1. Introduction
Phylogenetic networks are a generalization of evolutionary trees. They come in various forms, and are commonly used to represent the evolutionary history for a set of species in which events such as hybridization or recombination are suspected to have occurred [huson]. For this paper, a phylogenetic network (on species set ) is a connected directed acyclic graph, with a single root vertex and leafset in which every internal vertex has degree 3 except for the root which has outdegree 2. We call the number of vertices in a phylogenetic network with indegree 2 the network’s reticulation number, so that a phylogenetic tree is a network with reticulation number 0. We shall mainly focus on temporal treechild networks in which each nonleaf vertex has a child whose indegree is 1, whose vertices can be labelled with times that move strictly forward on treelike parts of the network and so that vertices with indegree 2 have parents with the same time label (also known as treechild, timeconsistent networks [CLRV09]).
Any phylogenetic network on the set displays a set of phylogenetic trees on , where a phylogenetic tree is displayed by a network if there is a subgraph of the network that is isomorphic to a subdivision of the tree [vi2010]. It is therefore natural to try to construct phylogenetic networks by reversing this process, i.e. by trying to find a network which displays a given set of trees. These trees are usually obtained from genomic data, by considering different genes (which leads to “gene trees”) or regions of the species’ genomes. For a given set of phylogenetic trees, this also leads to the concept of the (temporal) hybrid number, which is the minimum reticulation number taken over all (temporal treechild) networks that display each tree in the set [BS, HLS13]. While the hybrid number exists for any set of phylogenetic trees, it is worth noting that the temporal hybrid number does not always exist, i.e. there are sets of trees that cannot simultaneously be embedded in a temporal treechild network.
Several results have been presented in the literature concerning displaying phylogenetic trees and hybrid numbers, mainly for pairs of trees. These include structural information on how the hybrid number is related to the socalled maximum acyclic agreement forest for two phylogenetic trees [B05], characterizations for when collections of trees are displayed by special types of networks [HLS13, L19] and related algorithms/complexity results [BS05, BS07, Dl17, HL13, H16]. However, all of these results rely on the fact that the networks display the set of trees in question, a notion that may not appropriately model certain evolutionary scenarios, such as incomplete lineage sorting [L, degan].
A possible solution to this problem is to relax the displaying condition. Roughly speaking, a phylogenetic tree is weakly displayed by a network [huber16] if it can be embedded in the network in such a way that the tree follows along the directed paths in the network (see Section 3 for the definition). In this paper we will consider the problem of deciding when a pair of phylogenetic trees is weakly displayed by a temporal treechild network under the assumption that there exist simultaneous embeddings of both trees that do not permit more than three branches of the trees to come together at a reticulation vertex. In this case we shall say that the pair of trees is rigidly displayed by the network. Note that related problems were recently considered in [L] (the Parental Tree Network Problem) and in [IJ18] (The Beaded Tree Problem). In the Parental Tree Network Problem the aim is to find a network with a minimum number of reticulation vertices that weakly displays all trees in a given set of phylogenetic trees; in the Beaded Tree Problem, however, networks with parallel edges are permitted and a different concept of displaying is used which can lead to different solutions (see Section 4 for more details).
We now summarize the rest of the paper, including statements of our main results. After presenting some definitions in Section 2, in Section 3 we present the definition of weakly displaying, and we prove some basic facts concerning this concept and its relationship with displaying. In Section 4, we then consider the weak hybrid number of two trees. In particular, we determine the weak hybrid number for a specific pair of phylogenetic trees and show that for this pair of trees we get a different number to the analogous hybrid number defined in [IJ18]. This example shows that the beaded trees introduced in [IJ18] can lead to a quite different solution when aiming to find a network in which to embed the given trees.
