The Species Problem from the Modeler’s Point of View

Abstract

How to define a partition of individuals into species is a long-standing question called the species problem in systematics. Here, we focus on this problem in the thought experiment where individuals reproduce clonally and both the differentiation process and the population genealogies are explicitly known. We specify three desirable properties of species partitions: (A) Heterotypy between species, (B) Homotypy within species and (M) Genealogical monophyly of each species. We then ask: How and when is it possible to delineate species in a way satisfying these properties? We point out that the three desirable properties cannot in general be satisfied simultaneously, but that any two of them can. We mathematically prove the existence of the finest partition satisfying (A) and (M) and the coarsest partition satisfying (B) and (M). For each of them, we propose a simple algorithm to build the associated phylogeny out of the genealogy. The ways we propose to phrase the species problem shed new light on the interaction between the genealogical and phylogenetic scales in modeling work. The two definitions centered on the monophyly property can readily be used at a higher taxonomic level as well, e.g., to cluster species into monophyletic genera.

Appendices

Appendix

Some of the results stated in Sects. A and B are classical results in combinatorics for partially ordered sets (see Bóna 2011, chapter 16). For the sake of self-containment and because all readers may not be familiar with these notions, we nevertheless expose them here.

A: ‘Finer than’, a partial order relation on $\mathcal {X}$-partitions

Definition 1

Let $\mathscr {S}_1$ and $\mathscr {S}_2$ be two $\mathcal {X}$-partitions. We say that $\mathscr {S}_1$ is finer than $\mathscr {S}_2$, and we write $\mathscr {S}_1 \le \mathscr {S}_2$ if $~\forall S_1 \in \mathscr {S}_1, \forall S_2 \in \mathscr {S}_2, ~ S_1 \cap S_2 \in \{ \varnothing , S_1 \}$.

We detail here the three criteria that make the ‘finer than’ relation a partial order on the set of $\mathcal {X}$-partitions.

Proof

One must check the reflexivity, antisymmetry and transitivity properties.

• Reflexivity. Take any $\mathcal {X}$-partition $\mathscr {S}$. Then for all $S_1, S_2 \in \mathscr {S}$ we either have $S_1 \cap S_2 = S_1$ if $S_1 = S_2$, or $S_1 \cap S_2 = \varnothing$ otherwise. It follows that $\mathscr {S} \le \mathscr {S}$.

• Antisymmetry. Take two $\mathcal {X}$-partitions denoted $\mathscr {S}_1$ and $\mathscr {S}_2$, verifying $\mathscr {S}_1 \le \mathscr {S}_2$ and $\mathscr {S}_2 \le \mathscr {S}_1$. Then for all $(S_1, S_2) \in \mathscr {S}_1 \times \mathscr {S}_2, ~ S_1 \cap S_2 \in \{ \varnothing , S_1 \}$ and $S_1 \cap S_2 \in \{ \varnothing , S_2 \}$.

  If $S_1 \cap S_2 \ne \varnothing$, it follows that $S_1 = S_2$, and finally $\mathscr {S}_1 = \mathscr {S}_2$.

• Transitivity. Take now three $\mathcal {X}$-partitions denoted $\mathscr {S}_1, \mathscr {S}_2, \mathscr {S}_3$, verifying $\mathscr {S}_1 \le \mathscr {S}_2$ and $\mathscr {S}_2 \le \mathscr {S}_3$. Let $S_1 \in \mathscr {S}_1$ and $S_3 \in \mathscr {S}_3$ and assume that $S_1\cap S_3\not =\varnothing$. Then, there is $x\in S_1\cap S_3$ and we let $S_2$ be the unique element of $\mathscr {S}_2$ such that $x\in S_2$. Thus $S_1\cap S_2\not =\varnothing$ and $S_2\cap S_3\not =\varnothing$, which implies by assumption that $S_2\cap S_1 = S_1$ and $S_2 \cap S_3= S_2$. So we see that $S_1\subseteq S_2\subseteq S_3$, so that $S_1\cap S_3=S_1$.

