Proof. We prove \(\ell(L)\leq |\mathrm{JIrr}(L)|\) as for the meet-irreducibles it is dual. Let \(\phi: \phi(0)\lessdot \cdots \lessdot \phi(n)\) be a maximal chain of length \(n\). For all \(i<n\), there is always a join-irreducible that is below \(\phi(i+1)\) but not \(\phi(i)\). If it was not the case, then \(\phi(i+1)\), which is the join of the join-irreducibles below it, would be below \(\phi(i)\), which is absurd. This proves that we have at least \(n\) join-irreducibles. ◻

Proof. We only prove the statement related to the coatoms, as the rest is straightforward. Any element of the form \(\mathcal{C}_P^{\phi} \setminus \{m\}\), without assumptions on \(m\), is a coatom. Let \(k\geq 0\). The element \(\mathcal{C}_P^{\phi} \setminus \mathcal{C}_k\) is torclosed, and any element that covers \(\mathcal{C}_P^{\phi} \setminus \mathcal{C}_k\) contains \((\phi(k),k)\), which is the maximum of \(\mathcal{C}_k\). Suppose that every minimal element of \(I_P(\phi(k))\) is an element in \(I_P(\phi(k-1))\). In this case, the minimal torclosed set that contains both the elements of \(\mathcal{C}_{k-1}\) and the element \((\phi(k),k)\) is \(\mathcal{C}_{k-1} \sqcup \mathcal{C}_k\). This proves that \(\mathcal{C}_P^{\phi} \setminus \mathcal{C}_k\) is a coatom.

Conversely, suppose that \(T\) is a coatom. If there exist two elements \(i<j\) such that \(T_i\neq \mathcal{C}_i\) and \(T_j\neq \mathcal{C}_j\), then \(T < T\vee \mathcal{C}_j \neq \mathcal{C}_P^{\phi}\) since \(T\vee \mathcal{C}_j\) does not contain \(\mathcal{C}_i\), which proves that \(T\) is not a coatom. Thus there exists a unique \(k\) such that \(T_k \neq \mathcal{C}_k\). Minimal elements of \(I_P(\phi(k))\) that are not in \(I_P(\phi(k-1))\) cannot be obtained through a completion. Thus we must have \(T=\mathcal{C}_P^{\phi} \setminus \{m\}\) with \(m\) such an element if these minimal elements exist. But if there are no such elements, we saw that we must have \(T=\mathcal{C}_P^{\phi} \setminus \mathcal{C}_k\) since the minimal torclosed set containing \(T\cup \{(\phi(k),k)\}\) is the maximum \(\mathcal{C}_P^{\phi}\). ◻

Proof. Recall that \(T<T'\) means \(T\subsetneq T'\). Let \(k\) be the biggest nonnegative integer such that \(T'_k\setminus T_k \neq \emptyset\). Let \(R\) be the subset of \(T'\) obtained from \(T\) by replacing \(T_k\) with \(T'_k\). Then \(R\) is torclosed, and \(T<R\leq T'\). Since \(T\lessdot T'\), we have \(T'=R\). This proves that there exists \(k<n\) such that \(T'\setminus T \subseteq \mathcal{C}_k\). Now let us prove that \(T'\setminus T\) has a maximum. We know that \(T'\setminus T \subseteq \mathcal{C}_k\). Seeking a contradiction, suppose that \(m_1\) and \(m_2\) are two maximal elements of \(T'\setminus T\). It could be that one of them is of the form \((\phi(i),k)\) with \(i\leq k\). But since these latter elements form a chain in \(\mathcal{C}_k\), and \(m_1\) and \(m_2\) are incomparable, without loss of generality we can assume that \(m_1\) is not of the form \((\phi(i),k)\) with \(i\leq k\). Thus \(T\cup \{m_1\}\) is torclosed, but it is strictly between \(T\) and \(T'\), which is a contradiction with the fact that \(T\lessdot T'\) is a cover relation. ◻

Proof. It is immediate that these order filters are join-irreducible elements. We need to prove that these are the only join-irreducible elements. Seeking a contradiction, suppose that \(T\in \mathrm{JIrr}(\mathop{\mathrm{Tam}}(P,\phi))\) is not of the previous form. Let \(k\geq 0\) be minimal such that \(T_k \neq \emptyset\). If \(T_k \subseteq \mathcal{C}_k\) has two minimal elements \(a\) and \(b\), then \(T\) covers both \(T\setminus \{a\}\) and \(T\setminus \{b\}\), which is absurd since \(T\) is a join-irreducible element. Thus there exists \(a\) such that \(T_k\) is the principal order filter of \(\mathcal{C}_k\) generated by \(a\). Then the only torclosed set that \(T\) covers is \(T\setminus \{a\}\). By hypothesis \(T\neq T_k\), thus there exists \(j>k\) such that \(T_j\neq \emptyset\). The subset \(T\setminus T_j\) is torclosed, and \((T\setminus T_j) < T\). But \((T\setminus T_j) \not\leq T\setminus \{a\}\), which is absurd since the only element that \(T\) covers is \(T\setminus \{a\}\).

The other statements follow immediately from what we just proved. ◻

Proof. We start by proving that any meet-irreducible is of one of the two kinds. Let \(T\) be a meet-irreducible, covered by \(T^*\). Let \(k<n\) be the biggest nonnegative integer such that \(T_k\neq \mathcal{C}_k\). Since \(T<T\cup \mathcal{C}_k\), then \(T^{*}\setminus T \subseteq \mathcal{C}_k\). Moreover, by 8, \(T^{*}\setminus T\) has a maximum element \((x,k)\). In particular, \((x,k)\) is a maximal element of \(\mathcal{C}_k \setminus T_k\). In fact, for any maximal element \(m\) of \(\mathcal{C}_k \setminus T_k\) such that \(m\neq (\phi(i),k)\) for every \(i\leq k\), we have \(T\lessdot T\cup \{m\}\). Since \(T\) is meet-irreducible and the elements \((\phi(i),k)\) for \(i\leq k\) form a chain, then \((x,k)\) is the maximum of \(\mathcal{C}_k \setminus T_k\), or there exists \(j\leq k\) such that the maximal elements of \(\mathcal{C}_k \setminus T_k\) are \((x,k)\) and \((\phi(j),k)\). Let us prove that in the former case \(T=M_{(x,k)}\), whereas in the latter case \(T=M_{(x,k),(\phi(j),k)}\). In the remainder of this proof, for these two different cases we will refer to the former case and to the latter case.

Let \(A_i=\{(y,i)\in \mathcal{C}_P^{\phi} \mid y\not\leq x\}\) and \(B_i=\{(y,i) \in \mathcal{C}_P^{\phi} \mid y\not\leq x,\,y\not\leq \phi(j)\}\). In the former case we have \((\phi(l(x)+1),k)\in T\) but \((x,k)\not\in T\), and in the latter case we have \((\phi(j+1),k)\in T\) but \((x,k)\not\in T\) and \((\phi(j),k)\not\in T\). Since \(T\) is torclosed, then in the former case for all \(i\in \{l(x),l(x)+1,\dots,k\}\), we have \(T_i\subseteq A_i\), and in the latter case for all \(i\in \{j,j+1,\dots,k\}\), we have \(T_i\subseteq B_i\).

