In this section, we prove the second Thomas conjecture. It is stated as follows.
Theorem 5.1 Let L be a finite distributive lattice and F : L→ L. Let D be a cyclic attractor of G(F ). Then∪
x∈LΓ(F′(x)) contains a negative circuit.
As an immediate consequence of Theorem 5.1, we have
Theorem 5.2 Let L be a finite distributive lattice. If F : L → L has no fixed point, then∪
x∈LΓ(F′(x)) contains a negative circuit.
Proof. If F has no fixed point, then G(F ) contains a cyclic attractor. By Theorem 5.1,∪
x∈LΓ(F′(x)) contains a negative circuit. To prove Theorem 5.1, we need the following lemmas.
Lemma 5.3 Let L be a finite lattice and D be an attractor. For x, y ∈ D, there exists a path from x to y in G(F ).
Proof. Suppose that there exist x, y ∈ D such that there has no path from x to y in G(F ). Let D′ = {z : there exists a path from x to z in G(F )}.
Then D′ is a trap domain and y /∈ D′. So D′ D, which contradicts the
smallest trap domain D.
Lemma 5.4 Let L be a finite distributive lattice. If (x, y) ∈ G(F ), then {i : xi̸= yi} ⊂ IF(x).
Proof. If (x, y)∈ G(F ), then there exists i ∈ IF(x) such that y = ˜xi. We may assume that x < y. Let j∈ {i : xi̸= yi}, we have xj < yj.
Case 1. ai < aj. According to Birkhoff’s representation theorem, yj = (˜xi)j = 0L= xj, which is a contradiction.
Case 2. aiand aj are incomparable. According to Birkhoff’s representation theorem, yj = (˜xi)j = 0L= xj, which is a contradiction.
Case 3. aj < ai. Since fi(x) = ai, it implies that aj < ai ≤ F (x) =∨
{fi(x) : i = 1, . . . , n}.
Thus fj(x) = aj ̸= xj, j∈ IF(x). Let G1 and G2 be two directed graphs with the same set of vertices V and with set of edges A1 and A2 respectively. We say that G1 is a subgraph of G2 if A1 ⊂ A2. For x∈ L, we set
si(x) =
1 if fi(x) > xi, 0 if fi(x) = xi,
−1 if fi(x) < xi.
Lemma 5.5 Let L be a finite distributive lattice and F : L → L. Let {x0, x1, . . . , xk} be a path of G(F ) of length k ≥ 1, and let i ∈ IF(xk). If si(xp) ̸= si(xk) for all 0 ≤ p < k, then there exists j ∈ IF(x0) such that
∪
x∈LΓ(F′(x)) has a path from j to i with sign sj(x0)si(xk).
Proof. We prove the assertion by the induction on the length of the path k. In case k = 1, by hypothesis, si(x0) ̸= si(x1) and x1 = fx0l for some l∈ IF(x0).
Case 1. dp(x0, x1) = 1. We set l = j.
Case 1.1. i̸= j. Then x0i = x1i. We may assume that si(x1) = 1. Thus fi(fx0j) = fi(x1) > x1i = x0i ≥ fi(x0), it implies that fij(x0) = 1. If x0j = 0L, then sj(x0) = 1 and x0j = fi(x0) = 0L. It follows that Γ(F′(x0)) has an edge from j to i with positive sign. If x0j = aj, then sj(x0) = −1. Since fi(x0) = 0L, Γ(F′(x0)) has an edge from j to i with negative sign. Thus Γ(F′(x0)) has an edge from j to i with sign sj(x0)si(xk) when i̸= j.
Case 1.2. i = j. Then si(x0)si(x1) =−1. We may assume si(x1) = 1.
Thus fi(fx0i) = fi(x1) > x1i = fi(x0) and x0i ̸= fi(x0), it implies that fii(x0) = 1 and either x0i = 0L or fi(x0) = 0L. Thus G(F′(x0)) has an edge from j to i with negative sign. This completes the proof of Case 1.