$\square$

B: Proof of Theorem 1

Here, we will consider sets of partitions verifying one or two desirable properties. Hence the following definitions

$$\begin{aligned} \varSigma _A&:= \{ \mathcal {X}\text {-partitions satisfying (A)} \}\\ \varSigma _B&:= \{ \mathcal {X}\text {-partitions satisfying (B)} \}\\ \varSigma _M&:= \{ \mathcal {X}\text {-partitions satisfying (M)} \}\\ \varSigma _{AM}&:=\{ \mathcal {X}\text {-partitions satisfying (AM)} \} = \varSigma _A \cap \varSigma _M \\ \varSigma _{BM}&:= \{ \mathcal {X}\text {-partitions satisfying (BM)} \} = \varSigma _B \cap \varSigma _M \\ \varSigma _{AB}&:= \{ \mathcal {X}\text {-partitions satisfying (AB)} \} = \varSigma _A \cap \varSigma _B \end{aligned}$$

We will see that the collection of $\mathcal {X}$-partitions $\varSigma _M$ plays a singular role in Theorem 1. This is due to the characterization of $\varSigma _M$ by the fact that there is a hierarchy $\mathscr {H}$ (here the hierarchy associated with the genealogy T) such that

$$\begin{aligned} \mathscr {S}\in \varSigma _M \Longleftrightarrow \mathscr {S} \subseteq \mathscr {H} . \end{aligned}$$

Also recall that the collections of $\mathcal {X}$-partitions $\varSigma _A$ and $\varSigma _B$ can be defined as follows

$$\begin{aligned}&\mathscr {S}\in \varSigma _A \Longleftrightarrow \forall P \in \mathscr {P},\ \forall S \in \mathscr {S},\ P\cap S\in \{\varnothing ,P\} \\&\mathscr {S}\in \varSigma _B \Longleftrightarrow \forall P \in \mathscr {P},\ \forall S \in \mathscr {S},\ P\cap S\in \{\varnothing ,S\}. \end{aligned}$$

In this section, we aim at giving a proof of Theorem 1 which can now be restated as follows

$$\begin{aligned}&\exists \mathscr {S}_\text {loose} \in \varSigma _{AM}, \text { such that } \forall \mathscr {S} \in \varSigma _{AM}, ~ \mathscr {S}_\text {loose} \le \mathscr {S}\\&\exists \mathscr {S}_\text {lacy} \in \varSigma _{BM}, \text { such that } \forall \mathscr {S} \in \varSigma _{BM}, ~ \mathscr {S} \le \mathscr {S}_\text {lacy} \end{aligned}$$

The proof is divided into two parts. First, given a set of partitions $\varSigma$ (resp. $\varSigma \subseteq \varSigma _M$), we prove the existence of the finest (resp. coarsest) partition finer (resp. coarser) than any element of $\varSigma$, which we call $\inf \varSigma$ (resp. $\sup \varSigma$). Second, we show that $\inf \varSigma _{AM} \in \varSigma _{AM}$ and $\sup \varSigma _{BM} \in \varSigma _{BM}$, hence yielding the definitions $\mathscr {S}_\text {loose} := \inf \varSigma _{AM}$ and $\mathscr {S}_\text {lacy} := \sup \varSigma _{BM}$.

1.1 B.1: Defining the Supremum and the Infimum of a Set of $\mathcal {X}$-partitions

Definition 2

For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions, we define the two relations $\underline{\mathcal {R}}_\varSigma$ and $\overline{\mathcal {R}}_\varSigma$ on $\mathcal {X}$ by

$$\begin{aligned} \forall (x,y) \in \mathcal {X}^2, ~~&x~ \underline{\mathcal {R}}_\varSigma ~y \Longleftrightarrow \forall \mathscr {S} \in \varSigma , ~ \exists S \in \mathscr {S}, ~ x \in S \text { and } y \in S \\&x ~\overline{\mathcal {R}}_\varSigma ~y \Longleftrightarrow \exists \mathscr {S} \in \varSigma , ~ \exists S \in \mathscr {S}, ~ x \in S \text { and } y \in S. \end{aligned}$$

Lemma 1

For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions, $\underline{\mathcal {R}}_\varSigma$ is an equivalence relation. For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$, $\overline{\mathcal {R}}_\varSigma$ is an equivalence relation.

Proof

The reflexivity and symmetry of the two relations are easily seen. Now let us prove their transitivity. Let $\varSigma$ be a non-empty collection of $\mathcal {X}$-partitions, and $(x,y,z) \in \mathcal {X}^3$ such that $x~ \underline{\mathcal {R}}_\varSigma ~y$ and $y~ \underline{\mathcal {R}}_\varSigma ~z$. Let $\mathscr {S} \in \varSigma$. By definition,

$$\begin{aligned}&\exists S_1 \in \mathscr {S}, ~ x \in S_1 \text { and } y \in S_1\\&\exists S_2 \in \mathscr {S}, ~ y \in S_2 \text { and } z \in S_2 \end{aligned}$$

It follows that $y \in S_1 \cap S_2$, and because $\mathscr {S}$ is a partition, $S_1 = S_2$. Finally, with $S:=S_1= S_2$, there exists $S \in \mathscr {S}$ such that $x \in S$ and $z \in S$, so that $x~ \underline{\mathcal {R}}_\varSigma ~z$ and we can conclude that $\underline{\mathcal {R}}_\varSigma$ is transitive.