Then this forces in the former case to have \(T_{<l(x)} = \bigsqcup_{0\leq i<l(x)} \mathcal{C}_i\). Indeed, since \(T_i\subseteq A_i\) for all \(i\in \{l(x),l(x)+1,\dots,k\}\), then \((\phi(p),i)\not \in T\) for all \(p\leq l(x)\) and \(i\in \{l(x),l(x)+1,\dots,k\}\). Moreover, \(T_q=\mathcal{C}_q\) for all \(k<q<n\), which proves that the addition of elements of \(\sqcup_{i<l(x)} \,\mathcal{C}_i\) to \(T\) does not affect the others components of \(T\). Thus \(T\leq T\vee \big(\sqcup_{i<l(x)} \,\mathcal{C}_i\big) = T\cup \big(\sqcup_{i<l(x)} \,\mathcal{C}_i\big)\). If the inequality is strict, then \(T^{*}\setminus T \subseteq \sqcup_{i<l(x)} \,\mathcal{C}_i\), which is absurd. Thus, we have \(T=T\cup \big(\sqcup_{i<l(x)} \,\mathcal{C}_i\big)\). This proves that \(T_{<l(x)} = \bigsqcup_{0\leq i<l(x)} \mathcal{C}_i\). We would prove very similarly, by replacing \(l(x)\) by \(j\), that in the latter case \(T_{<j} = \bigsqcup_{0\leq i<j} \mathcal{C}_i\).

Moreover, it is immediate that \(\sqcup_{l(x)\leq i\leq k}\,A_i\) and \(\sqcup_{j\leq i\leq k}\,B_i\) are torclosed sets, and with the same argument as above we have that \(T_{<l(x)} \sqcup \big(\sqcup_{l(x)\leq i\leq k}\,A_i\big) \sqcup T_{>k}\) and \(T_{<j} \sqcup \big(\sqcup_{j\leq i\leq k}\,B_i\big) \sqcup T_{>k}\) are torclosed. But both of these torclosed sets contain \(T\), and since they can only differ in components that are not the \(k\)-th component, then in fact they are equal to \(T\) in the respective cases.

It remains to prove that the two torclosed sets \(M_{(x,k)}\) and \(M_{(x,k),(\phi(j),k)}\) are meet-irreducibles. Adding to these torclosed sets respectively an element \((y,i) \not\in A_i\), for \(i\in \{l(x),l(x)+1,\dots,k\}\), or \((y,i) \not\in B_i\), for \(i\in \{j,j+1,\dots,k\}\), will by completion add \((y,k)\) to each of these torclosed sets. Then any torclosed set that covers one of these two torclosed sets add only elements of the \(k\)-th component. Thus any element that covers \(M_{(x,k)}\) contains \((x,k)\), then \(M_{(x,k)}\) is meet-irreducible and its only cover is the minimal torclosed set containing \(M_{(x,k)}\cup \{(x,k)\}\). For \(M_{(x,k),(\phi(j),k)}\), the minimal torclosed set containing \(M_{(x,k),(\phi(j),k)}\cup \{(\phi(j),k)\}\) contains also \((x,k)\). Indeed, \((\hat{0},j-1) \in M_{(x,k),(\phi(j),k)}\) thus by completion \((\hat{0},k)\) is in this minimal torclosed set, then \((x,k)\) also. Thus \(M_{(x,k),(\phi(j),k)}\) has only one cover which is \(M_{(x,k),(\phi(j),k)} \cup \{(x,k)\}\). ◻

Proof. We use 3. Let \(T\lessdot T'\) in \(L\). By 8, there exists \(k<n\) such that \(T'\setminus T \subseteq \mathcal{C}_k\) and it has a maximum element \(m\). The torclosed set generated by \(m\), which is \(J_m\), is the minimum element of \(I_L(T')\setminus I_L(T)\). Using 3, this proves that \(L\) is join-semidistributive.

The second statement follows from 2, with the number of join-irreducibles given in 11 which is equal to the number of meet-irreducibles if and only if we do not have meet-irreducibles of the second kind as described in 12. ◻

Proof. If \(\phi\) has one element, then the torclosed sets are the order filters of \(I_P(\phi(0))\). Then the result follows from the fundamental theorem of finite distributive lattices. Suppose that \(|\phi|\geq 2\). By 16, \(\ell(\mathop{\mathrm{Tam}}(P,\phi))=|\mathcal{C}_P^{\phi}|\). But all maximal chains in the order filter of \(\mathop{\mathrm{Tam}}(P,\phi)\) generated by the atom \(\{(\phi(0),0)\}\) have length at most \(|\mathcal{C}_P^{\phi}|-2=\ell(\mathop{\mathrm{Tam}}(P,\phi))-2\). Indeed, the moment we will add \((\phi(1),1)\), the completion will also add with it at least another element \((\phi(0),1)\). Thus the maximal chains of \(\mathop{\mathrm{Tam}}(P,\phi)\) do not have all the same length, which means that \(\mathop{\mathrm{Tam}}(P,\phi)\) is not graded. ◻

Proof. Let \(T\in L\) be a torclosed set that is not \(\emptyset\) neither \(\mathcal{C}_P^{\phi}\). Let \(I=\{i<n \mid T_i\neq \emptyset\}\). Since \(\phi(0)=\hat{0}\) and \(T\neq \mathcal{C}_P^{\phi}\), then \(J:=\{0,1,\dots,n-1\}\setminus I\) is not empty. Indeed, if \(J=\emptyset\) then \((\phi(i),i)\in T\) for all \(i<n\), and successive completions would give \((\hat{0},i)\in T\) for all \(i<n\), but then \(T=\mathcal{C}_P^{\phi}\). Let \(T':=\bigsqcup_{i\in J} \mathcal{C}_i\). Then \(T'\) is torclosed and \(T\wedge T'=\emptyset\). Also, we have \(T\vee T' = \mathcal{C}_P^{\phi}\) for the same reason as why \(J\neq \emptyset\) using \(\phi(0)=\hat{0}\). This proves that \(T'\) is a complement of \(T\). Thus \(L\) is complemented. Moreover, it is immediate that \(T'=\max\,\{R\in L\mid T\wedge R=\emptyset \}\), which proves that \(L\) is meet-pseudocomplemented. We conclude that \(L\) is pseudocomplemented using for example [20], which proves that a complemented lattice is pseudocomplemented if and only if it is meet-pseudocomplemented. ◻