In Sections 5 and 6, we introduce the concepts of rigidly displaying and forkoperations, respectively and prove some results on these concepts which we use later on. Then in Section 7 we give a characterization for when a pair of phylogenetic trees can be rigidly displayed in terms of forkpicking sequences (Corollary 7.2), a generalization of cherrypicking sequences [HLS13]. In Section 8, we go on to show that when the rigid hybrid number of two trees exists, it is equal to the weight of a minimum forkpicking sequence (Corollary 8.2). These results can be regarded as analogues of [HLS13, Theorem 1] and [HLS13, Theorem 2], respectively. In Section 9 we show that there is a pair of phylogenetic trees on a set with arbitrarily large, so that the difference between the temporal and rigid hybrid numbers for these two trees is at least (Theorem 9.1). We conclude with a discussion on possible future directions in Section 10.
2. Preliminaries
Let denote a directed, acyclic graph with a single root vertex, i.e., a vertex with indegree 0. We let denote the vertex set of , the set of (directed) edges of , and the unique root of . A vertex in with indegree 1 and outdegree 0 is called a leaf; an edge of incident with a leaf of a pendant edge of . Furthermore, we denote the set of all leaves of by .
Suppose . We say that a vertex is above if there exists a directed path from the root of to that contains (note that could equal ). If is above , then we also write or simply if is clear from the context. Furthermore, we also say that is below . We call any vertex above an ancestor of and any vertex below a descendant of . Finally, we say that two distinct edges and of are comparable if is above or is above . Otherwise we say that and are incomparable.
Let be a finite set of size at least 2. Following e.g. [huber16, p.1764] a rooted, directed acyclic graph is called a phylogenetic network (on ) if the leafset of is , the root of has outdegree two and any nonroot, nonleaf vertex either has indegree one and outdegree two or zero (in which case is called a tree vertex) or has indegree two and its outdegree is one (in which case is called a reticulation vertex). The set of reticulation vertices of is denoted by . We put . Unless stated otherwise, phylogenetic networks do not contain parallel edges. Moreover, we call a directed path in a phylogenetic network of length one or more in which every vertex, except possibly the first vertex, is a tree vertex a treepath in .
A phylogenetic tree (on ) is a phylogenetic network on that does not have any reticulation vertices. We say that two phylogenetic trees and on are isomorphic, denoted by , if there exists a bijection that induces a graph isomorphism between and that is the identity on . If is a phylogenetic tree on , and , then the last common ancestor of , denote by , is the unique vertex of that is an ancestor of every element in and there is no vertex such that is a descendant of and is an ancestor of every element in . For any elements , we sometimes also write rather than . In addition, we denote by the minimal subtree of spanned by all leaves in and by the restriction of to , that is, the phylogenetic tree on obtained from by suppressing all resulting vertices of both indegree and outdegree one. Note that the root of is the last common ancestor of all elements in .
Suppose is a phylogenetic network on . Following [S16], we say that is temporal [M04] if there exists a map such that, for all , we have whenever is a reticulation vertex and , otherwise. In that case, we call a temporal labelling of . Unless of relevance to the discussion, we always omit the temporal labelling when depicting a temporal network. We say that is treechild [CRV09] if, for each nonleaf vertex at least one of the children of is a tree vertex. Note that a treechild network was called a phylogenetic network in [HLS13, p.1883]. Also note that a temporal treechild network (in our sense) has also been called a (binary) timeconsistent treechild network, or TCTCnetwork in [CLRV09]. Finally, we say that is normal if in addition to being treechild it does not contain a shortcut, that is, if there is a directed path from a vertex to a vertex with at least two edges, then there is no directed edge [MS15]. Note that any temporal treechild network is normal [S16, Proposition 10.12]
3. Weakly displaying two trees in a network
In this section, we derive some basic properties for the notion of weak displaying which will be useful later. In the following, assume that is a phylogenetic tree on and that is a phylogenetic network on .
We call a map that is the identity on a display map for in if the following additional properties hold