Now let $\varSigma \subseteq \varSigma _M$ be a non-empty collection of $\mathcal {X}$-partitions and $(x,y,z) \in \mathcal {X}^3$ such that $x~ \overline{\mathcal {R}}_\varSigma ~y$ and $y~ \overline{\mathcal {R}}_\varSigma ~z$. By definition,

$$\begin{aligned}&\exists \mathscr {S}_1 \in \varSigma , ~ \exists S_1 \in \mathscr {S}_1, ~ x \in S_1 \text { and } y \in S_1\\&\exists \mathscr {S}_2 \in \varSigma , ~ \exists S_2 \in \mathscr {S}_2, ~ y \in S_2 \text { and } z \in S_2 \end{aligned}$$

Because $\varSigma \subseteq \varSigma _M$, $\mathscr {S}_1 \subseteq \mathscr {H}$ and $\mathscr {S}_2 \subseteq \mathscr {H}$, so that $S_1 \in \mathscr {H}$ and $S_2 \in \mathscr {H}$. From the definition of hierarchy, we get $S_1 \cap S_2 \in \{ \varnothing , S_1, S_2 \}$. Since $y \in S_1 \cap S_2$, we have $S_1 \cap S_2 \ne \varnothing$.

Suppose that $S_1 \cap S_2 = S_2$. It follows that $\exists \mathscr {S}_1 \in \varSigma , ~ \exists S_1 \in \mathscr {S}_1, ~x \in S_1 \text { and } z \in S_1$.

Suppose that $S_1 \cap S_2 = S_1$. It follows that $\exists \mathscr {S}_2 \in \varSigma , ~ \exists S_2 \in \mathscr {S}_2, ~x \in S_2 \text { and } z \in S_2$. So $x~ \overline{\mathcal {R}}_\varSigma ~z$ and we can conclude that $\overline{\mathcal {R}}_\varSigma$ is transitive. $\square$

Definition 3

For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions, we call $\inf \varSigma$ the $\mathcal {X}$-partition induced by the equivalence relation $\underline{\mathcal {R}}_\varSigma$. For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$, we call $\sup \varSigma$ the $\mathcal {X}$-partition induced by the equivalence relation $\overline{\mathcal {R}}_\varSigma$.

Readers familiar with lattice theory will note that these definitions match the usual ‘meet’ and ‘join’ operators used for lattices, and in particular the lattice of partitions of a set, ordered by refinement. For the other readers, recall first that any equivalence relation on a set $\mathcal {X}$ induces an $\mathcal {X}$-partition obtained by placing all elements in relation in one cluster. Further, the following lemma justifies the notation $\inf$ and $\sup$.

Lemma 2

Let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions. Then for any $\mathscr {S} \in \varSigma$, $\inf \varSigma \le \mathscr {S}$. Let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$. Then for any $\mathscr {S} \in \varSigma$, $\mathscr {S} \le \sup \varSigma$.

Proof

Let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions and $S\in \inf \varSigma$. Let also $\mathscr {S} \in \varSigma$ and $S'\in \mathscr {S}$. We need to prove that $S \cap S' \in \{ \varnothing , S \}$. Assume that $S\cap S'\not =\varnothing$ and $S\cap S'\not = S$. Then there is $x\in S\cap S'$ and $y\in S$ such that $y\not \in S'$. Because $x,y\in S$, by definition of $\inf \varSigma$, we have $x~ \underline{\mathcal {R}}_\varSigma ~ y$ and by definition of $\underline{\mathcal {R}}_\varSigma$, $\exists S'' \in \mathscr {S}, ~ x, y \in S''$. So $S'$ and $S''$ are both elements of $\mathscr {S}$ containing x, which implies that $S'=S''$ and contradicts $y\not \in S'$.