Proof. By definition, a linear extension of the dual of \((\mathcal{C}_P^{\phi},\leq_{prod})\) is one that takes \((y,j)\) before taking \((x,i)\), for all \(j\geq i\) and \(y\geq x\) in \(P\), where \(x\leq \phi(i)\) and \(y\leq \phi(j)\). It follows that if \(e_1\prec e_2 \prec \cdots \prec e_{m}\) is a linear extension of the dual of \((\mathcal{C}_P^{\phi},\leq_{prod})\), then \(\{e_1,e_2,\dots ,e_k\}\) is torclosed for any \(k\in [m]\). Thus \(\emptyset \subsetneq \{e_1\} \subsetneq\{e_{1}, e_{2}\} \subsetneq\dots \subsetneq\{e_1,e_2,\dots , e_{m} \}=\mathcal{C}_P^{\phi}\) is a maximal chain of \(\mathop{\mathrm{Tam}}(P,\phi)\), and is obviously of longest length as we only add one new element at each step. Thus the longest chains are of length \(m=|\mathcal{C}_P^{\phi}|\), which is the length of \(\mathop{\mathrm{Tam}}(P,\phi)\). Conversely, let \(\emptyset \subsetneq \{e_1\} \subsetneq\{e_{1}, e_{2}\} \subsetneq\dots \subsetneq\{e_1,e_2,\dots , e_{m} \}=\mathcal{C}_P^{\phi}\) be a longest chain of \(\mathop{\mathrm{Tam}}(P,\phi)\) (now we know that they have this form). Let \(T(i)=\{e_1,\,e_2,\dots , e_i\}\). Then the longest chain is \(\emptyset \subsetneq T(1) \subsetneq T(2) \subsetneq \cdots \subsetneq T(m)\). Let us prove that the sequence \(e_1,\,e_2,\dots,\,e_m\) defines a linear extension of the dual of \((\mathcal{C}_P^{\phi},\leq_{prod})\). Seeking a contradiction, suppose that there exists \(i<j\) such that \(e_i \leq_{prod} e_j\). Note that since \(i<j\), then \(e_j\not\in T(i)\). Let \(e_i=(x,k)\) and \(e_j=(y,p)\). Then \(e_i \leq_{prod} e_j\) means \(x\leq y\) in \(P\) and \(k\leq p\). If \(p=k\), since \(e_i=(x,k)\leq (y,k)=e_j\) and \(e_i\in T(i)\) which is torclosed, then \(e_j\in T(i)\). This is absurd, thus \(p>k\).

Let \(e_l=(\phi(k+1),p)\). If \(e_l\in T(i)\), then by completion we have \((x,p)\in T(i)\). But \(e_j=(y,p)\geq (x,p)\), thus since \(T(i)\) is torclosed we have \(e_j\in T(i)\). This is absurd, thus \(e_l\not\in T(i)\). Then \(i<l\). Let us explain why this is impossible. Since \((x,p)<(\phi(k+1),p)=e_l\) then a torclosed set that contains \((x,p)\) also contains \(e_l\). Since both these elements are not in \(T(i)\) and following the chain \(\emptyset \subsetneq T(1) \subsetneq T(2) \subsetneq \cdots \subsetneq T(m)\) we add only a unique element at each step, we will add to a torclosed set that contains \(T(i)\) the element \(e_l\) before adding later the element \((x,p)\). But we proved that as soon as a torclosed set contains both \(T(i)\) and the element \(e_l\), then it contains \((x,p)\). This is absurd. This finishes the proof of the proposition. ◻

Proof. This follows from 11 and 16. ◻

Proof. This follows from 16 where we proved that the torclosed sets on a longest chain of \(\mathop{\mathrm{Tam}}(P,\phi)\) are the order ideals of the linear extensions of the dual of \((\mathcal{C}_P^{\phi},\leq_{prod})\). ◻

Proof. The case \(|\phi|=1\) is immediate. If \(\phi\) has two elements, then a torclosed set may or may not contain \((\phi(1),1)\). The number of torclosed sets that do not contain \((\phi(1),1)\) is \(|J(I_0)|\) since they correspond to the torclosed sets \(T\) such that \(T_1=\emptyset\). On the other hand, the number of torclosed sets containing \((\phi(1),1)\) is \(|J(\mathcal{C}_P^{\phi},\leq_{prod})|-1\). Indeed, in this case an order filter \(T\) of \(\mathcal{C}_P^{\phi}\) is torclosed if and only if \((x,0) \in T\) implies \((x,1)\in T\). ◻

Proof. We first count the torclosed sets that do not contain \((\phi(2),2)\). Among these, the number of torclosed sets that do not contain \((\phi(1),1)\) is \(2\), whereas we have \(|J(I_1)|\) such torclosed sets that contain \((\phi(1),1)\). Now, let us count the torclosed sets that contain \((\phi(2),2)\). Among these, the number of torclosed sets that do not contain \((\phi(0),0)\) is \(|J(\mathcal{C}_P^{\phi},\leq_{prod})|-2\), whereas we have \(|J(I_2\setminus I_1)|+1\) such torclosed sets that contain \((\phi(0),0)\). The result follows by summing the cardinalities of these disjoint families of torclosed sets. ◻

Proof. The first equality \(\gamma_{1,\psi_{\mathcal{L}}}=\omega_{\mathcal{L}}\) follows easily from the definitions of the labellings, since with our conventions \(\delta_{\psi_{\mathcal{L}}}(J_{e_i})=i\) for all \(i\in [m]\). The second equality requires our particular choice of linear extension \(\mathcal{L}\). Recall that \(\gamma_{2',\psi_{\mathcal{L}}}(T\lessdot T') = \min\{i\mid T\vee T(i)\geq T'\}\). Since \(T\lessdot T'\) and \(\max(T'\setminus T)=e_{\omega_{\mathcal{L}}(T\lessdot T')}\), we have \(T\vee T(i) \geq T'\) if and only if \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\in T\vee T(i)\). Thus \(\gamma_{2',\psi_{\mathcal{L}}}(T\lessdot T')\leq \omega_{\mathcal{L}}(T\lessdot T')\). Seeking a contradiction, suppose that there exists \(j<\omega_{\mathcal{L}}(T\lessdot T')\) such that \(\gamma_{2',\psi_{\mathcal{L}}}(T\lessdot T')=j\). Then \(T\vee T(j) \geq T'\), thus \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\in T\vee T(j)\). By definition \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\not\in T\), and since \(j<\omega_{\mathcal{L}}(T\lessdot T')\), we have \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\not\in T(j)\). Then \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\not\in T\cup T(j)\). But \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\in T\vee T(j)\), which proves that \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\) is obtained with a completion. By minimality of \(j\), we have \(T\vee T(j-1)\not\geq T'\), thus the completion that makes \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\in T\vee T(j)\) comes from the addition of \(e_j\) to \(T(j-1)\). Since by hypothesis \(j\neq \omega_{\mathcal{L}}(T\lessdot T')\), thus there exists \(p\leq k\) such that \(e_j=(\phi(p),k)\), where \(T'\setminus T \subseteq \mathcal{C}_k\). Moreover, \(e_j\) and \(e_{\omega_{\mathcal{L}}(T\lessdot T')}\) are incomparable. By definition of our particular choice of linear extension \(\mathcal{L}\), then \(e_{\omega_{\mathcal{L}}(T\lessdot T')} \prec e_j\), which means \(\omega_{\mathcal{L}}(T\lessdot T') < j\). This is absurd. ◻

Proof. From 1 and 20 follows the fact that \(\mathop{\mathrm{Tam}}(P,\phi)\) is left modular. Then \(\omega_{\mathcal{L}}\) is an EL-labelling by 2. ◻

