pt\text{vbox}to0.0pt\to \text{vss}}{r\text{lower}0.5pt{}^{\prime }}\wedge \stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s_0}\le \stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s_0},$$

$$O^{-}:=\{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\mid \{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}\}\in \mathcal{F}\}$$

$$O_1^{-}:=O^{-}\cup \{\stackrel{\text{raise}0.60275pt\text{vbox}to0.0pt\to \text{vss}}{r_1}\}=\{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\to \text{vss}}{s}\mid \{\stackrel{\text{raise}0.3014pt\text{vbox}to0.0pt\leftarrow \text{vss}}{s}\}\in {\mathcal{F}}_1\}$$

$${\mathcal{F}}^{\prime }:=\{\sigma \setminus {\vec{S}}^{-}\mid \sigma \in \mathcal{F}\},$$

Full text

We describe cohesive substructures in a unified way in the format of tangles: as orientations of low-order separations satisfying certain consistency axioms. These axioms can be expressed without reference to the underlying structure, such as a graph or matroid, but just in terms of the poset of the separations themselves. This makes it possible to identify tangles, and apply our tangle-tree duality theorem, in very diverse settings.

Our result implies all the classical duality theorems for width parameters in graph minor theory, such as path-width, tree-width, branch-width or rank-width. It yields new, tangle-type, duality theorems for tree-width and path-width. It implies the existence of width parameters dual to cohesive substructures such as $k$-blocks, edge-tangles, or given subsets of tangles, for which no width duality theorems were previously known.

Abstract separation systems can be found also in structures quite unlike graphs and matroids. For example, our theorem can be applied to image analysis by capturing the regions of an image as tangles of separations defined as natural partitions of its set of pixels. It can be applied in big data contexts by capturing clusters as tangles. It can be applied in the social sciences, e.g. by capturing as tangles the few typical mindsets of individuals found by a survey. It could also be applied in pure mathematics, e.g. to separations of compact manifolds.

1 Introduction

There are a number of theorems in the structure theory of sparse graphs that assert a duality between high connectivity present somewhere in the graph and an overall tree structure. For example, a graph either has a large complete minor or a tree-decomposition into torsos of essentially bounded genus, but not both [12, 40]. And it either has a large grid minor or a tree-decomposition into parts of bounded size, but not both [12, 38]. Let us loosely refer to such highly cohesive substructures of a graph, defined in terms of subsets of its vertices together with some required edges, as concrete highly cohesive substructures (HCSs).

An example of a concrete HCS for which no dual notion of global tree structure has been known is that of a $k$-block [6, 36]: a set of at least $k$ vertices no two of which can be separated in the graph by deleting fewer than $k$ vertices.

Conversely, there are a number of so-called width parameters for graphs, invariants whose boundedness asserts that the graph has some kind of global tree structure, for which there are no obvious dual concrete HCSs.

Amini, Mazoit, Nisse, and Thomassé [1] addressed this latter problem in a broad way: they showed how to construct, for many width parameters including all the then known ones, dual concrete HCSs akin to brambles (see [12]). For each parameter, the existence of such an HCS forces this parameter to be large, and conversely, whenever one of these width parameters is large there exists a concrete bramble-type HCS to witness this.

In one of their seminal papers on graph minors [39], Robertson and Seymour introduced a very different way to capture high cohesion somewhere in a graph, which they call tangles. The basic idea behind these is as follows. Given a concrete HCS $X$ in a graph $G$ and a low-order separation, most of $X$ will lie on one of its two sides: otherwise $X$ could not be highly cohesive. In this way $X$, whatever it is, orients each of the low-order separations of $G$ towards one of its sides, the side that contains most of $X$. These orientations of all the low-order separations will be ‘consistent’ in various ways, since they all point towards $X$: no two of them, for example, will point away from each other.

Robertson and Seymour [39] noticed that these orientations of all the low-order separations of $G$ captured most of what they needed to know about $X$. Consequently, they defined a tangle of order $k$ in a graph as a way to orient all its separations of order $<k$, consistently in some precise sense not relevant here.

The notion of a tangle brought with it a shift of paradigm in the connectivity theory of graphs [37]: we can now think of a consistent orientation of all the low-order separations of a graph as a ‘highly cohesive substructure’ in its own right: no longer a concrete one, but an abstract HCS. Such abstract HCSs, though maybe unfamiliar at first, are often ‘deeper’ than concrete ones, because they pick out only the essential information. But they are also easier to work with: one no longer has to worry about the details, say, of where exactly in the graph a subdivided grid has all its connecting paths. And most importantly, they are able to capture HCSs that are inherently fuzzy. For example, the additional detail that a subdivided grid contains over the tangle it defines is not only superfluous but can be misleading: each individual branch vertex can, and typically will, lie on the wrong side of some low-order separation, the side that does not contain most of the grid. (Consider, for example, the separation defined by the four neighbours of a given vertex in an actual grid.)

Our first aim in this paper is to do for abstract HCSs in graphs and matroids the converse of what Amini et al. did for concrete ones: starting from a unified definition of abstract HCSs, we prove a general duality theorem that describes corresponding tree structures to witness the nonexistence of these HCSs.

Generalizing the specific notion of a tangle from [39], we shall define types of abstract HCSs to be called ‘$\mathcal{F}$-tangles’, where $\mathcal{F}$ encodes some particular type of consistency. Thus, an $\mathcal{F}$-tangle in a graph will be a way to orient all its separations of order $<k$ (for some $k$) consistently in a sense specified by $\mathcal{F}$: different notions of consistency will give rise to different sets $\mathcal{F}$ and result in different $\mathcal{F}$-tangles. But we shall prove one unified duality theorem saying that, for every suitable $\mathcal{F}$, a given graph either has an $\mathcal{F}$-tangle or a global tree structure that clearly precludes the existence of an $\mathcal{F}$-tangle.

Our duality theorem will easily imply the two known tangle-type duality theorems from graph minor theory: the classical Robertson-Seymour one for tangles and branch-width in graphs [39], and its analogue for matroids [31, 39]. This has been shown in detail in [21].

It will also imply new, tangle-type, duality theorems for all the other classical width parameters, such as tree- and path-width: for each of these we shall find an $\mathcal{F}$, encoding some specific type of consistency, such that the graphs where this parameter is large are precisely those with an $\mathcal{F}$-tangle. The known duality theorems for these width parameters, in terms of concrete HCSs, will follow from our duality theorem in terms of abstract HCSs, but not conversely. This, too, has been shown in [21].

Our result will further imply duality theorems for $k$-blocks, the main concrete HCS for which no duality theorem has been known, and for any specified type of classical tangles (rather than all of them). This has been done in [16].

Finally, our duality theorem has recently found an unexpected more fundamental application. The emerging theory of abstract separation systems and their tangles – see [2, 3, 4, 5, 7, 8, 11, 14, 16, 17, 19, 21, 24, 25, 26, 27, 28, 29, 30, 32, 34, 33] – used to rest on two pillars: an abstract version [18] of Robertson and Seymour’s tree-of-tangles theorem for graphs [39], and the abstract tangle-tree duality theorem proved here. But, very recently, Elbracht, Kneip and Teegen [24] have been able to derive the abstract tree-of-tangles theorem from the tangle-tree duality theorem. With this deduction, only one pillar remains: the result proved in this paper, and re-proved in [24] in a slightly stronger form.

While the study of tangles as abstract HCSs marked a shift of paradigm from the earlier studies of concrete HCSs, there has since been another major shift of paradigm: from concrete to abstract separations. Separations in graphs – as well as traditional tangles and their dual branch decompositions – are defined in terms of the graph’s edges. But when we proved our duality theorem for graphs we found that, surprisingly, we needed to know only how these separations relate to each other, not how they relate to the graph which they separate.

Our main result, therefore, is now a duality theorem for abstract separation systems. Very roughly, these are partially ordered sets (reflecting the natural partial ordering between separations in graphs and matroids), with an order-reversing involution that reflects the flip $(A,B)\mapsto(B,A)$ of a graph separation.

Both tangles in graphs and their dual tree structure can be expressed in terms of just this partial ordering of their separations. Indeed, the consistency requirement for tangles, that no three ‘small’ sides of its oriented separations shall cover the graph, can be replaced by the requirement that whenever a tangle $\tau$ contains two oriented separations, $(A,B)$ and $(C,D)$ say, it also contains their supremum $(A\cup C,B\cap D)$ as long as this is oriented by $\tau$ at all, i.e., has order $<k$ if $\tau$ is a $k$-tangle. (Note the similarity to ultrafilters, a standard kind of abstract HCSs in infinite contexts.) And the tree-decompositions or branch-decompositions dual to graph tangles can be described purely in terms of the separations too, those that correspond to the edges of their decomposition trees, where the requirement that these edges form a tree can be replaced by requiring that those separations must be nested – which can in turn be expressed just in terms of our poset: two separations are nested if they have comparable orientations.

While our duality theorem for these abstract separation systems implies all the duality theorems mentioned so far, by applying it to separations in graphs or matroids, it can also be applied in very different contexts. These applications are surveyed in [9, 10, 15], in a style aimed at non-mathematicians in the sciences and in the quantitative social sciences, but also accessible to readers of this paper.

As a generic such application outside mathematics consider cluster analysis. The bipartitions of a (large data) set $D$ form a separation system: they are partially ordered by inclusion of their sides, and the involution of flipping the sides of the bipartition inverts this ordering. Depending on the application, some ways of cutting the data set in two will be more natural than others, which gives rise to a cost function on these separations of $D$. Taking this cost of a separation as its ‘order’ then gives rise to tangles: abstract HCSs signifying clusters. Unlike clusters defined by simply specifying a subset of $D$, clusters defined by tangles are allowed to be fuzzy (as in our earlier grid example) – which much enhances their real-world relevance. See [23] for more.