for all , is a tree vertex or the root of ,

for every edge of there exists a directed path having at least one edge from to in , and

for any two distinct edges and of that share the same tail the first edge of is not the first edge of .
Following [huber16], we say that is weakly displayed by if there exists a display map for in . To reduce notation, we will sometimes not explicitly refer to the display map. Note that if is weakly displayed by , then there could be more than one display map for in . In addition, note that if displays then also weakly displays (but not necessarily conversely).
The notion of weakly displayed was introduced in [huber16] in terms of a construction that allows the “unfolding” of a phylogenetic network on into a socalled “multilabelled tree on ” [HM06]. Such trees are similar to phylogenetic trees in that they have no vertices with in and outdegree one and the root has indegree zero. However the requirement that the leafset is is relaxed to the requirement that an element of can “label” more than one leaf (which is not allowed in the case of phylogenetic trees).
Note that although closely related, display maps are not weak embeddings sensu [IJ18]. Stated within our framework, such embeddings are maps from a multilabelled tree into a phylogenetic network such that (a) all leaves of that share the same “label” are mapped to the leaf of , (b) every edge of is mapped to either a vertex of or a directed path from to , and (c) for every nonleaf vertex of with outgoing edges and , the directed paths associated to and which each has at least one edge) start with different outgoing edges of . Since a phylogenetic tree is clearly also a multilabelled tree, it follows that the map that is the identity on and maps all nonleaf vertices of to the root of is a weak embedding of into as Property (c) vacuously applies. However is not a display map for in as Property (ii) does not hold. Even so, a display map is always a weak embedding.
Now, suppose that is a display map for in . For any edge , we denote by the set of all vertices in that lie on the path except for . Let , and let be an edge of . If , we say that the path ends at and if , but we say that passes through . In addition, we define the number
i.e., counts the number of edges in such that their image under either ends or passes through so that in particular . Finally, if is a further phylogenetic tree on that is also weakly displayed by via a map , then we put
(see e.g. Figure 1). To reduce notation we sometimes drop the subscript in as indicated when no confusion can arise about which maps are being used to weakly display and .
We now prove two lemmas about these concepts which will be useful later. The first concerns temporal treechild networks.
Lemma 3.1.
Let be a temporal treechild network on that weakly displays a phylogenetic tree on via a display map . Then displays if and only if for all , we have . In addition, so that, in particular, if displays then .
Proof: Consider the first statement. Suppose , then by the definition of displaying, it is straightforward to see that, for all , we have . Conversely, first note that if then, by assumption, there exists at most one edge such that either passes through or ends in . As is a reticulation vertex, cannot end at . Since has no shortcuts, by deleting each edge of that is directed into a reticulation vertex of and for which there exists no such that is an edge on and passes through , we obtain a subgraph of that is isomorphic to a subdivision of . Thus, displays .
To see that the last statement in the lemma holds, assume for contradiction that is not the root of . Since is temporal treechild and, therefore, normal [S16, Proposition 10.12], the two children and of must be distinct and tree vertices. Moreover, there must exist a treepath in from to some leaf and a treepath from to some leaf . Note that since and cannot intersect, we must have . Since is a display map for in and and are also leaves of it follows that is mapped to an ancestor of and in under . Extending the paths and to treepaths starting at implies that that ancestor must be . Thus, . ∎
To state the second lemma we require some further definitions. We call a subgraph of a pendant subnetwork of if there exists a tree vertex in such that when deleting the incoming edge of the network decomposes into two connected components such that the component that contains in its vertex set is a phylogenetic network on . A pendant subtree of is a pendant subnetwork of that is a phylogenetic tree. Note that a pendant subnetwork and therefore also a pendant subtree must have at least two leaves.
Lemma 3.2.
Suppose that is a phylogenetic network on that weakly displays two distinct phylogenetic trees and on via display maps and , respectively.

If is a tree vertex in and , then there is a vertex which is an ancestor of in .