Proof. Let \([T,T']\) be an interval of \(\mathop{\mathrm{Tam}}(P,\phi)\), and let \(I=\{i_1,\,i_2,\dots,\,i_k\}\) be the set of indices \(0\leq i_1<\cdots <i_k<n\) of the nonempty components of \(T'\setminus T\). Let us prove that we have at most one maximal decreasing chain in \([T,T']\) with respect to \(\omega_{\mathcal{L}}\). By our choice of linear extension \(\mathcal{L}\), if \(e_k\in\mathcal{C}_i\) and \(e_l\in\mathcal{C}_j\) with \(i<j\), thus \(l<k\). Then, since by 8 the cover relations happen between two torclosed sets that differ in only one component, any maximal decreasing chain of \([T,T']\) will contain the subchain \(T < T\cup T'_{i_1} < T\cup T'_{i_1} \cup T'_{i_2} <\cdots < T\cup T'_{i_1} \cup \cdots \cup T'_{i_k} =T'\). Seeking a contradiction, suppose that we have two different maximal decreasing chains \(C_1\) and \(C_2\) of \([T,T']\). As both chains contain the subchain described above, and two consecutive elements of this subchain differ in only one component, we look to the component \(\mathcal{C}_{i_j}\) with the least index where \(C_1\) and \(C_2\) differ. Suppose that the first time these chains differ, following them from bottom to top, is witnessed by \(R\lessdot S\) in \(C_1\) and \(R\lessdot U\) in \(C_2\). Since \(T\cup T_{i_1}'\cup \cdots \cup T_{i_j}'\) is torclosed and greater than both \(R\) and \(S\), the elements in components \(\mathcal{C}_p\) for \(p>i\) have no effect on which subsets of elements from \(\mathcal{C}_i\) can be added to \(R\) and \(S\). This just means that we add all the elements of \(T_{i_j}' \setminus R\) in \(C_1\) and \(C_2\) as if we did not have elements in components with bigger indices. By 8, if \(V\lessdot W\) is a cover relation in \(\mathop{\mathrm{Tam}}(P,\phi)\), then \(W\setminus V\) has a maximum element \(e_l\), and \(\omega_{\mathcal{L}}(V\lessdot W)=l\) by definition of \(\omega_{\mathcal{L}}\). Without loss of generality suppose that \(\omega_{\mathcal{L}}(R\lessdot S)<\omega_{\mathcal{L}}(R\lessdot U)\). Let \(V\lessdot W\) be the cover relation of \(C_1\) such that \(e_{\omega_{\mathcal{L}}(R\lessdot U)} \in W\setminus V\). Since \(e_{\omega_{\mathcal{L}}(R\lessdot U)}\) is a maximal element of \(T_{i_j}'\setminus R\), then \(\omega_{\mathcal{L}}(V\lessdot W)=\omega_{\mathcal{L}}(R\lessdot U)\). Thus the chain \(C_1\) is not decreasing. This is absurd.

Then the last sentence of the proposition follows by results recalled in 2.1 in the paragraph about poset topology. ◻

Proof. We suppose \(J_{(x,i)}\,D\,J_{(y,i)}\), which means that \(x\neq y\) and there exists \(p\in \mathop{\mathrm{Tam}}(P,\phi)\) such that \(J_{(x,i)}\leq J_{(y,i)}\vee p\) and \(J_{(x,i)}\not\leq (J_{(y,i)})_{*}\vee p\). Then \(J_{(x,i)}\not\leq J_{(y,i)}\), as noted at the end of 2.1. This means \((y,i)\not\leq (x,i)\), thus \(y\not\leq x\). It remains to prove that there exists \(k\leq i\) such that \(y=\phi(k)\). Seeking a contradiction, suppose that it is not the case. Then \(J_{(y,i)}\) and \((J_{(y,i)})_{*}=J_{(y,i)}\setminus \{(y,i)\}\) contain the same elements of the form \((\phi(k),i)\) for \(k\leq i\). Thus \((\,(J_{(y,i)})_{*}\vee p)_i = (\,J_{(y,i)}\vee p )_i\) or \((\,(J_{(y,i)})_{*}\vee p)_i = (J_{(y,i)}\vee p )_i \setminus \{(y,i)\}\). As \(J_{(x,i)}=(J_{(x,i)})_i\), we have \((J_{(x,i)})_i\leq (J_{(y,i)}\vee p)_i\) and \((J_{(x,i)})_i\not\leq (\,(J_{(y,i)})_{*}\vee p)_i\). In particular, \((\,(J_{(y,i)})_{*}\vee p)_i \neq (J_{(y,i)}\vee p )_i\), thus \((\,(J_{(y,i)})_{*}\vee p)_i = (J_{(y,i)}\vee p )_i \setminus \{(y,i)\}\). Then, we have \(J_{(x,i)}\leq (J_{(y,i)}\vee p)_i\) and \(J_{(x,i)}\not\leq (\,(J_{(y,i)})_{*}\vee p)_i= (J_{(y,i)}\vee p )_i \setminus \{(y,i)\}\). This means that \((x,i)=(y,i)\), which is absurd since \(y\not\leq x\).

Conversely, suppose that \(P\) has a minimum \(\hat{0}\), that \(\phi(0)=\hat{0}\), and let \((x,i)\) and \((y,i)\) be in \(\mathcal{C}_i\) such that there exists \(k\leq i\) such that \(y=\phi(k)\) and \(y\not \leq x\). Since \(y\not\leq x\), then \(y\neq \phi(0)=\hat{0}\). This proves that \(k\geq 1\). Let \(p=\mathcal{C}_{k-1} \in \mathop{\mathrm{Tam}}(P,\phi)\). We have \(J_{(x,i)} \subseteq \mathcal{C}_i \cup \mathcal{C}_{k-1} = J_{(\phi(k),i)} \vee p\) since this latter torclosed set contains \((\hat{0},k-1)\) and \((\phi(k),i)\), thus by completion contains \((\hat{0},i)=\min(\mathcal{C}_i)\). But \((J_{(\phi(k),i)})_{*} \vee p = (J_{(\phi(k),i)})_{*} \cup \mathcal{C}_{k-1}\) since \((\phi(k),i)\not\in (J_{(\phi(k),i)})_{*}\). Then \(J_{(x,i)}\not\leq (J_{(\phi(k),i)})_{*} \vee p\). This proves that \(J_{(x,i)}\,D\,J_{(y,i)}\). ◻

Proof. Let \((x,j)\) and \((y,i)\) be in \(\mathcal{C}_P^{\phi}\) such that \(i<j\), \(y\leq x\), \(\phi(i+1) \not \leq x\) and there is no \(y'\leq \phi(i)\) such that \(y<y'\leq x\). Let \(p=J_{(\phi(i+1),j)}\). We have both \((y,i)\) and \((\phi(i+1),j)\) in the torclosed set \(J_{(y,i)}\vee p\), thus by completion \((y,j)\in J_{(y,i)}\vee p\). Since \((x,j)\geq (y,j)\) and they are in the same component \(\mathcal{C}_j\), then \((x,j)\in J_{(y,i)}\vee p\). Thus \(J_{(x,j)}\leq J_{(y,i)}\vee p\).