If the cost function on the separations of our data set is submodular – which in practice is not normally a severe restriction – our duality theorem can be applied to these tangles [21]. For every integer $k$, the application will either find a cluster of order at least $k$ or produce a nested ‘tree’ set of bipartitions, all of order ${<k}$, which together witness that no such cluster exists. An example from image analysis, with a cost function chosen so that the clusters become the visible regions in a picture, is given in [22]. In an example from sociology, the yes/no questions of a political or social survey form a separation system whose tangles capture any existing mindsets: typical ways of answering its questions [10, 15]. Tangles can identify such mindsets in a quantitative and precise way even if there does not exist any one set of complete answers that occurs more often than others. Our theorem, in addition, determines how dominant or prevalent such opinions are in the population surveyed, by finding the maximum $k$ for which there exist mindsets that define a $k$-tangle of answers.

There are also potential applications in pure mathematics. For a very simple example, consider a triangulation of a topological sphere. This can be cut in two, in many ways, by closed paths along the edges of the triangulation, i.e., by cycles in its 1-skeleton. The lengths of these paths or cycles define a submodular order function on the separations of our sphere that they define. Our duality theorem then says that, for every integer $k$, there either exists a region dense enough that no closed path of length $<k$ can cut the sphere so as to divide this region roughly in half, or there exists a collection of non-crossing paths each of length $<k$ which, between them, cut up the entire sphere in a tree-like way (with a ternary tree) into single triangles. If we do this in a geometric disc, where the edges in these triangulations have lenths, the inverse limits of these triangulations under refinement will have tangles describing visible ‘blobs’, of varying granularity, of these geometric discs, and our theorem will tell us donw to which small granularity a given blob can be refined.

Our abstract tangle-tree duality theorem will come in two flavours, ‘weak’ and ‘strong’. Our weak duality theorem, presented in Section 3, will be easy to prove but has no direct applications. It will be used as a stepping stone for the strong duality theorem, our main result, which we prove in Section 4. In Section 5 we present a refinement of the strong duality theorem.

2 Terminology and basic facts

Our aim in this section is to introduce the reader to just enough of the terms and basic theory of abstract separation systems [13] to read this paper. In order to facilitate the build-up of enough intuition to flesh out the rather technical proof of our main theorem, however, we begin with a special case: some simple observations about separations of sets. The abstract definitions to follow will then be presented fairly concisely.

The ArXiv version [20] of this paper follows a slightly different approach here by seeking to motivate our abstract definitions before they are introduced, rather than just illustrating them afterwards by pointing out what they mean for separations of sets. Readers interested in why exactly our abstract notions are what they are, are invited to consults [20] for more insight.

2.1 Separations of sets

Let us start with a common generalization of separations in graphs on the one hand, and bipartitions of sets on the other. A separation of a set $V$ is a set $\{A,B\}$ such that ${A\cup B=V\!}$. Note that $A\cap B$ may be non-empty. It will usually be small, but technically also $\{V,V\}=\{V\}$ is allowed. The ordered pairs $(A,B)$ and $(B,A)$ are the two orientations of the (unoriented) separation $\{A,B\}$ of $V\!$. (The separation $\{V,V\}$ has only the one orientation $(V,V)$.) The oriented separations of $V$ are the orientations of its separations: the ordered pairs $(A,B)$ such that ${A\cup B=V\!}$. Later, we shall use the term ‘separation’ informally for either oriented or unoriented separations, as long as the context is clear.

Mapping every oriented separation $(A,B)$ to its inverse $(B,A)$ is an involution on the set of all oriented separations of $V\!$. Note that it reverses the partial ordering of this set given by

[TABLE]

since $(A,B)\leq(C,D)$ is equivalent to $(B,A)\geq(D,C)$; see Figure 1. Informally, we think of $(A,B)$ as pointing towards $B$ and away from $A$.

2.2 Separation systems

Generalizing these properties of separations of sets, we now give an axiomatic definition of abstract separations: these do not have to ‘separate’ anything, but their properties will resemble the properties of separations of sets we just outlined above. We shall define oriented (abstract) separations first, and then pair them with their inverses to form unoriented separations as quotients of oriented ones.

A separation system $({\vec{S}},\leq\,,\!{}^{*})$ is a partially ordered set ${\vec{S}}$ with an order-reversing involution *. Its elements are called oriented separations. When a given element of ${\vec{S}}$ is denoted as {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, its inverse {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}{}^{*} will be denoted as {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, and vice versa. The assumption that * be order-reversing means that, for all {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}},

[TABLE]

An (unoriented) separation is a set of the form \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, and then denoted by $s$. We call {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} the orientations of $s$. We say that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} points towards $s$, and {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} points away from $s$, if {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} or {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. In Figure 1, for example, the oriented separation $(E,F)$ points towards the separations $\{A,B\}$ and $\{C,D\}$.

The set of all unoriented separations s=\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\subseteq{\vec{S}} will be denoted by $S$. If {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, we call both {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and $s$ degenerate. A set $V\!$, clearly, has exactly one degenerate oriented separation: the separation $(V,V)$.

When a separation is introduced notationally ahead of its elements, and denoted by a single letter $s$, say, then its elements will subsequently be denoted as {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. Given a set $S^{\prime}\subseteq S$ of unoriented separations, we write {\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}:=\bigcup S^{\prime}\subseteq{\vec{S}} for the set of all the orientations of its elements. With the ordering and involution induced from ${\vec{S}}$, this is again a separation system.

Separations of sets, and their orientations, are clearly an instance of this if we identify $\{A,B\}$ with $\{(A,B),(B,A)\}$.

If there are binary operations $\vee$ and $\wedge$ on our separation system ${\vec{S}}$ such that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\vee{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is the supremum and {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\wedge{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} the infimum of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$, we call $({\vec{S}},\leq\,,\!{}^{*},\vee,\wedge)$ a universe of (oriented) separations. By (1), it satisfies De Morgan’s law:

[TABLE]

The oriented separations of a set $V$ form such a universe: if {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=(A,B) and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=(C,D), say, then {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\vee{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}:=(A\cup C,B\cap D) and {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\wedge{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}:=(A\cap C,B\cup D) are again oriented separations of $V$, and are the supremum and infimum of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, respectively. Similarly, the oriented separations of a graph (see [12]) form a universe of separations (see Figure 1). Its oriented separations of order $<k$ for some fixed $k$, however, form a separation system ${\vec{S}}_{k}$ inside this universe that may not itself be a universe (with the same definition of $\vee$ and $\wedge$). This is because the separations $(A\cup C,B\cap D)$ and $(A\cap C,B\cup D)$ may have an order greater than $k$, and then fail to lie in ${\vec{S}}_{k}$.

2.3 Small and trivial separations

A separation {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} is small if {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. The set of small separations is closed down in ${\vec{S}}$: if {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is small then so is any {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, because {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} by (1). The small separations of a set $V\!$ are those of the form $(A,V)$.

A separation {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} is trivial in ${\vec{S}}$, and its inverse {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} co-trivial, if there exists $s\in S$ such that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as well as {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. Such an $s$ is a witness of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and its triviality. The trivial separations of a set $V\!$, in the system ${\vec{S}}$ of all its separations, are those of the form {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=(X,V) for which there exists $s=\{A,B\}\in S\smallsetminus\{r\}$ with $X\subseteq A\cap B$.

All trivial separations are small: if $s$ witnesses the triviality of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, then {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as well as {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, and hence {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} by (1). As these inequalities are strict, trivial separations are never degenerate.

Small but nontrivial separations can exist but are rare: only the maximal small separations in ${\vec{S}}$ can be nontrivial. Indeed, if {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is small then every {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is not only small but in fact trivial, since {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. As trivial separations are small, this means that they, too, are closed down in ${\vec{S}}$. We thus have two down-closed subsets of ${\vec{S}}$: the set of trivial separations, and the (possibly) slightly larger set of small separations.

2.4 Nestedness and consistency

Two separations $r,s\in S$ are nested if they have comparable orientations; otherwise they cross. Two oriented separations {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} are nested if $r$ and $s$ are nested.111Terms introduced for unoriented separations may be used informally for oriented separations too if the meaning is obvious, and vice versa. Thus, two nested oriented separations are either comparable, or point towards each other, or point away from each other. A set of separations is nested if every two of its elements are nested. In Figure 1, the separations $\{A,B\}$ and $\{C,D\}$ cross but are nested with $\{E,F\}$.

A set $O\subseteq{\vec{S}}$ of oriented separations is antisymmetric if it does not contain the inverse of any of its nondegenerate elements. It is consistent if there are no distinct $r,s\in S$ with orientations {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} such that {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in O. In other words, a set of oriented separations is consistent if no two of its elements that are orientations of distinct separations point away from each other.

An orientation of a set ${S}$ of separations is a set $O\subseteq{{\vec{S}}}$ that contains for every $s\in{S}$ exactly one of its orientations {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. A partial orientation of ${S}$ is an orientation of a subset of ${S}$, i.e., an antisymmetric subset of ${\vec{S}}$.

Every consistent orientation of $S$ contains all separations {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} that are trivial in ${\vec{S}}$, because it cannot contain their inverse {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}: if the triviality of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is witnessed by $s\in S$, say, then {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} would be inconsistent with both {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. It is not hard to show that every consistent partial orientation of $S$ containing no co-trivial {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in{\vec{S}} extends to a consistent orientation of all of $S$; see [14].