Case 2. dp(x0, x1) > 1. Let {x0 = p0, p1,· · · , pm = x1} be a path which connects x0 and x1 with pr < pr+1 or pr > pr+1 for all r = 0, 1,· · · , m − 1 in the diagram of L.
Case 2.1. i /∈ {s : p0s ̸= pms }. There exists 0 ≤ q ≤ m − 1 such that si(pq) ̸= si(pq+1) = si(x1). By Lemma 3.5 and Lemma 3.6, there exists j such that epq j = pq+1. By lemma 5.4, j ∈ {s : pqs ̸= pq+1s } ⊂ {s : p0s ̸=
pms } = {s : x0s ̸= x1s} ⊂ IF(x0), and pqi = pq+1i . We may assume that
si(pq+1) = 1. Thus fi( epq j) = fi(pq+1) > pq+1i = pqi ≥ fi(pq), it implies that fij(pq) = 1. If sj(x0) = 1, then fj(x0) > x0j = 0L. Thus x0 < x1 and pq < pq+1, we have shown that Γ(F′(pq)) has an edge from j to i with positive sign. If sj(x0) = −1, then 0L = fj(x0) < x0j. Thus x1 < x0 and pq+1 < pq, we have shown that Γ(F′(pq)) has an edge from j to i with negative sign. Thus Γ(F′(pq)) has an edge from j to i with sign sj(x0)si(xk) when i /∈ {s : p0s̸= pms }.
Case 2.2. i∈ {s : p0s ̸= pms }. By lemma 5.4, i ∈ {s : p0s ̸= pms } = {s : x0s ̸= x1s} ⊂ IF(x0), and si(x0)si(x1) = si(p0)si(pm) =−1. We may assume that si(pm) = 1 and si(p0) = −1. Since fi(pm) > pmi and fi(p0) < p0i, it implies that p0i > pmi and pr > pr+1 for all r = 0, 1,· · · , m−1 in the diagram of L. There exists 0≤ q ≤ m − 1 such that fi(pq) = 0L and fi(pq+1) = ai. By Lemma 3.5 and Lemma 3.6, there exists j such that epq j = pq+1. It follows that Γ(F′(pq)) has an edge from j to i with negative sign. We have shown that Γ(F′(pq)) has an edge from j to i with sign sj(x0)si(xk) when i∈ {s : p0s ̸= pms }. This completes the proof of Case 2.
In case k > 1, let i∈ IF(xk). By the induction hypothesis,{xk−1, xk} is a path of G(F ) of length 1.
Case 1. dp(xk−1, xk) = 1. There exists l∈ IF(xk−1) such that xk = ]xk−1
l
. Thus Γ(F′(xk−1)) has an edge from l to i with sign sl(xk−1)si(xk).
Case 2. dp(xk−1, xk) > 1. There exists a path {xk−1 = p0, p1,· · · , pm = xk} which connects xk+1 and xk with pr < pr+1 or pr > pr+1 for all r = 0, 1,· · · , m − 1 in the diagram of L.
Case 2.1. i /∈ {s : p0s ̸= pms }. Since si(xk−1) = si(p0) ̸= si(xk) = si(pm), fi(pm) ̸= pmi = p0i. It implies that p0i = fi(p0). (If p0i ̸= fi(p0), it implies that si(p0) = si(pm), which is a contradiction.) There exists 0 ≤ q ≤ m − 1 such that pqi = fi(pq) and pq+1i ̸= fi(pq+1). By Lemma 3.5 and Lemma 3.6, there exists l such that epq l = pq+1. By lemma 5.4, l∈ {s : pqs̸= pq+1s } ⊂ {s : p0s̸= pms } = {s : xks−1̸= xks} ⊂ IF(xk−1) = IF(p0), and si(xk) = si(pm) = si(pq+1). We may assume that si(pq+1) = 1. Thus fi( epq j) = fi(pq+1) > pq+1i = pqi = fi(pq), it implies that fil(pq) = 1. If sl(p0) = 1, then fl(p0) > p0l = 0L. It implies that pq < pq+1. Thus Γ(F′(pq)) has an edge from l to i with positive sign. If sl(p0) = −1, then 0L = fl(p0) < p0l. It implies that pq+1 < pq. Thus Γ(F′(pq)) has an edge
from l to i with negative sign. We have shown that Γ(F′(pq)) has an edge from l to i with sign sl(xk−1)si(xk) when i /∈ {s : p0s̸= pms }.