If \(y=\phi(i)\), then \((J_{(y,i)})_{*}=\emptyset\). And since \(\phi(i+1)\not\leq x\), then \((x,j)\not\in J_{(\phi(i+1),j)} =(\,J_{(y,i)})_{*}\vee J_{(\phi(i+1),j)}\). This proves \(J_{(x,j)}\,D\,J_{(y,i)}\). Otherwise \(y\neq \phi(i)\). Then \((\,(J_{(y,i)})_{*}\vee J_{(\phi(i+1),j)})_j =\{(z,j)\mid \exists y'\in P,\, y< y'\leq \phi(i) \text{ and }y'\leq z\leq \phi(j)\}\). Then \((x,j)\) is not in this latter set since there is no \(y'\leq \phi(i)\) such that \(y<y'\leq x\). Thus \(J_{(x,j)}\not\leq (\,(J_{(y,i)})_{*}\vee J_{(\phi(i+1),j)})_j\), which proves that \(J_{(x,j)}\,D\,J_{(y,i)}\).

Conversely, suppose that we have \(J_{(x,j)}\,D\,J_{(y,i)}\), which means that there exists \(p\in \mathop{\mathrm{Tam}}(P,\phi)\) such that \(J_{(x,j)}\leq J_{(y,i)}\vee p\) and \(J_{(x,j)}\not\leq (J_{(y,i)})_{*}\vee p\). If \(j<i\), then \(J_{(y,i)}\) does not contribute in \(J_{(x,j)}\leq J_{(y,i)}\vee p\), which means that \(J_{(x,j)}\leq p\). Then \(J_{(x,j)}\leq (J_{(y,i)})_{*}\vee p\), which is absurd. This proves that \(i<j\). As observed, we do not have \(J_{(x,j)}\leq p\), thus \(x\not\in p\). Since \(J_{(x,j)}\leq J_{(y,i)}\vee p\), this means that \(J_{(y,i)}\) is essential when computing the join to contain \(J_{(x,j)}\), even though it is not in the same component. Thus we have completions involving \(J_{(y,i)}\), which proves both \((\phi(i+1),j)\in p\) and \((y,j)\leq (x,i)\). The latter gives \(y\leq x\), while the former gives \(\phi(i+1)\not\leq x\), because if it was not the case then \((x,j)\in p\) since \(p\) is torclosed, which is absurd. It remains to prove that there is no \(y'\leq \phi(i)\) such that \(y<y'\leq x\). Seeking a contradiction, suppose that such a \(y'\) exists. Then \((y',i)\in (J_{(y,i)})_{*}\), which with \((\phi(i+1),j)\in p\) gives by completion \((y',j)\in (J_{(y,i)})_{*}\vee p\). Since \(y'\leq x\), this gives \((x,i)\in (J_{(y,i)})_{*}\vee p\). Thus \(J_{(x,j)}\leq (J_{(y,i)})_{*}\vee p\), which is absurd. ◻

Proof. By 3, we need to prove that the \(D\) relation is acyclic. This follows from 9 and 10. ◻

Proof. Let \(K\) be a Kupisch-like series. Since \(\mathop{\mathrm{Tam}}(P,\phi)\) is join-congruence uniform (24), by 3 it suffices to prove that the set \(S\) of join-irreducibles that \(\equiv_K\) contracts satisfies that \(J_{(x,j)} D J_{(y,i)}\) and \(J_{(y,i)}\in S\) together imply \(J_{(x,j)}\in S\). We have \[J_{(y,i)}\in S \iff J_{(y,i)} \equiv_K (J_{(y,i)})_{*} \iff (y,i)\not\in R_K \iff y\not\geq \phi(K_i) .\]

Suppose that \(J_{(x,j)} D J_{(y,i)}\) and \(y\not\geq \phi(K_i)\). We need to prove that \(x\not \geq \phi(K_j)\). We consider two cases; first \(j=i\), then \(j\neq i\).

Suppose \(j=i\). By 9, \(J_{(x,i)} D J_{(y,i)}\) implies that there exists \(k\leq i\) such that \(y=\phi(k)\) and \(x\not\geq y\). Since \(\phi(k)=y\not\geq \phi(K_i)\), and \(\phi(k)\) is comparable to \(\phi(K_i)\), then \(\phi(K_i)> \phi(k)=y\). This proves that \(x\not \geq \phi(K_j)\), because if it was not the case then \(x\geq \phi(K_j)>y\), which is absurd since \(x\not\geq y\).

Suppose \(j\neq i\). By 10, \(J_{(x,j)} D J_{(y,i)}\) implies that \(i<j\), \(y\leq x\), \(\phi(i+1) \not \leq x\) and there is no \(y'\leq \phi(i)\) such that \(y<y'\leq x\). Seeking a contradiction, assume that \(x \geq \phi(K_j)\). Since \(x\not\geq \phi(i+1)\) and \(\phi(i+1)\) is comparable to \(\phi(K_j)\), then \(\phi(K_j)< \phi(i+1)\). Thus \(\phi(K_j)\leq \phi(i)\). Since \(y\not\geq \phi(K_i)\), either \(y<\phi(K_i)\), or \(\phi(K_i)\) and \(y\) are incomparable.

If \(y<\phi(K_i)\), then \(y<\phi(K_i)\leq \phi(K_j)\) since \(K\) is non-decreasing. Thus \(y<\phi(K_j)\leq x\), and we proved above that \(\phi(K_j)\leq \phi(i)\). This is absurd since we know that no such element \(y':=\phi(K_j)\) exists.

Otherwise \(\phi(K_i)\) and \(y\) are incomparable. Since \(P\) is a join-semilattice, we can define in \(P\) the element \(y':=\phi(K_j) \vee y\). We proved that \(\phi(K_j)\leq \phi(i)\), and by definition \(y\leq \phi(i)\). Then \(y'\leq \phi(i)\). By hypothesis, we have \(\phi(K_j)\leq x\) and \(y\leq x\), thus \(y'\leq x\). To have something absurd, it suffices to prove that \(y<y'\). By definition of \(y'\), we have \(y\leq y'\). If \(y=y'\), then \(\phi(K_j) \leq y\), and since \(\phi(K_i)\leq \phi(K_j)\) because \(K\) is non-decreasing, it would give \(\phi(K_i)\leq y\), which is absurd. Thus \(y<y'\). This finishes the proof of the proposition. ◻

Proof. The left modular property follows from 22. Using 13, it follows that \(\mathop{\mathrm{Tam}}(C_m,\phi)\) is semidistributive, and \(\mathop{\mathrm{Tam}}(C_m,\phi)\) is join-congruence uniform by 24. Thus \(\mathop{\mathrm{Tam}}(C_m,\phi)\) is congruence uniform since a semidistributive join-congruence uniform lattice is congruence uniform (this is an easy consequence of [21]). By 1, \(\mathop{\mathrm{Tam}}(C_m,\phi)\) is join-extremal, but since it is semidistributive then it is extremal. ◻

Proof. The word \(u\) corresponds to an order filter of \(\mathcal{C}_{C_m}^{\phi}\) if and only if for all \(i\in[n]\), \(u_i\leq |\mathcal{C}_{i}|=\phi(i)+1\). Denote \(T\) the order filter associated to \(u\). It remains to understand on the word \(u\) the completion operation. Let \(i< n\) and \(1\leq k< n-i\). We look at the completions between components \(\mathcal{C}_i\) and \(\mathcal{C}_{i+k}\). The completions between these two components happen if and only if \(T_i\neq \emptyset\) and \((\phi(i+1),i+k) \in T_{i+k}\), and in this case we need \((x,i)\in T_i\) implies \((x,i+k)\in T_{i+k}\). We have \(T_i\neq \emptyset \iff u_i\neq 0\) and \((\phi(i+1),i+k) \in T_{i+k} \iff u_{i+k} >\phi(i+k)-\phi(i+1)\). Moreover, \((x,i)\in T_i \Longrightarrow (x,i+k)\in T_{i+k}\) means \(u_{i+k}-u_i\geq \phi(i+k)-\phi(i)\). This finishes the proof of the proposition. ◻