2.5 Stars of separations

A family (\,{\mathop{\kern 0.0pts_{i}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\mid i\in I\,), possibly empty, of nondegenerate oriented separations is a multistar of separations if they point towards each other, that is, if {\mathop{\kern 0.0pts_{i}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{j}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} for all distinct $i,j\in I$ (Figure 2). Note that if a multistar contains a separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} more than once then {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} must be small.

Multistars in which each element occurs only once are called stars. To avoid notational hairsplitting, we think of stars as the obvious sets rather than as families, and say that a multistar (\,{\mathop{\kern 0.0pts_{i}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\mid i\in I\,) induces the star \{\,{\mathop{\kern 0.0pts_{i}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\mid i\in I\,\} obtained from it by forgetting the multiplicities of its elements.

Multistars of separations are clearly nested. They are also consistent: if {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} lie in the same multistar we cannot have {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, since also {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} by the multistar property.

Note that a multistar $\sigma$ need not, by definition, be antisymmetric. But if it is not, i.e. if \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\subseteq\sigma, say, then any other {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\sigma will be trivial, witnessed by $s$. Hence most of the stars we have to deal with will in fact be antisymmetric, but it is important to keep this example in mind as a pathological case that can, and will, occur.

Given a tree $T$ (which, by definition [12], has at least one node), there is a natural partial ordering on the set

[TABLE]

of its oriented edges defined by letting $(x,y)<(u,v)$ if $\{x,y\}\neq\{u,v\}$ and the unique $\{x,y\}$–$\{u,v\}$ path in $T$ joins $y$ to $u$ (see Figure 3). For each node $t$ of $T$, we call the set

[TABLE]

of its incoming oriented edges the oriented star at $t$ in $T$. This is also a star in the separation system $(\vec{E}(T),\leq,{}^{*})$, where $\leq$ comes from the natural partial ordering on $\vec{E}(T)$ and ${}^{*}\colon(x,y)\mapsto(y,x)$ flips the orientations of edges.

2.6 $S$-trees

Let $({\vec{S}},\leq\,,\!{}^{*})$ be a separation system. An ${S}$-tree is a pair $(T,\alpha)$ of a tree $T$ and a function $\alpha\colon\vec{E}(T)\to{\vec{S}}$ that commutes with the involutions, i.e., satisfies \alpha({\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)^{*} for all \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T). If $T$ has an edge and we consider it as rooted at a leaf $x$, then this implicitly defines its oriented edge emanating from $x$ as {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}.

For every node $t\in T$, the families (\,\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\mid\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{F}_{t}\,) and the sets $\alpha(\vec{F}_{t})$ of separations in ${\vec{S}}$ are said to be associated with $t$ in $(T,\alpha)$. If all the sets $\alpha(\vec{F}_{t})$ are elements of some set $\mathcal{F}$ we say that $(T,\alpha)$ is an $S$-tree over $\mathcal{F}$. If the elements of $\mathcal{F}$ are stars, we also say that $(T,\alpha)$ is an $S$-tree over stars.

If $\alpha$ preserves the natural ordering on $\vec{E}(T)$, i.e., if for all \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt,\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T) with \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\leq\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt we have \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\leq\alpha(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) in ${\vec{S}}$ (Fig. 3), we call $(T,\alpha)$ order-respecting.

Note that the map $\alpha$ in an order-respecting $S$-tree $(T,\alpha)$ need not preserve strict inequalities: it can happen for \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt<\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt that \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)=\alpha(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt). Similarly, while the sets $\vec{F}_{t}\subseteq\vec{E}(T)$ are stars, by definition, their images (\,\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\mid\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{F}_{t}\,) as families will, in general, only be multistars. The sets $\alpha(\vec{F}_{t})\subseteq{\vec{S}}$, then, are the stars in ${\vec{S}}$ which these multistars induce.

While order-respecting $S$-trees are clearly over stars, $S$-trees over stars need not be order-respecting. For example, let $T_{3}$ be obtained from the 3-star with centre $t$ and leaves $x_{1},x_{2},x_{3}$ by subdividing each edge $x_{i}t$ by a new vertex $y_{i}$. Let $\alpha_{3}$ map all the oriented edges $(y_{i},t)$ to the same separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, and the edges $(x_{i},y_{i})$ to separations {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, all nondegenerate. Then $(T_{3},\alpha_{3})$ is an $S$-tree over stars. In particular, \alpha_{3}(\vec{F}_{t})=\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} is a star, even though $(\alpha_{3}(y_{1},t),\alpha_{3}(y_{2},t),\alpha_{3}(y_{3},t))$ is not a multistar (unless {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is small). So the separations $r_{i}$ need not be nested with each other: it is easy to think of examples where three separations {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} ($i=1,2,3$) cross pairwise. And if the {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} are not nested, then $(T_{3},\alpha_{3})$ will not be order-respecting.

However, this example is essentially the only one. Indeed, let us call an $S$-tree $(T,\alpha)$ redundant if it has a node $t$ of $T$ with distinct neighbours $t^{\prime},t^{\prime\prime}$ such that $\alpha(t^{\prime},t)=\alpha(t^{\prime\prime},t)$; otherwise we call it irredundant. Figure 4 shows an irredundant $S$-tree over stars, so $\alpha(\vec{E}(T))\subseteq{\vec{S}}$ is nested.

Lemma 2.1**.**

[14]* Every irredundant $S$-tree $(T,\alpha)$ over stars is order-respecting. In particular, $\alpha(\vec{E}(T))$ is a nested set of separations in ${\vec{S}}$.*

Proof.

As $(T,\alpha)$ is irredundant, our assumption that the sets $\alpha(\vec{F}_{t})$ are stars is tantamount to saying that the families (\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\mid\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{F}_{t}) are multistars. In other words, the $S$-trees which $(T,\alpha)$ induces on its maximal stars, the subtrees consisting of a fixed node $t$ and the neighbours of $t$, are order-respecting. As the relation $\leq$ is transitive, this propagates through $\vec{E}(T)$ to make the entire $(T,\alpha)$ order-respecting. ∎

Note that Lemma 2.1 does not have a direct converse: order-respecting $S$-trees over stars can be redundant. Indeed, if the separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in our earlier example of $(T_{3},\alpha_{3})$ is small, i.e., satisfies {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, then $(T_{3},\alpha_{3})$ is an order-respecting but redundant $S$-tree over stars. See [14, Lemma 6.3] for more details on how the orderings on $\vec{E}(T)$ and its $\alpha$-image in ${\vec{S}}$ are related or not.

Two edges of an irredundant $S$-tree over stars cannot have orientations that point towards each other and map to the same separation, unless this is trivial:

Lemma 2.2**.**

[14]* Let $(T,\alpha)$ be an irredundant $S$-tree over a set $\mathcal{F}$ of stars. Let $e,f$ be distinct edges of $T$ with orientations \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt<\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt such that \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)=\alpha({\mathop{\kern 0.0ptf}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})=:{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Then {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial.*

In particular, $T$ cannot have distinct leaves associated with the same star \{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} unless {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial.

Note that the irredundancy assumption in Lemma 2.2 cannot be replaced by the weaker assumption that $(T,\alpha)$ be order-respecting: if $T$ is a 3-star whose edges oriented towards the centre map to the same small but nontrivial separation\,{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, then $(T,\alpha)$ is order-respecting but its three leaves are associated with the same star \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} with {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} nontrivial.

Redundant $S$-trees can clearly be pruned to irredundant ones over the same $\mathcal{F}$, by deleting entire branches that hang off edges causing a redundancy:

Lemma 2.3**.**

[14]* If $(T,\alpha)$ is an $S$-tree over $\mathcal{F}$, possibly redundant, then $T$ has a subtree $T^{\prime}$ such that $(T^{\prime},\alpha^{\prime})$ is an irredundant $S$-tree over $\mathcal{F}$, where $\alpha^{\prime}$ is the restriction of $\alpha$ to $\vec{E}(T^{\prime})$. If $(T,\alpha)$ is rooted at a leaf $x$ and $T$ has an edge, then $T^{\prime}$ can be chosen so as to contain $x$ and $e_{x}$.∎*

Recall that stars of separations need not, by definition, be antisymmetric. While it is important for our proofs to allow this, we can always contract an $S$-tree $(T,\alpha)$ over a set $\mathcal{F}$ of stars to an $S$-tree $(T^{\prime},\alpha^{\prime})$ over the subset $\mathcal{F}^{\prime}\subseteq\mathcal{F}$ of its antisymmetric stars. Indeed, if $T$ has a node $t$ such that $\alpha(\vec{F}_{t})$ is not antisymmetric, then $t$ has neighbours $t^{\prime},t^{\prime\prime}$ such that \alpha(t^{\prime},t)={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha(t,t^{\prime\prime}) for some {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}}. Let $T^{\prime}$ be the tree obtained from $T$ by deleting the component of $T-t^{\prime}t-tt^{\prime\prime}$ containing $t$ and joining $t^{\prime}$ to $t^{\prime\prime}$. Let \alpha^{\prime}(t^{\prime},t^{\prime\prime}):={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and \alpha^{\prime}(t^{\prime\prime},t^{\prime}):={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, and otherwise let $\alpha^{\prime}:=\alpha\!\restriction\!\vec{E}(T^{\prime})$. Then $(T^{\prime},\alpha^{\prime})$ is again an $S$-tree over $\mathcal{F}$. Since we can do this whenever some $\vec{F}_{t}$ maps to a star of separations that is not antisymmetric, but only finitely often, we must arrive at an $S$-tree over $\mathcal{F}^{\prime}$.