If is tree vertex in with and has a child that is the root of a pendant subtree, say , of then is a pendant subtree of both and .
Proof: (i) Suppose that is a tree vertex of and . Then, without loss of generality, we may assume that is such that . Hence, there are two distinct edges and in with tails and respectively, such that and . Note that may or may not hold. In either case, the definition of a display map combined with the fact that does not contain parallel edges implies that the heads of the outgoing edges of and , respectively, must be distinct. Thus, .
We claim that and are incomparable in . Indeed, assume for contradiction that and are comparable. Without loss of generality we may assume that is above in . Then the directed path in starting at and ending at the head of is mapped by to a directed path in with edge set . Since it follows that , a contradiction which yields the claim.
In particular, since is weakly displayed by there must be a vertex in with and such that (a) the two directed paths in from up to and including the heads of and , respectively, are mapped by to two directed paths and in , (b) the first edge on is different from the first edge on , and (c), is a vertex on both and . Hence there must be some vertex in which lies on and and which is an ancestor of .
(ii) Note that every leaf in must be contained in the image under of some directed path in from the root of to , and similarly for . Since, by assumption, , there can be at most one such path in and , respectively which has this property for every leaf in . Hence must be a pendant subtree of both and . ∎
4. The weak hybrid number
Given two phylogenetic trees and on , we define the weak hybrid number of and as
Note that for any two phylogenetic trees and there always exists a phylogenetic network that displays and and so is welldefined. In addition, the weak hybrid number has been implicitly considered in The Parental Tree Network Problem [L, Definition 5]. In this section, we give an example which shows that the weak hybrid number is different from the related beaded hybrid number [IJ18], whose definition we next recall.
A beaded tree on is a phylogenetic network on in which parallel edges are allowed, and in which each reticulation has a unique parent such that there are two parallel edges from to [IJ18, Definition 7]. Now for two phylogenetic trees and on , we define the beaded hybrid number for and to be
Note that [IJ18, Lemma 9] implies that any phylogenetic network on that weakly displays two phylogenetic trees and on can be transformed into a beaded tree on such that there exist weak embeddings of and into for which (so in particular exists for any pair of trees ). Hence, .
We now use Lemma 3.2 to show in Proposition 4.1 that there exist phylogenetic trees and such that . First, consider the two phylogenetic trees and depicted in Figure 2. Note that since and are not isomorphic, and there exist weak embeddings of and into the pictured beaded tree , respectively. Hence, . We now show that .
To this end, we call two leaves and of a phylogenetic tree with a cherry of , denoted by , if and share a parent.
Proposition 4.1.
Let and denote the two phylogenetic trees on pictured in Figure 2. Then .
Proof: As and are not isomorphic we have . Moreover, as the phylogenetic network pictured in Figure 2 is also on and weakly displays and we have . We now show that , from which the proposition follows.
Suppose to the contrary that . Then there exists a phylogenetic network that weakly displays and such that . Let be the unique vertex in . Let be a parent of . Note that since has no parallel edges, must exist. Also note that must be a tree vertex of as is the sole reticulation vertex of . Finally, note that the other child of cannot be as does not contain parallel edges.
Denoting that child by we next claim that must be a leaf of . Assume for contradiction that is not a leaf. Let the subtree of rooted at . Let be a leaf of and, thus, of . Then since weakly displays via a map say, and is the sole reticulation vertex of we obtain . Similarly, as weakly displays via a map say, must hold. Thus, . Since Lemma 3.2(i) implies that as is the sole reticulation vertex of , it follows that . Hence, by Lemma 3.2(ii), is also a pendant subtree of and of ; a contradiction as and are the two trees depicted in Figure 2. Thus, is a leaf of , as claimed.
Since every element in is contained in a cherry of either or , we may choose some such that is a cherry in either or . Without loss of generality, assume that is that tree. Since the only two cherries of are and we may assume without loss of generality that . Let denote the parent of and in .
Let be the parent of which must exist as . Then as otherwise the fact that is a cherry of but not of implies that is below . But then is not weakly displayed by because is an edge of and is not an edge in ; a contradiction.
We next claim that cannot be an edge in . To see this, assume for contradiction that is an edge in . Then since is weakly displayed by and is contained in a cherry of but not of it follows that must be a leaf of below . If there existed another leaf of below then that leaf would have to be “5”. Since is a cherry of and is weakly displayed by it follows that that cherry must also be below ; a contradiction as is the phylogenetic trees depicted in Figure 2. Thus, is in fact the sole leaf of below . But then must also be a cherry of ; a contradiction since and are the phylogenetic trees depicted in Figure 2. Thus, cannot be an edge in , as claimed. Hence, the other child of , call it , must either be a leaf of or is the root of a pendant subtree of .
Note first that arguments similar to the case of imply that must be a leaf of . Let denote the parent of . We next distinguish between the cases that and that .
If then . To see this, assume for contradiction that . Then is an edge in . Since is a cherry of and the parent of and is not adjacent with it follows that is not weakly displayed by ; a contradiction. Thus, , as required.
We next claim that also cannot be an edge of . Assume for contradiction that is an edge of . Then since is a cherry of and is weakly displayed by , similar arguments as before imply that “5” must be the sole leaf of below and that the unique directed path from to leaf “2” does not cross . Since is one of the two phylogenetic trees depicted in Figure 2 it follows that is not weakly displayed by ; a contradiction. Thus, cannot be an edge of either.
Let denote the other child of . Then similar arguments as in the case of imply that must also be a leaf of . Hence, is a pendant subtree of . Since is a tree vertex of , Lemma 3.2(ii) implies that is a pendant subtree of and of ; a contradiction in view of Figure 2. Hence, in case .
Assume for the remainder that . Then is a descendant of in . Since is a leaf of , it follows that is a pendant subtree of . Since we must have . But then as is a cherry of and is weakly displayed by ; a final contradiction. Hence, and, so, the proposition follows. ∎
This example is important as it indicates that the beaded hybrid number could potentially underestimate the number of reticulations required to weakly display two phylogenetic trees in a network. It would be interesting to understand how large the difference between and could be in general.
5. Rigidly displaying
We now introduce and present some basic properties of the notion of rigidly displaying. We begin with a lemma which will help to motivate our definition. In Figure 1 we present an example where two phylogenetic trees and are weakly displayed by the depicted phylogenetic network , for all but and are not both displayed by . So, in general, it does not suffice to insist that for all for two phylogenetic trees to be displayed by a phylogenetic network. However, if we insist that the network is temporal treechild, we now show that this condition actually suffices.
Lemma 5.1.
Suppose that is a temporal treechild network on and that and are two phylogenetic trees on that are weakly displayed by via display maps and , respectively. Then the following statements are equivalent.