Case 2.2. i∈ {s : p0s ̸= pms }. By lemma 5.4, i ∈ {s : p0s ̸= pms } = {s : xks−1 ̸= xks} ⊂ IF(xk−1) = IF(p0), and si(xk−1)si(xk) = si(p0)si(pm) =−1.
We may assume that si(pm) = 1 and si(p0) =−1. Since fi(pm) > pmi and fi(p0) < p0i, it implies that p0i > pmi and pr > pr+1 for all r = 0, 1,· · · , m − 1 in the diagram of L. There exists 0 ≤ q ≤ m − 1 such that fi(pq) = 0L and fi(pq+1) = ai. By Lemma 3.5 and Lemma 3.6, there exists l such that peq l = pq+1. It follows that Γ(F′(pq)) has an edge from l to i with negative sign. We have shown that Γ(F′(pq)) has an edge from l to i with sign sl(xk−1)si(xk) when i∈ {s : p0s ̸= pms }. This completes the proof of Case 2.
According to Cases 1 and 2, there exist l∈ IF(xk−1) and a point x∈ L such that Γ(F′(x)) has an edge from l to i with sign sl(xk−1)si(xk). Consider the smallest 0≤ p < k such that sl(xp) = sl(xk−1). If p = 0, then l∈ IF(x0) and Γ(F′(]xk−1)) has an edge from l to i with sign sl(x0)si(xk). Let j = l.
Then ∪
x∈LΓ(F′(x)) contains a path from j to i with sign sj(x0)si(xk). If p > 0, then sl(xi′)̸= sl(xp) for all 0≤ i′ < p. The path{x0, . . . , xp} satisfies the conditions of the lemma for l ∈ IF(xp). By the induction hypothesis, there exists j ∈ IF(x0) such that ∪
x∈LΓ(F′(x)) has a path from j to l with sign sj(x0)sl(xp). Since Γ(F′(xk−1)) has an edge from l to i with sign sl(xk−1)si(xk), the graph∪
x∈LΓ(F′(x)) contains a path from j to i with sign sj(x0)sl(xp)sl(xk−1)si(xk). Since sl(xp) = sl(xk−1),∪
x∈LΓ(F′(x)) contains a path from j to i with sign sj(x0)si(xk).
Lemma 5.6 L is a finite distributive lattice and F : L → L. Let D be a cyclic attractor of G(F ). If there exists x∈ D such that |IF(x)| = 1, then
∪
x∈LΓ(F′(x)) contains a negative circuit.
Proof. Suppose that there exists x0 ∈ D such that IF(x) ={i}. We have dp(x, ˜xi) = 1 by Lemma 3.6. (If dp(x, ˜xi)≥ 2, then there exists j such that dp(x, ˜xj) = 1 and xj ̸= ˜xij. If xi = 0L, we have xj = 0L and aj < ai. Since fi(x) = ai, it implies that fj(x) = aj and j∈ IF(x), which is a contradiction.
Similarly, if xi = ai, then xj = aj and ai < aj. Since fi(x) = 0L, it implies that fj(x) = 0L and j∈ IF(x), which is a contradiction.)
We need only to prove that the case si(x0) = 1, the other case being similar. Let x1 = fx0i. Then G(F ) has an edge from x0 to x1 and x0i < x1i. Since x0 ∈ D, we have x1 ∈ D and there exists a path {x1, . . . , xk = x0} from x1 to x0, and xi ∈ D for all i = 1, . . . , k. If si(xq)≥ 0 for all 1 ≤ q < k, then xqi ≤ xq+1i for all 1≤ q < k. We have x0i < x1i ≤ xki = x0i, which is a contradiction. Thus, there exists a smallest 1 ≤ p < k such that fi′(xp) =
−1. We have {x1, . . . , xp} is a path in G(F ) and i ∈ IF(xp), si(xq)̸= si(xp) for all 0 ≤ q < p. According to Lemma 5.5, there exists j ∈ IF(x0) such that ∪
x∈LΓ(F′(x)) has a path from j to i with sign sj(x0)si(xp). Since IF(x0) = {i}, we have j = i and sj(x0)si(xp) = si(x0)si(xp) = −1. Thus
∪
x∈LΓ(F′(x)) has a negative circuit from i to i. We proceed now to prove Theorem 5.1.