An $S$-tree $(T,\alpha)$ is called tight if all the sets $\alpha(\vec{F}_{t})$ for nodes $t\in T$ are antisymmetric. The reduction described above thus turns an arbitrary $S$-tree over $\mathcal{F}$ into a tight $S$-tree over $\mathcal{F}^{\prime}$.

Lemma 2.4**.**

[14]* Let $(T,\alpha)$ be an $S$-tree over a set $\mathcal{F}$ of stars, rooted at a leaf $x$. Assume that $T$ has an edge, and that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) is nontrivial. Then $T$ has a minor $T^{\prime}\!$ containing $x$ and $e_{x}$ such that $(T^{\prime},\alpha^{\prime})$, where $\alpha^{\prime}=\alpha\!\restriction\!\vec{E}(T^{\prime})$, is a tight and irredundant $S$-tree over $\mathcal{F}$.*

For every such $(T^{\prime},\alpha^{\prime})$ the edge {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is the only edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T^{\prime}) with \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}.

Proof.

By Lemma 2.3 we may assume that $(T,\alpha)$ is irredundant and contains both $x$ and $e_{x}$. We now apply to $(T,\alpha)$ the reduction described before this lemma to obtain a tight and irredundant $S$-tree $(T^{\prime},\alpha^{\prime})$ over $\mathcal{F}$.

Let us show that $T^{\prime}$ still contains $x$ and $e_{x}$. When, in the said reduction process, we formally deleted an edge $e$ at a node $t$ (with \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt oriented towards $t$, say), then \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) was trivial, witnessed by $s$. (As $(T,\alpha)$ is irredundant, we have \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\notin\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}.) But every other edge $e^{\prime}$ deleted at that step had an orientation {\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt for such an edge $e$ at $t$, making \alpha({\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) trivial too. As $(T,\alpha)$ was order-respecting (Lemma 2.1), we thus never deleted $e_{x}$, because {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) was nontrivial by assumption.

It remains to show that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is the only edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T^{\prime}) with \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. If there is another such edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt, then $e\neq e_{x}$, since otherwise \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt={\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} and hence \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)={\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} as well as \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, which would make $r$ degenerate, contradicting the fact that $(T,\alpha)$ is an $S$-tree over stars.

By Lemma 2.2 we cannot have {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, so {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt since {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} issues from a leaf. By Lemma 2.1, every edge {\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} with {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt satisfies {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\leq\alpha({\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\leq\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, so \alpha({\mathop{\kern 0.0pte\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Our assumption of \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\neq{\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} thus implies that $(T^{\prime},\alpha^{\prime})$ is not tight, a contradiction. ∎

3 Weak duality

Our paradigm in this paper is to capture the notion of ‘highly cohesive substructures’ in a given combinatorial structure by orientations $O$ of a set $S$ of separations of this structure that satisfy certain consistency rules laid down by specifying a set $\mathcal{F}$ of ‘forbidden’ sets of oriented separations that $O$ must not contain.

Let us say that a partial orientation $P$ of $S$ avoids $\mathcal{F}\subseteq 2^{{\vec{S}}}$ if $2^{P}\cap\mathcal{F}=\emptyset$.

Theorem 3.1** (Weak Duality Theorem).**

Let $({\vec{S}},\leq\,,\!{}^{*})$ be a finite separation system and $\mathcal{F}\subseteq 2^{\vec{S}}\!$ a set of stars. Then exactly one of the following assertions holds:

1. (i)

There exists an ${S}$-tree over $\mathcal{F}$. 2. (ii)

There exists an orientation of ${S}$ that avoids $\mathcal{F}$.

We remark that, by Lemma 2.3, the $S$-tree in (i) can be chosen irredundant, in which case it will be order-respecting by Lemma 2.1.

For our proof of Theorem 3.1 we need the following simple lemma, whose proof uses the fact that every orientation of a finite tree has a sink. To find one, just follow a maximal directed path.

Lemma 3.2**.**

Let $({\vec{S}},\leq\,,\!{}^{*})$ be a separation system and $\mathcal{F}\subseteq 2^{\vec{S}}$. If there exists an ${S}$-tree over $\mathcal{F}$, then no orientation of ${S}$ avoids $\mathcal{F}$.

Proof.

Let $(T,\alpha)$ be an ${S}$-tree over $\mathcal{F}$, and let $O$ be an orientation of ${S}$. Let $t\in V(T)$ be a sink in the orientation of the edges of $T$ that $O$ induces via $\alpha$. Then $\alpha(\vec{F}_{t})\subseteq O$. Since $\alpha(\vec{F}_{t})\in\mathcal{F}$, as $(T,\alpha)$ is an $S$-tree over $\mathcal{F}$, this means that $O$ does not avoid $\mathcal{F}$. ∎

Before we launch into the proof of Theorem 3.1, let us sketch its idea. Our aim will be to find either an $\mathcal{F}$-avoiding orientation $O$ of $S$ or construct an $S$-tree over $\mathcal{F}$ to witness that no such orientation exists (by Lemma 3.2). If $\mathcal{F}$ contains any singleton sets \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, then every $\mathcal{F}$-avoiding orientation of $S$ must contain {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} rather than {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. We think of these {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as ‘forced’ by $\mathcal{F}$, and will apply induction on the number of separations in $S$ neither of whose orientations is forced.

In the induction step, we shall consider some such $s$, call it $s_{0}$, and see what happens if we force one of its orientations by adding either \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} or \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} to $\mathcal{F}$. Then the induction will give us either an orientation of $S$ that avoids one of these augmented $\mathcal{F}$, and hence also the original $\mathcal{F}$, or two $S$-trees, one over each augmented $\mathcal{F}$. If one of these is an $S$-tree even over the original $\mathcal{F}$, we are again done, so we assume not.

Then one of these $S$-trees contains a leaf associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, while the other contains a leaf associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}. Assume, for simplicity, that these are the only such leaves in their respective $S$-trees. We can then combine these two trees into a single $S$-tree over our original $\mathcal{F}$ by identifying those two leaves and then suppressing the identified node, completing the proof.

It will help, also with the more difficult proof of our strong duality theorem in Section 4, to visualize the outline above once more for the case when $S$ consists of separations of a graph $G$. Then $s_{0}$ will be a separation $\{A,B\}$ of $G$, and the two $S$-trees we obtain from the induction hypothesis will essentially be $S$-trees over $\mathcal{F}$ of the two sides of this separation, of the graphs $G[A]$ and $G[B]$. Only ‘essentially’, because they will each have one additional leaf, associated with $(A,B)$ or $(B,A)$, respectively. In the tree-decompositions naturally associated with these $S$-trees, these leaf nodes will correspond to the bag $B$ or the bag $A$, respectively. The rest of these $S$-trees will decompose the other side of $s_{0}=\{A,B\}$, the graph $G[A]$ or $G[B]$.

Proof of Theorem 3.1.

By Lemma 3.2, at most one of (i) and (ii) holds. We now show that at least one of them holds. Let

[TABLE]

Then any $\mathcal{F}$-avoiding orientation of $S$ must include $O^{-}$ as a subset. As $\mathcal{F}$ consists of stars, $O^{-}$ contains no degenerate separations.

If O^{-}\supseteq\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} for some $s\in S$, then $(T,\alpha)$ with $T=K_{2}$ and {\rm im}\,\alpha=\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is an $S$-tree over $\mathcal{F}$. So we may assume that $O^{-}$ is antisymmetric: a partial orientation of $S\smallsetminus D$, where $D$ is the set of degenerate elements of $S$. We apply induction on $|S\smallsetminus D|-|O^{-}|$ to show that, whenever $\mathcal{F}$ is such that $O^{-}$ is antisymmetric, either (i) or (ii) holds.

If $|S\smallsetminus D|=|O^{-}|$, then ${O^{-}\cup\vec{D}}$ is an orientation of all of ${S}$. If (ii) fails then ${O^{-}\cup\vec{D}}$ has a subset $\sigma\in\mathcal{F}$. As $\mathcal{F}$ consists of stars we have $\sigma\cap\vec{D}=\emptyset$, so $\sigma\subseteq O^{-}$. By definition of $O^{-}$, and since $O^{-}$ is antisymmetric, $\sigma$ is not a singleton set (though it may be empty). Let $T$ be a star of $|\sigma|$ edges with centre $t$, say, and let $\alpha$ map its oriented edges $(x,t)$ bijectively to $\sigma$. Then $(T,\alpha)$ satisfies (i).

We may thus assume that ${S\smallsetminus D}$ contains a separation $s_{0}$ such that neither {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} nor {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}{} is in ${O^{-}}$. Let

[TABLE]

and put O^{-}_{i}:=\{\,{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\mid\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\in\mathcal{F}_{i}\,\} for $i=1,2$. Note that $|O^{-}_{i}|>|O^{-}|$, and $O^{-}_{i}$ is again antisymmetric.

Since any $\mathcal{F}_{i}$-avoiding orientation of $S$ also avoids $\mathcal{F}$, we may assume for both $i=1,2$ that no orientation of $S$ avoids $\mathcal{F}_{i}$. By the induction hypothesis, there are ${S}$-trees $(T_{i},\alpha_{i})$ over $\mathcal{F}_{i}$. Unless one of these is in fact over $\mathcal{F}$, the tree $T_{1}$ has a leaf $x_{1}$ associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}, while $T_{2}$ has a leaf $x_{2}$ associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}. Use Lemma 2.3 to prune the $(T_{i},\alpha_{i})$ to irredundant $S$-trees $(T^{\prime}_{i},\alpha^{\prime}_{i})$ over $\mathcal{F}_{i}$ containing $x_{i}$ and $e_{x_{i}}$. Suppose first that $s_{0}$ has no trivial orientation. Then Lemma 2.2 implies that $x_{1}$ and $x_{2}$ are the only leaves of $T^{\prime}_{1}$ and $T^{\prime}_{2}$ associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} and \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, respectively.