Now let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$ and $S\in \sup \varSigma$. Let also $\mathscr {S} \in \varSigma$ and $S'\in \mathscr {S}$. We need to prove that $S \cap S' \in \{ \varnothing , S' \}$. Assume that $S\cap S'\not =\varnothing$ and $S\cap S'\not = S'$. Then there is $x\in S\cap S'$ and $y\in S'$ such that $y\not \in S$. Because $x\in S$ and $y\not \in S$, by definition of $\sup \varSigma$, x and y are not in relation $\overline{\mathcal {R}}_\varSigma$ and by definition of $\overline{\mathcal {R}}_\varSigma$, either $x\not \in S'$ or $y\not \in S'$ and we get a contradiction. $\square$

Note that, in general, we can have $\inf \varSigma \notin \varSigma$ and $\sup \varSigma \notin \varSigma$. Here are two examples to provide the reader with some intuition.

Example 1

Take

$$\begin{aligned} \mathcal {X}&= \{ 1, 2, 3, 4 \} \\ \mathscr {S}&= \{ \{1\}, \{2\}, \{3,4\} \} \\ \mathscr {S}'&= \{ \{1,2\}, \{3\}, \{4\} \} \\ \varSigma&= \{ \mathscr {S}, \mathscr {S}' \} \end{aligned}$$

In this case, we get $\inf \varSigma = \{ \{1\}, \{2\}, \{3\}, \{4\} \}$, which does not belong to $\varSigma$. Moreover, if we define the hierarchy $\mathscr {H} := \{ \{1,2,3,4\}, \{1,2\}, \{3,4\}, \{1\}, \{2\}, \{3\}, \{4\} \}$, we have $\varSigma \subseteq \varSigma _M$, which allows us to consider $\sup \varSigma = \{ \{1,2\}, \{3,4\} \}$ which again does not belong to $\varSigma$.

Example 2

Take

$$\begin{aligned} \mathcal {X}&= \{ 1, 2, 3, 4 \}\\ \mathscr {S}&= \{ \{1, 3, 4\}, \{2\} \} \\ \mathscr {S}'&= \{ \{1,2\}, \{3,4\} \} \\ \varSigma&= \{ \mathscr {S}, \mathscr {S}' \} \end{aligned}$$

In this case, we get $\inf \varSigma = \{ \{1\}, \{2\}, \{3, 4\} \}$, which does not belong to $\varSigma$. Moreover, there is no $\mathcal {X}$-hierarchy $\mathscr {H}$ such that $\mathscr {S}, \mathscr {S}' \in \mathscr {H}$. Then we can see that the relation $\overline{\mathcal {R}}_\varSigma$ is not an equivalence relation on $\mathcal {X}$, because $1 ~\overline{\mathcal {R}}_\varSigma ~ 2$ and $1 ~\overline{\mathcal {R}}_\varSigma ~ 3$, but we do not have $2 ~\overline{\mathcal {R}}_\varSigma ~ 3$. Thus, $\sup \varSigma$ is not defined.

1.2 B.2: Proving that $\inf \varSigma _{AM} \in \varSigma _{AM}$ and $\sup \varSigma _{BM} \in \varSigma _{BM}$

In order to prove that $\inf \varSigma _{AM} \in \varSigma _{AM}$ and $\sup \varSigma _{BM} \in \varSigma _{BM}$, we will rely on properties of $\inf \varSigma$ and $\sup \varSigma$ presented in the following lemma.

Lemma 3

For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions, for any $S \in \inf \varSigma$, S can be written in the form of the following non-empty intersection

$$\begin{aligned} S = \underset{ \mathscr {S} \in \varSigma : S \subseteq S^* \in \mathscr {S}}{ \bigcap } S^* \end{aligned}$$

(S1)

For any non-empty collection $\varSigma$ of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$, for any $S \in \sup \varSigma$, S can be written in the form of the following non-empty union

$$\begin{aligned} S = \underset{ \mathscr {S} \in \varSigma : S \supseteq S^* \in \mathscr {S}}{ \bigcup } S^* \end{aligned}$$

(S2)

In addition,

$$\begin{aligned} \exists \mathscr {S} \in \varSigma , ~ \exists S^* \in \mathscr {S}, ~ S^* = S. \end{aligned}$$

(S3)