If there exists x∈ D such that |IF(x)| = 1, then∪
x∈LΓ(F′(x)) contains a negative circuit by Lemma 5.6. We may assume that |IF(x)| ≥ 2 for all x∈ D. Let J =∪
x∈DIF(x) and i1 ∈ J be such that ai ≤ ai1 or ai and ai1 are incomparable for all i∈ J with i ̸= i1. Let H : L→ L be defined by
H(x) =∨
{xi1, fi(x) : i̸= i1}.
If x∈ D, then IH(x)⊂ IF(x). Let i ∈ IH(x) and hi(x)̸= xi. If i /∈ IF(x), then fi(x) = xi. If xi = ai, then ai ≤ H(x) and hi(x) = ai = xi, which is a contradiction. If xi = 0L, then j /∈ IF(x) for all j such that ai ≤ aj. Thus, ai H(x) and hi(x) = 0L = xi, which is a contradiction. It follows that IH(x)⊂ IF(x).
We may assume that xi1 = 0L. If ii ∈ I/ F(x), then H(x) = F (x) and hence IH(x) = IF(x). If ii ∈ IF(x), then i1 ∈ H(x) by definition of H. If/ i∈ IF(x)\ {i1} and xi = 0L, then fi(x) = ai and hence hi(x) = ai. We have i∈ IH(x). If i∈ IF(x)\ {i1} and xi= ai, then fi(x) = 0L. Thus fj(x) = 0L
for all j such that ai ≤ aj. We have ai H(x) and hence hi(x) = 0L̸= xi
and i∈ IH(x). In particular, if xi1 = 0L, then IH(x) = IF(x)\ {i1} when ii ∈ IF(x) or IH(x) = IF(x) when ii∈ I/ F(x).
We now prove that D is a trap domain of G(H). Let x∈ D and IH(x)⊂ IF(x). Since D is a trap domain of G(F ). If i ∈ IH(x) ⊂ IF(x), then
˜
xi ∈ D. It follows that D is a trap domain of G(H). By definition, G(H) contains at least one attractor D′ ⊂ D. We may assume that xi1 = 0L for
all x ∈ D′. Then there exists x ∈ D satisfies xi1 = 0L. Since |IF(x)| ≥ 2 and IH(x) = IF(x)\ {i1} when ii ∈ IF(x) or IH(x) = IF(x) when ii∈ I/ F(x) for all x ∈ D′, we have |D′| ≥ 2. It follows that D′ is a cyclic attractor of G(H). For x ∈ D′, consider the principle submatrix M (x) = (mij(x)) of H′(x) defined by
mij(x) = {
hij(x) i, j∈ J, 0 otherwise.
Since xi1 = 0L, we have hi(x) = fi(x) for all i∈ J and M(x) is the principle submatrix of F′(x). We have Γ(M (x)) is a subgraph of Γ(F′(x)). If there exists x ∈ D′ such that |IH(x)| = 1, then ∪
x∈LΓ(M (x)) has a negative circuit by Lemma 5.6. Thus∪
x∈LΓ(F′(x)) has a negative circuit.
If|IH(x)| ≥ 2 for all x ∈ D′, consider J′ =∪
x∈LIH(x) and i2∈ J′ such that ai≤ ai2 or ai and ai2 are incomparable for all i∈ J′\ {i2}. Repeating the above process, we obtain a function ¯H and a cyclic attractor ¯D such that ¯D⊂ D and |IH¯(x)| = 1 for some x ∈ ¯D. It follows that∪
x∈LΓ(F′(x)) has a negative circuit and the proof is completed.