Let $T$ be the tree obtained from the disjoint union of $T^{\prime}_{1}-x_{1}$ and $T^{\prime}_{2}-x_{2}$ by joining the neighbour $y_{1}$ of $x_{1}$ in $T^{\prime}_{1}$ to the neighbour $y_{2}$ of $x_{2}$ in $T^{\prime}_{2}$. Let $\alpha\colon\vec{E}(T)\to{\vec{S}}$ map $(y_{1},y_{2})$ to {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and $(y_{2},y_{1})$ to {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}{} and otherwise extend $\alpha^{\prime}_{1}$ and $\alpha^{\prime}_{2}$. Then $\alpha^{\prime}_{1}(y_{1},x_{1})=\alpha(y_{1},y_{2})=\alpha^{\prime}_{2}(x_{2},y_{2})$, so $\alpha$ maps the oriented stars of edges at $y_{1}$ and $y_{2}$ to the same multistars of separations in ${\vec{S}}$ as $\alpha^{\prime}_{1}$ and $\alpha^{\prime}_{2}$ did. The stars they induce lie in $\mathcal{F}$, so $(T,\alpha)$ is an ${S}$-tree over $\mathcal{F}$.

Suppose now that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, say, is trivial. Then {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} is nontrivial, and $x_{1}$ is the only leaf of $T^{\prime}_{1}$ associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}, by Lemma 2.2 as before. Let $x_{2}^{1},\dots,x_{2}^{n}$ be the leaves of $T^{\prime}_{2}$ associated with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, and let $T$ be obtained from the union of $T^{\prime}_{2}-\{x_{2}^{1},\dots,x_{2}^{n}\}$ with $n$ copies of $T^{\prime}_{1}-x_{1}$ by joining, for all $i=1,\dots,n$, the neighbour $y_{1}^{i}$ of $x_{1}$ in the $i$th copy of $T^{\prime}_{1}-x_{1}$ to the neighbour $y_{2}^{i}$ of $x_{2}^{i}$ in $T^{\prime}_{2}$. Define $\alpha\colon\vec{E}(T)\to{\vec{S}}$ as earlier, mapping $(y_{1}^{i},y_{2}^{i})$ to {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and $(y_{2}^{i},y_{1}^{i})$ to {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} for all $i$, and otherwise extending $\alpha^{\prime}_{1}$ and $\alpha^{\prime}_{2}$. ∎

4 Strong duality

Theorem 3.1, alas, has a serious shortcoming: there are few, if any, sets $S$ and $\mathcal{F}\subseteq 2^{\vec{S}}$ such that $\mathcal{F}$ consists of stars in ${\vec{S}}$ and the $\mathcal{F}$-avoiding orientations of $S$ (all of them) capture an interesting notion of highly cohesive substructure found in the wild. The reason for this is that we are not, so far, requiring these orientations $O$ to be consistent: we allow that $O$ contains separations {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} when {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, which will not usually be the case when $O$ is induced by a meaningful highly cohesive substructure in the way discussed earlier. (We cannot simply add such sets \{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} to $\mathcal{F}$, since in order to be able to use Lemma 2.2 we must assume that $\mathcal{F}$ consists of stars of separations.)

So what happens if we strengthen (ii) so as to ask for a consistent orientation of $S$? Let us call a consistent $\mathcal{F}$-avoiding orientation of $S$ an $\mathcal{F}$-tangle. Since all consistent orientations of $S$ will contain all trivial {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}}, we may then add all co-trivial singletons \{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} to $\mathcal{F}$ without impeding the existence of an $\mathcal{F}$-tangle; this might help us find an $S$-tree over $\mathcal{F}$ if no such orientation exists.

Still, our proof breaks down as early as the induction start: we now also have to ask that $O^{-}$ – indeed, $O^{-}\cup\vec{D}$ – should be consistent. It is not even unnatural to ensure that $O^{-}$ is closed down in $({\vec{S}},\leq)$ (which implies consistency), by requiring that if \{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\in\mathcal{F} and {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} then also \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\in\mathcal{F}. For if a singleton star \{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} is in $\mathcal{F}$, the idea is that the part of our structure to which {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} points is too small to contain a highly cohesive substructure; and then the same should apply to all {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} with {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}.

But now we have a problem at the induction step: when forming the $\mathcal{F}_{i}$, we now have to add not only \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} or \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}{}\} to $\mathcal{F}$, but all singleton stars \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} with {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} or {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}{}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, respectively, to keep the $O_{i}^{-}$ closed down. This, then, spawns more problems: now both $T_{i}$ can have many leaves associated with a singleton star of $\mathcal{F}_{i}$ that is not in $\mathcal{F}$. Even if each of these occurs at most once, there is no longer an obvious way of how to merge $T_{1}$ and $T_{2}$ into a single $S$-tree over $\mathcal{F}$.

We shall deal with this problem as follows. Rather than adding singletons of the orientations of some fixed separation $s_{0}$ to $\mathcal{F}$ to form the $\mathcal{F}_{i}$ for $i=1,2$, we shall provisionally add, separately for $i=1,2$, some \{{\mathop{\kern 0.0ptr_{i}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} such that {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is minimal in ${\vec{S}}\smallsetminus(O^{-}\cup\vec{D})$. This will most likely mean that $r_{1}\neq r_{2}$. But \{{\mathop{\kern 0.0ptr_{i}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} can be associated with at most one leaf of $(T^{\prime}_{i},\alpha^{\prime}_{i})$, because {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\notin O^{-} will still be nontrivial (cf. Lemma 2.2).

If $r_{1}\neq r_{2}$, however, we shall no longer be able to combine the two $S$-trees over \mathcal{F}\cup\{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} and \mathcal{F}\cup\{{\mathop{\kern 0.0ptr_{2}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} into a single $S$-tree over $\mathcal{F}$, as we did in the proof of the weak duality theorem: this step hinged on the fact that the two oriented separations whose singleton sets we added to $\mathcal{F}$ were orientations of the same separation $s_{0}$, which enabled us to think of each of these $S$-trees as decomposing one ‘side’ of $s_{0}$. (We illustrated this for the case of graph separations, where for $s_{0}=\{A,B\}$ the two $S$-trees decomposed the two sides $A$ and $B$ of the graph separately and could thus be joined into a single decomposition of the entire graph.) However, we shall be able to adapt that idea as follows.

Our two separations {\mathop{\kern 0.0ptr_{i}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} will be chosen nested, so that both point to some $s_{0}$ between them. We shall then modify the two $S$-trees over the \mathcal{F}_{i}=\mathcal{F}\cup\{{\mathop{\kern 0.0ptr_{i}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} into $S$-trees over \mathcal{F}\cup\{\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\} and \mathcal{F}\cup\{\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\}, respectively, by ‘shifting’ the separations to which they map their edges to either side of $s_{0}$, and then merge these shifted $S$-trees as before to obtain one over $\mathcal{F}$.

To illustrate this, let us again consider the case that $S$ consists of separations of a graph, with $s_{0}=\{A,B\}$ say. We shall have to turn the separations of $G$ to which the first $S$-tree maps its edges into separations essentially of $G[A]$ (though formally still of $G$), those for the other $S$-tree into separations essentially of $G[B]$ (but formally still of $G$). The modified separations decomposing $G[A]$ will be nested since they came from the first $S$-tree, and will in addition be nested with $\{A,B\}$ (by definition: this is the result of shifting them). Likewise, the modified separations decomposing $G[B]$ will be nested with each other and also with $\{A,B\}$. And the separations from the first collection cannot cross those from the second because they ‘lie on’ different sides of $\{A,B\}$. (In abstract terms: {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} will point to the first lot, and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} to the second.) So the union of these two nested sets of separations, one from each of the two $S$-trees provided by the induction hypothesis, will also be nested, and in addition nested with $\{A,B\}$. The two $S$-trees from which they come can thus be merged along the (unique) edge mapping to $\{A,B\}$ into a single $S$-tree over $\mathcal{F}$.

Let us now define this shifting operation. Consider a separation system $({\vec{S}},\leq\,,\!{}^{*})$ contained in some universe of separations $({\vec{U}},\leq\,,\!{}^{*},\vee,\wedge)$, the ordering and involution on ${\vec{S}}$ being induced by those of ${\vec{U}}$. Let $(T,\alpha)$ be an $S$-tree rooted at a leaf $x$, and let {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} be nondegenerate and such that \alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Then let \alpha^{\prime}=\alpha_{x,{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.21529pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}\colon\vec{E}(T)\to{\vec{U}} be defined by setting

[TABLE]

and letting \alpha^{\prime}({\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}):=\alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)^{*} for these \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt. For example, we have \alpha^{\prime}({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})={\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and \alpha^{\prime}({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})={\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}.

Figure 5 illustrates this for the case that $S$ consists of separations of a graph $G$. If {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=(A,B), say, we would like to turn the separation $s$ of $G$ into a separation ‘essentially’ of $G[B]$, the side of $G$ to the right of $s_{0}$ in the picture, that is nested with $s_{0}$. There are two candidates for such a shift of $s$: the separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, which is indeed chosen, but potentially also the separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. As long as $s$ is considered without a default orientation, it will not be clear which of these should be its shift. But, fortunately, $s$ has a default orientation in our scenario: we need shifts only of separations $s\in S$ that label an edge of our $S$-tree rooted at $x$; and every such $s\in S$ can be oriented by default, as {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) with \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in T pointing away from $x$, say. This is the orientation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} of $s$ whose supremum with {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} we use in our definition of $\alpha^{\prime}$, a choice which determines the shift not only of {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} but also of {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}.

