U p to this point, it has been shown that not all (sets of) relations are tree representable and that those, which are tree representable, can
Theorem 7.2. Let RT be a strict dense triple set on L with |L| > 3 The set RT is consistent if and only if cl(RT
)⊆RT holds for all RT0⊆RT with |RT0| = 2. Proof. ⇒: IfRT is strict dense and consistent, then for any triple (ab|c) /∈ RT holdsRT ∪ (ab|c) is inconsistent as either (ac|b) or (bc|a) is already contained in RT. Hence, for each a, b, c ∈ L exactly one RT ∪ {(ab|c)}, RT ∪ {(ac|b)}, RT ∪ {(bc|a)} is consistent, and this triple is already contained in RT. Hence, RT is closed. Therefore, for any subset RT0 ⊆RT holds cl(RT0
) ⊆ cl(RT) = RT. In particular, this holds for all RT0⊆RT with |RT0| = 2.
⇐ (Induction on|L|):
If|L| 6 4, then Lemma 7.6 implies that if for any two-element subset RT0 ⊆ RT holds that cl(RT0
) ⊆ RT, then RT is consistent. Assume therefore, the assumption is true for all strict dense triple setsRT on L with |L| = n.
LetRT be a strict dense triple set on L with |L| = n + 1 such that for each RT0 ⊆ RT with |RT0| = 2 it holds cl(RT0
) ⊆ RT. Moreover, let L0 = L\ {x} for some x ∈ L and RT|L0 ⊂ RT denote the subset of all triples r ∈ RT
with L(r) ⊂ L0. Lemma 7.7 implies that RT|L0 is strict dense and for each
RT0 ⊆ RT
|L0 with |RT0| = 2 it follows that cl(RT0) ⊆ RT|L0. Hence, the
induction hypothesis can be applied for any suchRT|L0 implying thatRT|L0
is consistent. Moreover, since RT|L0 is strict dense and consistent, for any
triple (xy|z) /∈ RT|L0it holds thatRT|L0∪ (xy|z) is inconsistent. But this implies
thatRT|L0 is closed, i.e., cl(RT|L0) =RT|L0. Lemma 7.4 implies that the Aho
graph [RT|L0,L] has exactly two connected components C1 and C2 for each
L ⊆ L0 with |L| > 1. In the following, let L
i = V(Ci), i = 1, 2 denote the set of vertices of the connected component Ci in [RT|L0,L]. Clearly, L =
L1∪L2 and L1∩L2 = ∅. It is easy to see that [RT, L] ' [RT|L0,L] for any
L ⊆ L0, since none of the graphs contain vertex x. Hence, [RT, L] is always disconnected for any L ⊆ L0. Therefore, it remains to show that, for all L ∪ {x} with L ⊆ L0the following statement is true: If for anyRT0⊆RT with |RT0| = 2 holds cl(RT0
) ⊆RT, then [RT, L ∪ {x}] is disconnected and hence, RT is consistent.
To proof this statement the different possibilities forL are considered sep- arately. It is frequently used that [RT|L0,L] is a subgraph of [RT, L] for every
L ⊆ L (Lemma 7.5).
Case 1. If|L| = 1, then L ∪ {x} implies that [RT, L ∪ {x}] has exactly two vertices and clearly, no edge. Thus, [RT, L ∪ {x}] is disconnected.
Case 2. Let |L| = 2 with L1 = {a} and L2 = {b}. Since RT is strict dense, exactly one of the triples (ab|x), (ax|b), or (xb|a) is contained in RT. Hence, [RT, L ∪ {x}] has exactly three vertices where two of them are linked by an edge. Thus, [RT, L ∪ {x}] is disconnected.
Case 3. Let |L| > 3 with L1 = {a1, . . . , an} and L2 = {b1, . . . , bm}. Since RT|L0 is consistent and strict dense and by construction of L1 and L2 it
holds ∀ai, aj ∈ L1, bk ∈ L2, i 6= j : (aiaj|bk) ∈ RT|L0 ⊆ RT and ∀ai ∈
L1, bk, bl ∈ L2, k 6= l : (bkbl|ai) ∈ RT|L0 ⊆ RT. Therefore, since RT is
strict dense, there cannot be any triple of the form (aibk|aj) or (aibk|bl) with ai, aj ∈ L1, bk, bl ∈L2 that is contained RT. It remains to show that RT is consistent. The following three subcases can occur.
and b ∈L2. Hence, in order to prove that RT is consistent, it has to be shown that there is no triple (cx|d) contained RT for all c, d ∈ L, which would imply that [RT, L ∪ {x}] stays disconnected, i.e., [RT, L ∪ {x}] then consists of two components, one containing the vertices V(C1)∪ V(C2) and the other containing the single vertex x.