displays and .

for all .

for all .
Proof: (i) (ii) We show first that must hold for all . Assume for contradiction that there exists some vertex such that . Then one of or must hold. Without loss of generality we may assume that . Then there exists no edge such that either passes through or ends in . But then there cannot exist a leaf of that can be reached from via a treepath. Thus, is not treechild; a contradiction. Since, by assumption, displays both and , Lemma 3.1 implies that for all . Thus, must hold for all .
(ii) (iii) This is trivial.
(iii) (i) By Lemma 3.1 it suffices to show that and holds for all . Assume for contradiction that there exists some and some tree in , say , such that . Then . In view of the assumptions on and , the last statement in Lemma 3.1 implies that there must be a directed path in that starts at such that the image under of the last edge in this path passes through or ends at . Hence, must hold too. Thus, . By assumption, it follows that must be a tree vertex of . In view of Lemma 3.2(i), there must exist some vertex that is an ancestor of . Without loss of generality, we may assume that is such that no vertex in distinct from that is above and below is contained in . By assumption, it follows that ; a contradiction to the choice of and the fact that . ∎
Motivated in part by this lemma, we say that a phylogenetic network on rigidly displays two phylogenetic trees and on if weakly displays and via display maps , respectively, and, for all we have and, for each parent of , we have . For example, the network pictured in Figure 2 rigidly displays the two phylogenetic trees depicted in that figure.
Note that, in contrast to the definitions of displaying and weakly displaying which refer to a single tree, rigidly displaying always refers to two trees. In addition, by Lemma 5.1 it follows that if and are two phylogenetic trees on that are displayed by a temporal treechild network on , then also rigidly displays and . We also have the following:
Lemma 5.2.
Suppose is a temporal treechild network on and that rigidly displays two phylogenetic trees and on via display maps and . Then for all .
Proof: Suppose . Then since rigidly displays and it also weakly displays and . Since it follows that and that . Hence, .
For the remainder, assume for contradiction that there exists some such that . Then must be a tree vertex of as rigidly displays and and . Let be a longest directed path of tree vertices in that ends at . Note that , for all . Also note that since is not a tree vertex of , we cannot have . Let denote the parent of . Note that . Hence, we cannot have . Since rigidly displays and it follows that must be a tree vertex of . But then the extension of by results in a directed path of tree vertices of that ends in and that is longer than ; a contradiction. ∎
Note that the converse of the last lemma does not hold in general (see e. g. Figure 3). We conclude this section with one more lemma that will be useful later.
Lemma 5.3.
Suppose that is a treechild network on that rigidly displays two phylogenetic trees and on , and that is weakly displayed via the display map . If is an edge of such that passes through a vertex , then must be a parent of in .
Proof: Suppose in is such that passes through a vertex . Assume for contradiction that is not a parent of . Let be the parent of in such that lies on . Then . As is treechild, there must be a treepath in starting at and ending at some leaf . So, as is a leaf of and , there must be some edge in such that passes through . Moreover, considering the leaf again, there must be an edge in which maps to a path in via that either ends at or passes through . It follows that ; a contradiction as rigidly displays and and is the parent of a vertex in . ∎
6. Fork operations
In the next section we shall characterize when two trees are rigidly displayed by a temporal treechild network in terms of sequences of certain operations on these trees. The basis for these sequences are forkoperations which we shall now introduce.
By a fork we mean a 2leaved tree (i. e. a cherry), a 3leaved rooted tree (a 3fork) or a 4leaved fullybalanced rooted tree (a 4fork). The following basic fact concerning forks is straightforward to show.
Lemma 6.1.
Suppose is a phylogenetic tree with leaves. If then is a 3fork and if then must contain a pendant subtree that is either a 3fork or a 4fork.
A forkoperation for a pair of phylogenetic trees and on consists of a leaf , together with a fork in each of and containing as depicted in the second and third columns of Figure 4. In case the type of forkoperation is relevant to the discussion we also write , , where is a type operation. In addition, we shall call the leaf associated to the operation. When we apply an operation to some element , we remove the leaf from both trees, and suppress any resulting vertices of degree 2 (removing the root and both edges incident with it in case ).
Now, given two phylogenetic trees and on the set , , we call a sequence of , forkoperations a special sequence for and if is a type1 operation on and and, in case , the following properties hold:

There exists some such that each , is a type2 or a type3 operation applied to and the associated 3 or 4fork is a pendant subtree of ,

the lastbutone operation is a type2 operation with fork and cherry some , and the last operation is a type1 operation with cherries and , some distinct, and

if then must hold for all for the tree in (i).
To illustrate this definition, consider the phylogenetic network on depicted in Figure 5. Then is a special sequence for the two phylogenetic trees on also pictured in that figure where, for example, is a forkoperation of type3 and the tree with cherry is the tree mentioned in the definition. Note that an application of a special sequence always results in phylogenetic trees with at least two leaves. The following proposition will be key to the proof of our main results.
Proposition 6.2.
Suppose that is a temporal treechild network on , , that rigidly displays two phylogenetic trees and on . If no type0 operation can be applied to and , then there is a special sequence for and . Moreover, the two phylogenetic trees resulting from applying can be rigidly displayed by a temporal treechild network with .
Proof: Note first that as otherwise would be a phylogenetic tree that is isomorphic with both and implying that a type0 operation can be applied to and ; a contradiction.
Now, let denote a temporal labelling for and pick some whose value is maximum under . Let and be the parents of . Note that as does not contain parallel edges. Also note that since does not contain shortcuts as it is normal, cannot be an ancestor of and cannot be an ancestor of . In particular, this implies that and must be tree vertices. Let be the child of that is not and, similarly, let be the child of that is not . We claim that is a leaf of .
To see that this claim holds, assume for contradiction that is the root of a pendant subgraph of . Note that the choice of implies that is in fact a pendant subtree of . Moreover, Lemma 5.2 implies that there are display maps for and in such that . By Lemma 3.2(ii) it follows that is a pendant subtree of both and . Hence, and have a common cherry and, so, we can apply a type0 operation to and ; a contradiction. Thus must be a leaf of . Applying similar arguments to implies that must also be a leaf of .
Since is temporal, the choice of implies that the child of is a leaf of or the root of a pendant subtree of . Assume first that is a leaf of . Then since and are rigidly displayed by and and do not contain a common cherry, it is straightforward to see using Lemma 5.3 that, without loss of generality, and must contain the cherries and , respectively. Hence we can apply a type1 operation to . This gives a special sequence of length 1 for and , from which the first part of the proposition follows.
So, suppose that is the root of a pendant subtree of , so that has at least two leaves. Note first that . Indeed since and are rigidly displayed by we obtain in view of Lemma 5.2. If held then and would have a common cherry which implies that a type0 operation can be applied to and ; a contradiction. We can therefore assume without loss of generality that is a pendant subtree of , and that this tree together with the leaf also forms a pendant subtree of .
In case has only two leaves and , say, then since it follows that contains the 3fork and contains, without loss of generality, the cherries and . Hence we can apply a type2 operation to and then apply a type1 operation to (since and must contain the cherries and , respectively). This gives a special sequence of length 2, from which the first part of the proposition again follows.
Assume for the remainder that has at least three leaves. We claim that we can perform a sequence of type2 and type3 operations involving the removal of an element from one at a time, and at no stage creating a common cherry, followed by a type1 operation which, when applied, results in a special sequence for and . We prove the claim by induction on the number of leaves of . Note that we have just shown that the claim holds for the base case . So suppose the claim holds for all , , and that contains leaves. Note that as , Lemma 6.1 implies that contains either a 3fork or a 4fork.
Suppose contains a 3fork where are distinct. Then must be a pendant subtree of . As and have no cherries in common and , it follows that we may assume without loss of generality that contains the cherry . Hence, we can apply the type2 operation . Note that this creates a cherry in
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