Informally, we think of a separation \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) as what becomes of \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) when we ‘shift it across’ {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. If \alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) is nontrivial and $(T,\alpha)$ is irredundant, which will usually be the case, then these shifts \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\mapsto\alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) can be expressed by a map purely between separations in ${\vec{S}}$, without any reference to $(T,\alpha)$. Our next aim is to define such a map.

Consider, instead of {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\vec{E}(T) or \alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}), an arbitrary nondegenerate and nontrivial separation {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}}, and pick {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} with {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as before. As {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is nontrivial and nondegenerate, so is {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Let S_{\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}} be the set of all separations $s\in S$ that have an orientation {{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}. (Note that if {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}), with $(T,\alpha)$ an irredundant $S$-tree over stars rooted at $x$, then {\vec{S}}_{\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}} includes $\alpha(\vec{E}(T))$ by Lemma 2.1.) Since {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is nontrivial, only one of the two orientations {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} of every s\in S_{\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}\smallsetminus\{r\} satisfies {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Letting

[TABLE]

for all {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in {\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} thus defines a map {\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\to{\vec{U}}, the shifting map f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, (Fig. 5, right). Note again that f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})={\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, since {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, and hence f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})={\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. In the case of {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, with $(T,\alpha)$ an irredundant $S$-tree over stars rooted at $x$, we then have

[TABLE]

since for every edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T) oriented away from $x$ we have \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\in{\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, by Lemmas 2.1 and 2.2, so f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, maps \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) to \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) and \alpha({\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)^{*} to its inverse \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)^{*}=\alpha^{\prime}({\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}).

Our aim will be to show that $(T,\alpha^{\prime})$ is another $S$-tree, and order-respecting if $(T,\alpha)$ is. But we will need some assumptions to ensure this.

Lemma 4.1**.**

Let $(T,\alpha)$ be an order-respecting $S$-tree, rooted at some leaf $x$, and let {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} be nondegenerate and such that \alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Assume that \alpha^{\prime}=\alpha_{x,{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.21529pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}} maps $\vec{E}(T)$ to ${\vec{S}}$. If {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is nontrivial in ${\vec{S}}$, or if the supremum in ${\vec{U}}$ of two separations that are trivial in ${\vec{S}}$ is never degenerate, then $(T,\alpha^{\prime})$ is an order-respecting $S$-tree.

Proof.

By definition, $\alpha^{\prime}$ commutes with the involutions on $\vec{E}(T)$ and ${\vec{S}}$. Let us show that if $(T,\alpha)$ is order-respecting then $\alpha^{\prime}$, too, respects the ordering of $\vec{E}(T)$. Consider edges \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt<\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt in $\vec{E}(T)$. Suppose first that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt. Then \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)=\alpha(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt), as desired. We now assume that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\not\leq\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt.

If {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptf}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} then {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptf}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, which reduces to the case above on renaming {\mathop{\kern 0.0ptf}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} as \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt and {\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} as \kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt. We may thus assume that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\not\leq{\mathop{\kern 0.0ptf}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, so that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt.

Now \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)=(\alpha({\mathop{\kern 0.0pte}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})^{*}=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\land{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\leq\alpha(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\leq\alpha(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0ptf}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) by (2).

In order for $(T,\alpha^{\prime})$ to be an $S$-tree it remains to show that \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) is nondegenerate for every \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T). But if \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is degenerate, it is distinct from both \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} (which are nondegenerate by assumption) and hence strictly greater than these. But then \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} are both trivial in ${\vec{S}}$, so their supremum \alpha^{\prime}(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) is nondegenerate by assumption. ∎

In the premise of Lemma 4.1 we assumed that $\alpha^{\prime}$ maps $\vec{E}(T)$ to ${\vec{S}}$. Let us now define some conditions that ensure this. Let us say that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{U}} in ${\vec{S}}$ if {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and every {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} with {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} satisfies {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\vee{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}}. Applied with {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) for {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T), these conditions will help ensure that $\alpha^{\prime}$ takes $\vec{E}(T)$ to ${\vec{S}}$; see Lemma 4.2 below.

Finally, we need a condition on $\mathcal{F}$ to ensure that the shifts of multistars of separations associated with nodes of $T$ are not only again multistars but also induce stars in $\mathcal{F}$. Given any set $\mathcal{F}\subseteq 2^{\vec{U}}\!$ of stars, let us say that a separation {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{U}} in ${\vec{S}}$ for $\mathcal{F}$ if {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ and for any star \sigma\subseteq{\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} in $\mathcal{F}$ that has an element {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} we also have f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,(\sigma)\in\mathcal{F}.222In fact, we could make do with less: that f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, is defined (with image in ${\vec{S}}$, as now) only on some symmetric subset of {\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}} that contains $\bigcup\mathcal{F}$, and that for some fixed $S^{\prime}\subseteq S$ and every $\sigma$ as above we have f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,(\sigma)\cap{\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\mathcal{F} if f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\in{\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} for the unique {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in $\sigma$. Informally, we need not insist that our $S$-trees over $\mathcal{F}$ shift to $S$-trees over $\mathcal{F}$ in their entirety, as long as those shifts of their separations \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt) that lie in some specified set {\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\subseteq{\vec{S}} contain an $S$-tree over $\mathcal{F}$ between them. It will be easy to adapt the proof of Theorem 4.3 should this ever be necessary. We shall get back to this right at the end, in Section 5.

We now have all the ingredients needed to shift a suitably prepared $S$-tree:

Lemma 4.2**.**

Let $\mathcal{F}\subseteq 2^{\vec{U}}\!$ be a set of stars. Let $(T,\alpha)$ be a tight and irredundant ${S}$-tree over $\mathcal{F}$ with at least one edge, rooted at a leaf $x$. Assume that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}:=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) is nontrivial and nondegenerate, let {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} emulate {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$, and consider \alpha^{\prime}:=\alpha_{x,{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.21529pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}. Then $(T,\alpha^{\prime})$ is an order-respecting $S$-tree over \mathcal{F}\cup\{\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\}, in which \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is a star associated with $x$ but with no other leaf of $T$.

Proof.

By Lemma 2.1, $(T,\alpha)$ is order-respecting. For every \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\in\vec{E}(T) with {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt we therefore have \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\geq\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, as well as \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt)\in{\vec{S}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} by Lemma 2.2. As {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$, these two facts imply that $\alpha^{\prime}$ takes $\vec{E}(T)$ to ${\vec{S}}$. As {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}=\alpha({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}) is nontrivial and nondegenerate, {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is nondegenerate and nontrivial. By Lemma 4.1, therefore, $(T,\alpha^{\prime})$ is an order-respecting $S$-tree.

By definition of $\alpha^{\prime}$, we have \alpha^{\prime}({\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Hence, $x$ is associated in $(T,\alpha^{\prime})$ with \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, which is a star since $s_{0}$ is nondegenerate. The other nodes $t$ of $T$ are associated in $(T,\alpha^{\prime})$ with stars in $\mathcal{F}$ because {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} for $\mathcal{F}$: note that $\alpha$ maps the stars $\vec{F}_{t}$ to {\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} by Lemmas 2.1, 2.2 and 2.4, that $\alpha(\vec{F}_{t})$ contains a separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} since $\vec{F}_{t}$ contains an edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt\geq{\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, and that \alpha^{\prime}=f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,\circ\ \alpha by (3).

Suppose finally that \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is associated in $(T,\alpha^{\prime})$ also with another leaf $y\neq x$ of $T$, with incident edge $(t,y)$ say. Let \{{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} be associated with $y$ in $(T,\alpha)$. Then {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} since $(T,\alpha)$ is order-respecting. If {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} then $\alpha^{\prime}=\alpha$, so \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}=\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is associated with $x$ and $y$ also in $(T,\alpha)$, contradicting Lemma 2.2. Hence

[TABLE]

where the first ‘=’ holds by definition of $y$, and the third by the definition of {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and of $\alpha^{\prime}$ based on the fact that {\mathop{\kern 0.0pte_{x}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq(t,y). Thus, $s_{0}$ witnesses that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial in ${\vec{S}}$, contrary to assumption. ∎

Let us say that $\mathcal{F}$ forces the separations {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} for which \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\in\mathcal{F}. We say that ${\vec{S}}$ is $\mathcal{F}$-separable if for every two nontrivial and nondegenerate {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in{\vec{S}} that are not forced by $\mathcal{F}$ and satisfy {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} there exists an $s_{0}\in S$ with an orientation {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} that emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$ and such that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$. As earlier, any such $s_{0}$ also be nondegenerate and have no trivial orientation.333However it can happen that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\ (\leq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}). Then {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as well as, by assumption, {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, so the nontriviality of {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} implies that $r=r^{\prime}$. Then {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} with equality in both cases, giving {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}.

We can now strengthen our weak duality theorem so as to yield consistent orientations, provided that ${\vec{S}}$ is $\mathcal{F}$-separable. Recall that for a separation system $({\vec{S}},\leq\,,\!{}^{*})$ and a set $\mathcal{F}$, an orientation $O$ of $S$ is called an $\mathcal{F}$-tangle if it is consistent and avoids $\mathcal{F}$, that is, if $2^{O}\cap\mathcal{F}=\emptyset$. Let us call $\mathcal{F}$ standard for ${\vec{S}}$ if it forces all {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{\vec{S}} that are trivial in ${\vec{S}}$, i.e., contains the singleton stars \{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} of their inverses.