Proof. The fact that \(B_p(u)\) and \(E_p(u)\) are both elements of \(\mathop{\mathrm{Tam}}(C_{np},\phi_p)\) follows easily from the diagrams using the fact that \(u_j-u_i\geq j-i\) for all \(i<j\) since \(u\) is a Tamari word. This can also be seen directly on the chains forming \(\mathcal{C}_{C_{np}}^{\phi}\) (see the right of 9). From this latter point of view follows the result about the partition of \(\mathop{\mathrm{Tam}}(C_{np},\phi_p)\), since for a torclosed set \(T\) of \(\mathop{\mathrm{Tam}}(C_{np},\phi_p)\) the unique \(u\) such that \(T\in I_p(u)\) is the word whose \(i\)-th letter is the number of elements of \((\phi(j),i)\) in \(T\) for \(j\leq i\). ◻

Proof. By definition of this subposet, denoted by \(P_{\phi}\), and of the completions, it is immediate that \(P_{\phi}\) is a sublattice of \(\mathop{\mathrm{Tam}}(P,\phi)\). Since, for each \(j<n\), the elements \((\phi(i),j)\) for \(i\leq j\) form a chain in \(\mathcal{C}_j\), an element \(T\) of \(P_{\phi}\) can be identified as the word \(u=u_0 u_1\dots u_{n-1}\) where \(u_j\) is the number of elements of the chain \((\phi(j),j)>(\phi(j-1),j)>\cdots >(\phi(0),j)\) in \(T\). Then \(P_{\phi} \cong \mathop{\mathrm{Tam}}_{n+1}\) follows from 30. ◻

Proof. We consider the diagrams of \(B_p(u)\) and \(E_p(u)\) drawn together as is done on the left of 7. Passing from \(B_p(u)\) to \(E_p(u)\) corresponds to adding \(p-1\) to the height of all the nonempty segments (this corresponds to adding the blue segments in 7). The elements of \(I_p(u)\) correspond to the diagrams with segments whose heights are between these of \(B_p(u)\) and these of \(E_p(u)\). We can increase freely the height of a given segment of \(x\)-coordinate \(i\) independently of the others before reaching the maximum height \(p\, u_i+(p-1)\), unless the top of this segment in \(u\) is touched by the line of slope \(p\) from the top of another segment (in 7 the segments which we cannot freely increase the height independently from the other segments are the first and fourth segments).

If we have \(k\geq 1\) segments in \(u\) of \(x\)-coordinate \(x_1,x_2,\dots,x_k\) whose tops are on the same line of slope \(p\) (and there are no other segments whose top is on that line), then just looking at these segments without changing the others we have as many elements in \(I_p(u)\) as the number of sequences \(i_1,i_2,\dots,i_k\) satisfying \(0\leq i_1\leq i_2 \leq \cdots \leq i_k \leq p-1\). Here \(i_j\) corresponds to the number we add to the height of the segment of \(x\)-coordinate \(x_j\). The number of such sequences is \(\binom{p-1+k}{k} = \binom{k+(p-1)}{p-1}\). The line of slope \(p\) we consider reaches a point on the \(x\)-axis whose \(x\)-coordinate is an integer. This point corresponds to a \(0\) (empty segment) of the weight sequence of the binary tree \(\varphi(u)\) (see the left of 8). By 32, in \(\psi(u)\) the top of the segments of \(x\)-coordinates \(x_1,x_2,\dots,x_k\) are identified, thus the resulting internal node that it creates has \(k+1\) children. Thus there are as many elements of \(I_p(u)\) as number of trees \(\psi(u)\) where each internal node is labeled by a positive integer less than or equal to \(\binom{(\text{number of children of x})-1+(p-1)}{p-1}\). This finishes the proof of the proposition. ◻

Proof. By 11, we have \(d_{n,p} = \sum_{u \text{ a Tamari word of length }n} |I_p(u)|\). With 33, it follows \[d_{n,p} =\sum_{u \text{ a Tamari word of length }n} \quad \prod_{x \text{ an internal node of } \varphi(u)} \binom{(\text{number of children of x}) -1+(p-1)}{p-1}.\] Thus \(d_{n,p}\) is the number of planar rooted trees on \(n+1\) vertices where each internal node is labeled by a positive integer less than or equal to \(\displaystyle \binom{(\text{number of children of x}) -1+(p-1)}{p-1}\). It follows: \[\begin{align} D_p(x) &= 1 + \sum_{k\geq 1} \binom{k +p-2}{p-1} \big(x\, D_p(x) \big)^k = 1 +x\,D_p(x)\,\sum_{k\geq 0} \binom{k+p-1}{p-1} \big(x\,D_p(x)\big)^k . \end{align}\] Using \(\frac{1}{(1-X)^p} = \sum_{k\geq 0} \binom{k+p-1}{p-1} X^k\), we have \(D_p(x)=1+\frac{x\,D_p(x)}{(1-x\,D_p(x))^p}\). ◻

Proof. This follows from 34 by using the Lagrange inversion formula. Let \(R(x)=x\,D_p(x)\). The formula of 34 can be written \(R(x)=x\, \big(1+ \frac{R(x)}{(1-R(x))^p} \big)\). Let \(\Phi(u)=1+\frac{u}{(1-u)^p}\). Thus we have \(R(x)=x\,\Phi(R(x))\) with \(R(0)=0\). By the Lagrange inversion formula, we have \[d_{n,p}=[x^{n+1}]\, R(x) =\frac{1}{n+1} [u^n] \Phi(u)^{n+1} = \frac{1}{n+1} [u^n] \Big(1+\frac{u}{(1-u)^p} \Big)^{n+1}.\] By the binomial theorem and the expansion of \(\frac{1}{(1-X)^p}\) recalled in the proof of 34, we have \[\Big(1+\frac{u}{(1-u)^p} \Big)^{n+1} = \sum_{k=0}^{n+1} \binom{n+1}{k} \frac{u^k}{(1-u)^{p\,k}}= \sum_{k=0}^{n+1} \binom{n+1}{k} u^k \sum_{m\geq 0} \binom{m+p\,k-1}{p\,k-1} u^m.\] This finishes the proof of the corollary by taking \(m=n-k\) and using the above formula for \(d_{n,p}\). ◻