3.b) The connected component C1 of [RT|L0,L] is connected to x in [RT, L ∪
{x}]. Hence, there must be a triple (ax|c) ∈ RT with a ∈ L1, c ∈ L. Hence, in order to prove that RT is consistent, it has to be shown that there are no triples (bkx|ai) and (bkx|bl) for all ai ∈ L1, bk, bl ∈ L2, which would imply that [RT, L ∪ {x}] stays disconnected.
3.c) As in Case 3.b), the connected component C2 of [RT|L0,L] might be
connected to x in [RT, L ∪ {x}] and it has to be shown that there are no triples (aix|bk)and (aix|aj)for all ai, aj ∈L1, bk∈L2 in order to prove that RT is consistent.
Case 3.a Let (ab|x) ∈ RT, a ∈ L1, b ∈L2. First it is shown that for all ai∈L1 it holds that (aib|x) ∈ RT. Clearly, if L1 = {a} the statement is trivially true. If |L1| > 1 then {(ab|x), (aia|b)} ` (aib|x) for all ai ∈ L1. Since the closure of all two-element subsets ofRT is contained in RT and (ab|x), (aia|b) ∈ RT one can conclude that (aib|x) ∈ RT. Analogously one shows that for all bk ∈L2 holds (abk|x) ∈ RT.
Since{(aia|bk), (abk|x)} ` (aibk|x) and (aia|bk), (abk|x) ∈ RT one can conclude that (aibk|x) ∈ RT for all ai ∈ L1, bk ∈L2. Furthermore, {(aiaj|b), (aib|x)} ` (aiaj|x) for all ai, aj ∈L1 and again, (aiaj|x) ∈ RT for all ai, aj ∈L1. Analo- gously, one shows that (bkbl|x) ∈ RT for all bk, bl∈L2.
Thus, it has to be shown that for all c, d ∈L it holds that (cd|x) ∈ RT. Since RT is strict dense, there is no triple (cx|d) contained in RT for any c, d ∈ L. Hence, [RT, L ∪ {x}] is disconnected.
Case 3.b Let (ax|c) ∈ RT with a ∈ L1, c ∈L. Assume first that c ∈ L1. Then there is triple (ac|b) ∈ RT. Moreover, {(ax|c), (ac|b)} ` (ax|b) and thus, (ax|b) ∈ RT. This implies that there is always some c0 = b ∈ L
2 with (ax|c0) ∈ RT. In other words, w.l.o.g. one can assume that for (ax|c) ∈ RT, a ∈ L1 it holds that c ∈L2.
Since {(ax|b), (aai|b)} ` (aix|b) and (ax|b), (aai|b) ∈ RT one can conclude that (aix|b) ∈ RT for all ai ∈ L1. Moreover, {(aix|b), (bbk|ai)} ` (aix|bk) and by similar arguments, (aix|bk) ∈ RT for all ai ∈ L1, bk ∈ L2. Finally, {(aix|bk), (blbk|ai)} ` (bkbl|x), and therefore, (bkbl|x) ∈ RT for all bk, bl ∈ L2. To summarize, for all ai ∈ L1, bk, bl ∈ L2 it holds that (aix|bk) ∈ RT and (bkbl|x) ∈ RT. Since RT is strict dense there cannot be triples (bkx|ai) and (bkx|bl)for any ai∈L1, bk, bl∈L2, and hence, [RT, L ∪ {x}] is disconnected. Case 3.c By similar arguments as in Case 3.b) and interchanging the role of L1 andL2, one can show that [RT, L ∪ {x}] is disconnected.
In summary, it has been shown that [RT, L ∪ {x}] is disconnected in all cases. Therefore, RT is consistent.
7.5 constructing a least resolved species tree 127
Since Theorem 7.2, it is possible to check the consistency of a strict dense triple set RT, by repeatedly applying the inference rules of order two and ensuring that only triples are inferred that were already contained inRT.
It remains to show that, given a (possibly inconsistent) set of specie triples S and a maximal consistent subset S∗ ⊆ S, there always exists a consistent strict dense setS0 withS∗⊆S0.
Lemma 7.8. LetRT be a consistent set of triples on L. Then there is a strict dense