Theorem 4.3** (Strong Duality Theorem).**

Let $({\vec{U}},\leq\,,\!{}^{*},\vee,\wedge)$ be a universe of separations containing a finite separation system $({\vec{S}},\leq\,,\!{}^{*})$. Let $\mathcal{F}\subseteq 2^{\vec{U}}\!\!$ be a set of stars, standard for ${\vec{S}}$. If ${\vec{S}}$ is $\mathcal{F}$-separable, exactly one of the following assertions holds:

1. (i)

There exists an $\mathcal{F}$-tangle of $S$. 2. (ii)

There exists an ${S}$-tree over $\mathcal{F}$.

We remark that, by Lemma 2.3, the $S$-tree in (ii) can be chosen irredundant, in which case it will be order-respecting (Lemma 2.1).

Proof.

Since replacing $\mathcal{F}$ with $\mathcal{F}\cap 2^{\vec{S}}$ leaves the validity of both (i) and (ii) unchanged we may, and shall, assume that $\mathcal{F}\subseteq 2^{\vec{S}}$. By Lemma 3.2, (i) and (ii) cannot both hold; we show that at least one of them holds.

Since $\mathcal{F}$ is standard, the set

[TABLE]

of separations that $\mathcal{F}$ forces contains all the trivial separations in ${\vec{S}}$. But it contains no degenerate ones, because the \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\in\mathcal{F} are stars. If O^{-}\supseteq\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} for some $s\in S$, then $(T,\alpha)$ with $T=K_{2}$ and {\rm im}\,\alpha=\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is an $S$-tree over $\mathcal{F}$. We may thus assume that $O^{-}$ is antisymmetric: a partial orientation of $S\smallsetminus D$, where $D$ is the set of degenerate elements of $S$.

Let us show that $O^{-}$ is consistent. If not, then $O^{-}$ contains some {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} and {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} such that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. As $O^{-}$ is antisymmetric, it then does not contain their inverses {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. So $\mathcal{F}$ does not force these; in particular, they are nontrivial. Since ${\vec{S}}$ is $\mathcal{F}$-separable, there exists an $s_{0}\in S$ with orientations {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} such that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$ and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$. Since $r$ is not degenerate and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} for $\mathcal{F}$, the singleton star \{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\in\mathcal{F} shifts to \{f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\}=\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\in\mathcal{F}, so {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in{O^{-}}. Likewise, since $r^{\prime}$ is not degenerate and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} we have {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in{O^{-}}. This contradicts our assumption that ${O^{-}}$ is antisymmetric.

Let us show that $O^{-}\cup\vec{D}$ is still consistent. Suppose $r\neq s$ are such that {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}},{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in O^{-}\cup\vec{D} and {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Then $r$ and $s$ are not both in $D$, since that would imply {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}={\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. Since $O^{-}$ is consistent, we may thus assume that {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in O^{-} and {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\vec{D} (or vice versa, which is equivalent by (1)). Then {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial, as {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. Hence {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in O^{-} as well as, by assumption, {\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in O^{-}. This contradicts our assumption that $O^{-}$ is antisymmetric.

Let $R$ be the set of separations in $S\smallsetminus D$ of which neither orientation lies in $O^{-}$. We shall apply induction on $|R|$ to show that (i) or (ii) holds whenever $\mathcal{F}$ is such that $O^{-}$ is antisymmetric. If $|R|=0$, then $O^{-}\cup\vec{D}$ is either itself an $\mathcal{F}$-tangle of $S$ or contains a star $\sigma\in\mathcal{F}$. Then $\sigma\subseteq O^{-}$, since stars have no degenerate elements. By definition of $O^{-}$, and since $O^{-}$ is antisymmetric, $\sigma$ is not a singleton subset of $O^{-}$ (though it may be empty). Let $T$ be a star of $|\sigma|$ edges with centre $t$, say, and let $\alpha$ map its oriented edges $(x,t)$ bijectively to $\sigma$. Then $(T,\alpha)$ satisfies (ii).

For the induction step, pick {\mathop{\kern 0.0ptr_{0}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\vec{R}. Then neither {\mathop{\kern 0.0ptr_{0}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} nor {\mathop{\kern 0.0ptr_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} lies in $O^{-}\cup\vec{D}$; let {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{0}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0ptr_{2}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} be minimal in ${\vec{S}}\smallsetminus(O^{-}\cup\vec{D})$. Then {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{0}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{2}}\limits^{\kern 1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. As $\mathcal{F}$ forces neither {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} nor {\mathop{\kern 0.0ptr_{2}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} and ${\vec{S}}$ is $\mathcal{F}$-separable, there exists an $s_{0}\in S\smallsetminus D$ with nontrivial orientations {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} such that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$ and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr_{2}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}$.

Since $O^{-}$ is antisymmetric, it does not contain both {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. Let us assume that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\notin O^{-}, i.e. that \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\notin\mathcal{F}. Then \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\notin\mathcal{F}, because {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} for $\mathcal{F}$ and f\!\!\downarrow\!{}^{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, maps the star \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\subseteq{\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} to \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}. Thus, {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} both lie outside $O^{-}\cup\vec{D}$, so $r_{1}\in R$.

We can now hope to apply the induction hypothesis to \mathcal{F}_{1}:=\mathcal{F}\cup\{\{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\}, because

[TABLE]

is again antisymmetric, and the set $R_{1}$ of separations in $S\smallsetminus D$ with neither orientation in $O^{-}_{1}$ is smaller than $R$. Also, $\mathcal{F}_{1}$ is a standard set of stars, because $\mathcal{F}$ is and $r_{1}\notin D$. But we still have to check that ${\vec{S}}$ is $\mathcal{F}_{1}$-separable.

To do so, consider (nontrivial and) nondegenerate separations {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\in{\vec{S}} not forced by $\mathcal{F}_{1}$ such that {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. We have to find an $s_{1}\in S$ with an orientation {\mathop{\kern 0.0pts_{1}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} that emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}_{1}$ and such that {\mathop{\kern 0.0pts_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} in ${\vec{S}}$ for $\mathcal{F}_{1}$. By assumption, there is such an $s_{1}\in S$ for {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} with respect to $\mathcal{F}$; let us take this $s_{1}$, with orientations {\mathop{\kern 0.0pts_{1}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}},{\mathop{\kern 0.0pts_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} such that {\mathop{\kern 0.0pts_{1}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} emulates {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} for $\mathcal{F}$ and {\mathop{\kern 0.0pts_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} emulates {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} for $\mathcal{F}$. We have to show that this emulation extends to $\mathcal{F}_{1}$, i.e., that for the unique star \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} in $\mathcal{F}_{1}\smallsetminus\mathcal{F}$ we have \{f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{1}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})\}\in\mathcal{F}_{1} if {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\neq{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} (so that \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\subseteq{\vec{S}}_{\raise 0.3014pt\vbox to0.0pt{\vss\hbox{\scriptstyle\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\smallsetminus\{{\mathop{\kern 0.0ptr}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}), and \{f\!\!\downarrow\!{}^{{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.04306pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{1}}\limits^{\kern 0.0pt\raise 0.30138pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\leftarrow$}\vss}}}}}}\,({\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}})\}\in\mathcal{F}_{1} if {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\leq{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\neq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. In either case, the image {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} of {\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} under the relevant map is either equal to {\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} (in which case we are done) or greater, by definition of the shift operator$\,\downarrow$. If {\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, then $s\notin D$, since otherwise {\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} would be trivial and hence in $O^{-}$. And {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} by (1), so {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in O^{-}\cup\vec{D} by the minimality of {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} in ${\vec{S}}\smallsetminus(O^{-}\cup\vec{D})$. Thus {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in O^{-}, and hence {\{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\in\mathcal{F}\subseteq\mathcal{F}_{1}}, by the definition of $O^{-}$. This completes our proof that ${\vec{S}}$ is $\mathcal{F}_{1}$-separable.

We can thus apply the induction hypothesis to $\mathcal{F}_{1}$. If it returns an $\mathcal{F}_{1}$-tangle of $S$, then this is our desired $\mathcal{F}$-tangle. So we may assume that it returns an $S$-tree $(T_{1},\alpha)$ over $\mathcal{F}_{1}$. If this $S$-tree is even over $\mathcal{F}$, our proof is complete. We may thus assume that $T_{1}$ has a leaf $x_{1}$ associated with \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}. We now apply Lemma 2.4 to prune and contract $(T_{1},\alpha)$ to a tight and irredundant $S$-tree over $\mathcal{F}_{1}$ that still contains $x_{1}$. In this $S$-tree, which for simplicity we continue to call $(T_{1},\alpha)$, no leaf other than $x_{1}$ is associated with \{{\mathop{\kern 0.0ptr_{1}}\limits^{\kern-1.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} (Lemma 2.2). Let \alpha^{\prime}_{1}:=\alpha_{x_{1},{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.21529pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}. By Lemma 4.2, $(T_{1},\alpha^{\prime}_{1})$ is an $S$-tree over \mathcal{F}\cup\{\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\}, in which the star \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} is associated with $x_{1}$ but with no other leaf of $T_{1}$. All the other nodes of $T_{1}$ are therefore associated with stars in $\mathcal{F}$.

If \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\in\mathcal{F}, then $(T_{1},\alpha^{\prime}_{1})$ is in fact an $S$-tree over $\mathcal{F}$, completing our proof. We may thus assume that \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}\notin\mathcal{F}, or equivalently that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\notin O^{-}. We can now use the induction hypothesis exactly as above (where we assumed that {\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\notin O^{-}), considering {\mathop{\kern 0.0ptr_{2}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}} in the same way as we just treated {\mathop{\kern 0.0ptr_{1}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, to obtain an irredundant $S$-tree $(T_{2},\alpha^{\prime}_{2})$ over \mathcal{F}\cup\{\{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}\} in which \{{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\} is associated with a unique leaf $x_{2}$, and all the other nodes are associated with stars in $\mathcal{F}$.