Proof. The first two formulas follow from 18. Now we suppose that \(n\geq 3\). We consider the lattice \(\mathop{\mathrm{Tam}}(C_n,\phi)\) and denote \(k_0=\phi(0)+1\) and \(k_q=\phi(q)-\phi(q-1)\) for all \(q\in [n-1]\). Let \(i<n\) and \(j\in [k_i]\). Let \(S(i,j)\) be the set of the torclosed sets of \(\mathop{\mathrm{Tam}}(C_n,\phi)\) that contain \((\phi(i)-j+1,n-1)\) but not \((\phi(i)-j,n-1)\), and \(S(n)\) be the set of torclosed sets that do not contain \((\phi(n-1),n-1)\). Since all the sets that we just introduced form a partition of the elements of \(\mathop{\mathrm{Tam}}(C_n,\phi)\), then \[c_{k_0,k_1,\dots,k_{n-1}}=|S(n)| + \sum_{i=0}^{n-1} \sum_{j=1}^{k_i} |S(i,j)|.\]

We have \(|S(n)|=c_{k_0,k_1,\dots,k_{n-2}}\). Now we want to know \(|S(i,j)|\). A torclosed set in \(S(i,j)\) is the disjoint union of a torclosed set of \(\mathop{\mathrm{Tam}}(C_n,\phi_{|\{0,1,\dots,i-2\}})\) and of a specific torclosed set \(T\) of \(\mathop{\mathrm{Tam}}(C_n,\phi)\) with elements only in the components \(\mathcal{C}_{i}\) through \(\mathcal{C}_{n-2}\), where \(T_i\) has at most \(j\) elements. More precisely, these latter torclosed sets in \(S(i,j)\) correspond bijectively to the torclosed sets of \(\mathop{\mathrm{Tam}}(C_{n-\phi(i-1)},\phi')\) where \(\phi'(0)=j\) and \(\phi'(p)-\phi'(p-1)=k_{i+p}\) for all \(p\in [n-i-2]\). See 9 for an illustration of \(S(i,j)\). This gives for all \(i<n-2\), \(|S(i,j)|= c_{k_0,k_1,\dots,k_{i-2}} \times c_{j,k_{i+1},k_{i+2},\dots,k_{n-2}}\), and \(|S(n-2,j)|= c_{k_0,k_1,\dots,k_{n-4}} \times c_{j}\), and \(|S(n-1,j)|= c_{k_0,k_1,\dots,k_{n-3}}\). The formula of the proposition for \(n\geq 3\) then follows from the above formula for \(c_{k_0,k_1,\dots,k_{n-1}}\) by extracting the terms of the sum corresponding to \(i=n-1\) and \(i=n-2\). ◻

Proof. Let \(T\subseteq os_n^{d+1}\). We prove that \(T\) satisfies both [cond1] and 1 if and only if it satisfies both [cond1] and [cond2]. We suppose that \(T\) satisfies [cond1].

Suppose that \(T\) satisfies [cond2]. Let \(i<j<n\). Suppose that \((x_1,\dots,x_d,i)\in T\) and \((x_2+1,x_3+1,\dots,x_d+1,i+1,j)\in T\). Since \((x_1,\dots,x_d,i) \rightsquigarrow \tau_d (x_2+1,x_3+1,\dots,x_d+1,i+1,j)\), by [cond2] we have \((x_1,\dots,x_d,j)\in T\). This proves that \(T\) satisfies 1 .

Conversely, suppose that \(T\) satisfies 1 . Let \(x,z\in T\). Suppose \(x\rightsquigarrow \tau_d(z)\), meaning \(x_1\leq z_1-1\leq x_2 \leq z_2-1\leq \dots \leq x_{d+1} \leq z_{d+1}-1\). We have that any \(y\in os_n^{d+1}\) with \(y_i\in \{x_i,z_i\}\) for each \(i\) is above \((x_1,\dots,x_d,z_{d+1})\) for the product order. Thus, using [cond1], it suffices to prove \((x_1,\dots,x_d,z_{d+1})\in T\). Since \(x=(x_1,\dots,x_d,x_{d+1})\in T\), it suffices to prove that \((x_2+1,x_3+1,\dots,x_d+1,x_{d+1}+1,z_{d+1})\in T\), as 1 would then imply that \((x_1,\dots,x_d,z_{d+1})\in T\). Since we have \(x_1\leq z_1-1\leq x_2 \leq z_2-1\leq \dots \leq x_{d+1} \leq z_{d+1}-1\), we know that \(z\leq (x_2+1,x_3+1,\dots,x_{d+1}+1,z_{d+1})\). Since \(z\in T\) and \(T\) satisfies [cond1], we have \((x_2+1,x_3+1,\dots,x_{d+1}+1,z_{d+1})\in T\). This finishes the proof of the lemma. ◻

Proof. Let \(T\subseteq os_n^{d+1}\). We prove that \(T\) satisfies both [cond1] and ?? if and only if it satisfies both [cond1] and 1 . Using 13, this will prove the proposition. We suppose that \(T\) satisfies [cond1].

Suppose that \(T\) satisfies ?? . Let \(i<j<n\). Suppose that \((x_1,\dots,x_d,i)\in T\) and \((x_2+1,x_3+1,\dots,x_d+1,i+1,j)\in T\). We need to prove that \((x_1,\dots,x_d,j)\in T\). Since \((x_1,\dots,x_d,i) \in os_n^{d+1}\), then \(x_p\leq i\) for all \(p\in [d]\). Thus \((i+1,\dots,i+1,j)\geq (x_2+1,x_3+1,\dots,x_d+1,i+1,j) \in T\). Since \(T\) satisfies [cond1], then \((i+1,\dots,i+1,j)\in T\). Since \(T\) satisfies ?? , then \((x_1,\dots,x_d,j)\in T\). Thus \(T\) satisfies 1 .

Conversely, suppose that \(T\) satisfies 1 . Let \(i<j<n\). Suppose that \((x_1,\dots,x_d,i)\in T\) and \((i+1,\dots,i+1,j)\in T\). We need to prove that \((x_1,\dots,x_d,j)\in T\). For \(k \in \{0,1,\dots ,i\}\), let \[\begin{align} X_k &:=(\max(x_1,i-k),\max(x_2,i-k), \dots , \max(x_d,i-k), i), \\ L_k &:=(\max(x_2,i-k)+1, \max(x_3,i-k)+1, \dots , \max(x_d,i-k)+1, i+1, j), \\ Y_k &:=(\max(x_1,i-k),\max(x_2,i-k), \dots , \max(x_d,i-k), j). \end{align}\]

See 10 for an illustration of these elements and of the following arguments. Since \(T\) satisfies 1 , having both \(X_k\in T\) and \(L_k\in T\) implies \(Y_k\in T\). Let \(P_k\) be the property \((X_k,L_k) \in T^2\). Let us prove by induction that \(P_k\) is true for all \(k \in \{0,1,\dots ,i\}\).

Since \((x_1,\dots,x_d,i) \in os_n^{d+1}\), then \(x_p\leq i\) for all \(p\in [d]\). Thus \(X_0=(i,i,\dots,i)\geq (x_1,\dots,x_d,i)\in T\). Since \(T\) satisfies [cond1], then \(X_0\in T\). We already know that \(L_0=(i+1,\dots,i+1,j)\in T\). Thus \(P_0\) is true.