Proof

We begin with proving (S1). Let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions and consider $S \in \inf \varSigma$. Now set

$$\begin{aligned} S' := \underset{ \mathscr {S} \in \varSigma : S \subseteq S^* \in \mathscr {S}}{ \bigcap } S^* \end{aligned}$$

and let us prove that $S=S'$. According to Lemma 2 that for any $\mathscr {S} \in \varSigma$, $\inf \varSigma \le \mathscr {S}$ so $\exists ! S^* \in \mathscr {S}$ such that $S \subseteq S^*$. This proves that the intersection in the definition of $S'$ is not empty. Now by definition of $S'$ we have $S \subseteq S'$, which also implies $S'\not =\varnothing$. We need to show now that $S' \subseteq S$. Let x be any element of $S'$ and y be any element of S. Then for any $\mathscr {S} \in \varSigma$, there is (a unique) $S^*\in \mathscr {S}$ such that $S\subseteq S^*$ and by definition of $S'$, we have $x\in S^*$. But since $S\subseteq S^*$ we also have $y\in S^*$. This shows that for any $\mathscr {S} \in \varSigma$ there is $S^* \in \mathscr {S}$ such that $x \in S^*$ and $y \in S^*$. This can be expressed equivalently as $x~ \underline{\mathcal {R}}_\varSigma ~y$, so that x and y are in the same element of $\inf \varSigma$, that is $x\in S$.

Now let us prove (S2). Let $\varSigma$ be any non-empty collection of $\mathcal {X}$-partitions such that $\varSigma \subseteq \varSigma _M$ and let $S \in \sup \varSigma$. Set

$$\begin{aligned} S' := \underset{ \mathscr {S} \in \varSigma : S \supseteq S^* \in \mathscr {S}}{ \bigcup } S^* \end{aligned}$$

and let us prove that $S=S'$. According to Lemma 2 that for all $\mathscr {S} \in \varSigma$, $\mathscr {S}\le \sup \varSigma$, so $\exists S^*\in \mathscr {S}$ such that $S^*\subseteq S$. In particular, the intersection in the definition of $S'$ is not empty and $S'\not =\varnothing$. Now by definition of $S'$ we have $S' \subseteq S$. We need to show now that $S \subseteq S'$. Let x be any element of S and y be any element of $S'$. Since $S'\subseteq S$, $y\in S$ so that x and y are in the same element of $\sup \varSigma$, which can be expressed equivalently as $x~ \overline{\mathcal {R}}_\varSigma ~y$. Now by definition of $\overline{\mathcal {R}}_\varSigma$, there is $\mathscr {S} \in \varSigma$ and $S^*\in \mathscr {S}$ such that $x,y\in S^*$. Now since $S^*\cap S\not =\varnothing$, we have $S^*\subseteq S$, which shows by definition of $S'$ that $x\in S'$.

It remains to show (S3)

$$\begin{aligned} \exists \mathscr {S} \in \varSigma , ~ \exists S^* \in \mathscr {S}, ~ S^* = S. \end{aligned}$$

Let us prove by induction on $n \ge 1$ that for any $F \subseteq S$ of cardinality n, there is $\mathscr {S} \in \varSigma$ and $S^* \in \mathscr {S}$ such that $F \subseteq S^* \subseteq S$. The result will follow by taking $F = S$. For $n = 1$, the property holds due to (S2). Let $n\ge 1$ strictly smaller than the cardinality of S and assume that the property holds for all integers smaller than or equal to n. Let F be any subset of S of cardinality $n+1$ and write $F = F_1 \cup \{x\}$, where $x \notin F_1$. Since $F_1$ is of cardinality n there is $\mathscr {S}_1 \in \varSigma$ and $S_1 \in \mathscr {S}_1$ such that $F_1 \subseteq S_1 \subseteq S$. Let $y \in F_1$. There is also $\mathscr {S}_2 \in \varSigma$ and $S_2\in \mathscr {S}_2$ such that $\{x,y\}\subseteq S_2\subseteq S$. Now because $\mathscr {S}_1, \mathscr {S}_2 \in \varSigma \subseteq \varSigma _M$, we have $\mathscr {S}_1, \mathscr {S}_2 \subseteq \mathscr {H}$, so that $S_1\in \mathscr {H}$ and $S_2 \in \mathscr {H}$. From the definition of hierarchy, we get $S_1 \cap S_2 \in \{ \varnothing , S_1, S_2 \}$. Since $y \in S_1 \cap S_2$, we have $S_1 \cap S_2 \ne \varnothing$, so one of the two, denoted $S^*$ contains the other one. In particular, $F_1 \subseteq S^* \subseteq S$ and $\{x,y\} \subseteq S^* \subseteq S$, which shows that $F = F_1 \cup \{x\} \subseteq S^* \subseteq S$ and terminates the proof. $\square$

We can now go back to the proof of Theorem 1.