These trees can now be combined to the desired $S$-tree $(T,\alpha^{\prime})$ over $\mathcal{F}$ as in the proof of Theorem 3.1: add to the disjoint union $(T_{2}-x_{2})\cup(T_{1}-x_{1})$ the edge $y_{2}y_{1}$ between the neighbour $y_{2}$ of $x_{2}$ in $T_{2}$ and the neighbour $y_{1}$ of $x_{1}$ in $T_{1}$, put \alpha^{\prime}(y_{2},y_{1}):={\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} and \alpha^{\prime}(y_{1},y_{2}):={\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}, and otherwise let $\alpha^{\prime}$ extend $\alpha^{\prime}_{1}$ and $\alpha^{\prime}_{2}$. ∎

5 Essential $S$-trees and $\mathcal{F}$-tangles: a refinement

Let us return to the question of how much of a restriction is our condition in the premise of the strong duality theorem that $\mathcal{F}$ must be standard for ${\vec{S}}$, i.e., contain all co-trivial singletons, the stars \{{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\} for which {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial in ${\vec{S}}$. As noted before, any consistent orientation of $S$, and hence any $\mathcal{F}$-tangle, will contain all trivial separations and hence avoid all these singletons. So adding them to $\mathcal{F}$ will not change the set of $\mathcal{F}$-tangles.

But neither would removing them. Which thus seems like a good idea, if only to avoid unnecessary clutter.

Removing the co-trivial singletons from $\mathcal{F}$ would, however, have an impact on the set of $S$-trees over $\mathcal{F}$. The leaves of an $S$-tree over a standard $\mathcal{F}$ can be associated with any co-trivial singleton star, but if we remove these stars from $\mathcal{F}$ then such an $S$-tree will no longer be over $\mathcal{F}$.

We might try to repair this by removing those leaves from our $S$-tree $(T,\alpha)$. The edge which such a leaf sends to its neighbour $t$, however, maps to a separation {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} that would then be missing from the star $\alpha(\vec{F}_{t})\in\mathcal{F}$ associated with $t$, perhaps knocking it out of $\mathcal{F}$. But as {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial, its membership in $\alpha(\vec{F}_{t})$ will not be the reason why we put $\alpha(\vec{F}_{t})$ in $\mathcal{F}$ in the first place: if the role of an $S$-tree over $\mathcal{F}$ is to witness the nonexistence of an $\mathcal{F}$-tangle, then only the nontrivial separations in its stars are essential for that role. So let’s try to delete all trivial separations from stars in $\mathcal{F}$, and see if we can retain an $S$-tree over the modified $\mathcal{F}$.

Given a separation system $({\vec{S}},\leq\,,\!{}^{*})$ and $\mathcal{F}\subseteq 2^{\vec{S}}$, define the essential core $\mathcal{F}^{\prime}$ of $\mathcal{F}$ as

[TABLE]

where ${\vec{S}}^{-}\subseteq{\vec{S}}$ is the set of all separations that are trivial in ${\vec{S}}$. Note that if $\mathcal{F}$ is standard then so is $\mathcal{F}^{\prime}$, since inverses of trivial separations are never trivial. Let us call an $S$-tree $(T,\alpha)$ essential if it is irredundant, tight, and $\alpha(\vec{E}(T))$ contains no trivial separation.

Theorem 5.1**.**

[14]* Let $({\vec{S}},\leq\,,\!{}^{*})$ be a separation system and $\mathcal{F}\subseteq 2^{\vec{S}}\!$ a set of stars.*

1. (i)

The $\mathcal{F}^{\prime}$-tangles of $S$ are precisely its $\mathcal{F}$-tangles. 2. (ii)

If $(T,\alpha)$ is any $S$-tree over $\mathcal{F}$, there is an essential $S$-tree $(T^{\prime},\alpha^{\prime})$ over $\mathcal{F}^{\prime}$ such that $T^{\prime}$ is a minor of $T$ and $\alpha^{\prime}=\alpha\!\restriction\!\vec{E}(T^{\prime})$. Conversely, from any essential $S$-tree over $\mathcal{F}^{\prime}$ we can obtain an $S$-tree over $\mathcal{F}$ by adding leaves, if $\mathcal{F}$ is standard for ${\vec{S}}$.

Proof.

(i) is immediate from the fact that $\mathcal{F}$-tangles, being consistent, contain all trivial separations and hence also avoid $\mathcal{F}^{\prime}$.

For the first statement in (ii), let us start by making the given $S$-tree $(T,\alpha)$ irredundant by pruning it, as in Lemma 2.3. We then contract edges violating tightness, as explained before Lemma 2.4. We finally make the resulting tree essential by deleting all its edges that $\alpha$ maps to trivial separations. This can be done recursively by deleting leaves associated with a co-trivial singleton: since $\mathcal{F}$ consists of stars, Lemma 2.1 implies that any $S$-tree over $\mathcal{F}$ with an edge mapping to a trivial separation will also have such an edge issuing from a leaf. (Recall that if {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} is trivial then so is every {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}.) Pruning off leaves recursively in this way will leave a well-defined tree at the end, which has the properties desired for $(T^{\prime},\alpha^{\prime})$.

For the second statement in (ii), let $(T,\alpha)$ be an essential $S$-tree over $\mathcal{F}^{\prime}$, and consider a node $t\in T$. As $\alpha(\vec{F}_{t})\in\mathcal{F}^{\prime}$, there exists $\sigma\in\mathcal{F}$ such that $\sigma\smallsetminus\alpha(\vec{F}_{t})$ is a set of trivial separations {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. For each of these add a new leaf, joining it to $t$ by an edge \kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt with \alpha(\kern-1.0pt{\mathop{\kern 0.0pte}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\kern-1.0pt):={\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}. ∎

Theorem 5.1 allows us to strengthen each of the two alternatives in the strong duality theorem from its current version with the given set $\mathcal{F}$ of stars to an ‘essential’ version with $\mathcal{F}^{\prime}$ instead. So why didn’t we prove this stronger version directly?

The answer is pragmatic: this would have been possible, and we shall indicate in a moment how to do it. But it would have made the proof notationally more technical. As the proof stands, we need to allow inessential $S$-trees, because they can arise in the induction step when we combine two shifted $S$-trees, even if these were essential before the shift.

Indeed, recall what happens to the leaves of an $S$-tree when we shift it, by f\!\!\downarrow\!{}^{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, say. A leaf, associated with \{{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\}, say, where $r\neq r^{\prime}\neq s_{0}$ for simplicity, will be associated in the shifted tree with the star \{{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}\lor{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\}, because {\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\leq{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}. But if {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}<{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, as will frequently happen, then this star is a co-trivial singleton, because {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\land{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}} as well as {\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\land{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0pts_{0}}\limits^{\kern 0.0pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}<{\mathop{\kern 0.0ptr\lower 0.5pt\hbox{{}{}^{\prime}}}\limits^{\kern 0.0pt\raise 0.06029pt\vbox to0.0pt{\hbox{\scriptscriptstyle\leftarrow}\vss}}}.

The way to overcome this problem is indicated in Footnote 2, with $S^{\prime}$ the set of separations in $S$ that have no trivial orientation in ${\vec{S}}$. When we shift a star $\sigma\in\mathcal{F}$, its shift may contain trivial separations, but we could simply delete these to make the shifted $S$-tree essential, as in the proof of Theorem 5.1(ii). To ensure that it is again over $\mathcal{F}$, we would need to replace the current requirement in the definition of $\mathcal{F}$-separable, that f\!\!\downarrow\!{}^{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\, should map to $\mathcal{F}$ any $\sigma\in\mathcal{F}$ that contains a separation {\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\geq{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.60275pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}, with the requirement that for any such $\sigma\in\mathcal{F}$ we have f\!\!\downarrow\!{}^{{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,(\sigma)\cap{\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}\in\mathcal{F} if f\!\!\downarrow\!{}^{\!{\mathop{\kern 0.0ptr}\limits^{\kern 1.5pt\raise 0.43054pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}}_{\raise 0.03012pt\vbox to0.0pt{\vss\hbox{\scriptstyle{\mathop{\kern 0.0pts_{0}}\limits^{\kern 1.0pt\raise 0.1507pt\vbox to0.0pt{\hbox{$\scriptscriptstyle\rightarrow$}\vss}}}}}}\,({\mathop{\kern 0.0pts}\limits^{\kern 1.5pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}})\in{\mathop{\kern 0.0ptS\lower-1.0pt\hbox{{}{}^{\prime}}}\limits^{\kern 2.0pt\raise 0.3014pt\vbox to0.0pt{\hbox{\scriptscriptstyle\rightarrow}\vss}}}.

Acknowledgement

We enjoyed a visit by Frédéric Mazoit to the University of Hamburg in the summer of 2013, when he gave a series of lectures on the bramble-based duality theorems for width parameters developed in [1, 35]. We then tried, unsuccessfully, to apply these to obtain a duality theorem for $k$-blocks. This triggered the development of a similar unified duality theory for tangles instead of brambles, first in graphs and later more generally. Our original goal, a duality theorem for $k$-blocks, was eventually achieved in [16], as an application of Theorem 4.3.

The second author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Science, ICT & Future Planning (2011-0011653) and also by the Institute for Basic Science (IBS-R029-C1).