Proof

• (i) $\inf \varSigma _{AM} \in \varSigma _A$: Consider $S \in \inf \varSigma _{AM}$ and $P \in \mathscr {P}$. From Lemma 3, we get

  $$\begin{aligned} S \cap P = P \cap \left( \underset{ \mathscr {S} \in \varSigma _{AM}: S \subseteq S^* \in \mathscr {S} }{ \bigcap } S^* \right) = \underset{ \mathscr {S} \in \varSigma _{AM}: S \subseteq S^* \in \mathscr {S} }{ \bigcap } (P \cap S^*) \end{aligned}$$

  Now for each $S^* \in \mathscr {S} \in \varSigma _{AM}, ~ P \cap S^* \in \{ \varnothing , P \}$, thus leading to $S \cap P \in \{ \varnothing , P \}$, that is $\inf \varSigma _{AM} \in \varSigma _A$.

• (ii) $\inf \varSigma _{AM} \in \varSigma _M$: Consider $S \in \inf \varSigma _{AM}$. From Lemma 3, we get

  $$\begin{aligned} S = \underset{ \mathscr {S} \in \varSigma _{AM}: S \subseteq S^* \in \mathscr {S} }{ \bigcap } S^* \end{aligned}$$

  Now for each $S^* \in \mathscr {S} \in \varSigma _{AM}, ~ S^* \in \mathscr {H}$. Moreover, the hierarchy $\mathscr {H}$ is closed under finite, non-disjoint intersections, thus leading to $S \in \mathscr {H}$, that is $\inf \varSigma _{AM} \in \varSigma _M$.

• (iii) $\sup \varSigma _{BM} \in \varSigma _B$: Consider $S \in \sup \varSigma _{BM}$ and recall from Lemma 3 that there is $\mathscr {S} \in \varSigma _{BM}$ and $S^*\in \mathscr {S}$ such that $S=S^*$. Now for any $P \in \mathscr {P}$,

  $$\begin{aligned} S \cap P = S^* \cap P \in \{ \varnothing , S^* \} = \{ \varnothing , S \}, \end{aligned}$$

  so that $\sup \varSigma _{BM} \in \varSigma _B$.

• (iv) $\sup \varSigma _{BM} \in \varSigma _M$: Consider $S \in \sup \varSigma _{BM}$ and $S^*=S$ as previously. Since $S^* \in \mathscr {H}$, $S \in \mathscr {H}$, so that $\sup \varSigma _{BM} \in \varSigma _M$.

This shows that $\inf \varSigma _{AM} \in \varSigma _{AM}$ and $\sup \varSigma _{BM} \in \varSigma _{BM}$, which completes the proof of Theorem 1. $\square$

C: Construction of the Lacy and Loose Phylogenies

This section aims at formalizing mathematically the construction of the lacy and loose phylogenies presented in the main text.

Recall that an interior node is convergent if there are two tips, one in each of its two descending subtrees, carrying the same phenotype, otherwise this node is said to be divergent. We will say that the two monophyletic groups subtended by a convergent (resp. divergent) node are convergent (resp. divergent). We define $\mathscr {H}_d$ as the collection of divergent monophyletic groups, that is

$$\begin{aligned} \mathscr {H}_d&= \{ h \in \mathscr {H}: ~ \exists h',h'' \in \mathscr {H}, ~h'= h \cup h'' , ~ \forall P \in \mathscr {P}, ~ h \cap \\ P&= \varnothing \text { or } h'' \cap P = \varnothing \} \cup \mathcal {X} \end{aligned}$$

We similarly consider phylogenetic and non-phylogenetic monophyletic groups for either the loose or the lacy definition. We call $\mathscr {H}_{\text {loose}}$ and $\mathscr {H}_{\text {lacy}}$ the collection of phylogenetic monophyletic groups for the loose and lacy definitions respectively. The procedure described in the main text amounts to defining

$$\begin{aligned} \mathscr {H}_{\text {loose}}&= \mathscr {H} \backslash \{h \in \mathscr {H}: ~ \exists h_{c} \in \mathscr {H} \backslash \mathscr {H}_d, ~ h \subseteq h_{c} \} \\ \mathscr {H}_{\text {lacy}}&= \{ h \in \mathscr {H}:~ \exists h' \in \mathscr {H},~ h \cup h' \in \mathscr {H}, ~h \cap h' \in \{ \emptyset , h' \}, ~ \exists h_d \in \mathscr {H}_d, ~ h_d \subseteq h' \} \end{aligned}$$