Let \(k \in \{0,1,\dots ,i-1\}\) and assume that \(P_k\) is true. It means that \((X_k,L_k) \in T^2\), which implies \(Y_k\in T\) since \(T\) satisfies 1 . We have \(X_{k+1}\geq (x_1,x_2,\dots,x_d,i)\in T\), thus \(X_{k+1}\in T\) since \(T\) satisfies [cond1]. Let \(l\in [d-1]\). Since \(x_{l+1} \geq x_l\), we have \(\max(x_{l+1},i-k-1)+1 \geq \max(x_l,i-k)\). This proves \(L_{k+1} \geq Y_k\), and since \(Y_k\in T\), then \(L_{k+1}\in T\) since \(T\) satisfies [cond1]. Then \((X_{k+1},L_{k+1}) \in T^2\), meaning \(P_{k+1}\) is true.

By induction, this proves in particular that \(P_i\) is true. Thus \((X_i,L_i) \in T^2\). This implies that \(Y_i\in T\) since \(T\) satisfies 1 . Thus \(Y_i=(x_1,\dots,x_d,j) \in T\). This finishes the proof of the proposition. ◻

Proof. Both \(L_n^d\) and \(\mathop{\mathrm{Tam}}\big((os_n^d,\leq),\phi\big)\) are lattices ordered by inclusion, thus it suffices to prove that the \(d\)-torsion classes \(T\in L_n^d\) are the torclosed sets of \(\mathop{\mathrm{Tam}}\big((os_n^d,\leq),\phi\big)\). Recall the definition of \(\mathcal{C}_P^{\phi}\) in 3.1. By definition, \(T\in L_n^d\) satisfies [1aus] if and only if \(T\) is an order filter of \(\mathcal{C}_P^{\phi}\). Since \((\phi(i+1),j)=(i+1,\dots,i+1,j)\), then \(T\in L_n^d\) satisfies ?? if and only if \(T\) is a torclosed set of \(\mathop{\mathrm{Tam}}\big((os_n^d,\leq),\phi\big)\). This proves the theorem using 37. ◻

Proof. It is immediate that \(|os_n^d|=\binom{n+d-1}{d}\). We only need to prove the formula for the number of meet-irreducibles. We use 12. The number of the first kind of meet-irreducibles is \(\binom{n+d}{d+1}\). Counting the second kind of meet-irreducibles amounts to count, for any \(i<n\) and any \(j\leq i\), the number of elements of \(\mathcal{C}_i\) incomparable to \((\phi(j),i)\). Since \(\phi(0)=\hat{0}_{os_n^d}\), no such elements exist if \(i\in \{0,1\}\), nor if \(j=\{0,i\}\) since \((\phi(0),i)\) and \((\phi(i),i)\) are respectively the minimum and maximum elements of \(\mathcal{C}_i\). Then \[|\mathrm{MIrr}(L_n^d)|=\binom{n+d}{d+1}+\sum_{i=2}^{n-1} \,\,\sum_{j=1}^{i-1}\,\, |\{(x,i)\in \mathcal{C}_i \mid (x,i) \not\sim (\phi(j),i) \}|\]

Let \(i\) and \(j\) such that \(2\leq i<n\) and \(0<j<i\). The integer in the sum above is also the number of elements of \(os_n^d\) less than \(\phi(i)=(i,\dots,i)\) that are incomparable to \(\phi(j)=(j,\dots,j)\). Such an element needs to start with a \(k<j\), and finish with a \(l\in \{j+1,j+2,\dots,i\}\), and between the two it should be non-decreasing. Thus when \(k\) and \(l\) are fixed, the number of such elements is \(|os_{l-k+1}^{d-2}|=\binom{l-k+d-2}{d-2}\). This proves that \[|\mathrm{MIrr}(L_n^d)|=\binom{n+d}{d+1}+\sum_{i=2}^{n-1} \,\,\sum_{j=1}^{i-1}\,\, \sum_{k=0}^{j-1}\,\, \sum_{l=j+1}^{i} \binom{l-k+d-2}{d-2} .\] By successive use of the Hockey-stick identity, this reduces to \[|\mathrm{MIrr}(L_n^d)|=\binom{n+d}{d+1}+\sum_{i=2}^{n-1} \,(i-1)\, \binom{i+d}{d} -2\,\binom{n+d}{d+2} +\frac{(n-1)(n+2)}{2}\] Using the identity \(i\,\binom{i+d}{d}=(d+1)\,\binom{i+d}{d+1}\) and again the Hockey-stick identity, we obtain the desired formula. ◻

Proof. Seeking a contradiction, suppose that there exists \(\sigma\in \mathfrak{S}_d\) such that \(x_{\sigma}\) is not a maximal element of \(I\). Then there exists \(y\in I\) such that \(x_{\sigma} <_{prod} y\). By applying \(\sigma^{-1}\) to both sides, which keeps the product order inequality, we obtain \(x<_{prod} y_{\sigma^{-1}}\). But \(y_{\sigma^{-1}} \in I\) and \(x\) is a maximal element of \(I\). This is absurd. ◻

Proof. The function which maps \((x_1,x_2,\dots,x_d)\in os_n^d\) to the non-decreasing sequence of size \(n-1\) with \(n-1-x_d\) zeros, \(x_d-x_{d-1}\) ones, and so on where the number of \(i\)’s is \(x_{d-i+1}-x_{d-i}\) for all \(i\in [d]\), is an isomorphism of posets from \(os_n^d\) to \(os_{d+1}^{n-1}\). Indeed this bijection comes from reading row by row or column by column the Ferrers diagram of \((x_1,x_2,\dots,x_d)\in os_n^d\). The fact that it is an isomorphism of posets is easily seen with this representation.

We now prove the combinatorial interpretation for \(|J(os_n^d)|\). By definition, we can identify a totally symmetric \(d\)-dimensional partition with its set of maximal elements. By 14, we in fact just need to specify the maximal elements that are non-decreasing. This shows that the totally symmetric \(d\)-dimensional partitions which fit in a \(d\)-dimensional box whose sides all have length \(n\) correspond to the antichains of \((os_n^d,\leq)\). These latter are counted by \(|J(os_n^d)|\). This proves what we wanted.

Finally, from this combinatorial interpretation the formulas for \(|J(os_3^d)|\) and \(|J(os_4^d)|\) follow. ◻

Proof. These order filters are identified with their minimal elements, which are the antichains of \(P\) that contain \(x\). Thus the number of these order filters is the number of antichains of \(P\) formed by elements that are all incomparable to \(x\). The results follow. ◻

Proof. The formulas for \(|L_n^d|\) with \(n\leq 3\) follow from 38 and 19, using for \(|L_3^d|\) the formula \(|J(os_3^d)|=2^{d+1}\) obtained in 40. For \(|L_4^d|\) it requires more work. By 38, \(L_4^d= \mathop{\mathrm{Tam}}(os_4^d,\phi)\) with \(\phi\) defined by \(\phi(i)=(i,\dots,i)\) for all \(i<n\). Let:

• \(R_1 :=\{T\in L_4^d\mid (\phi(3),3)\not\in T\}\)

• \(R_2 :=\{T\in L_4^d\mid (\phi(3),3)\in T,\, (\phi(2),3)\not\in T\}\)

• \(R_3 :=\{T\in L_4^d\mid (\phi(2),3)\in T,\,(\phi(1),3)\not\in T\}\)

• \(R_4 :=\{T\in L_4^d\mid (\phi(1),3)\in T,\,(\phi(0),3)\not\in T,\,(\phi(2),2)\not\in T\}\)

• \(R_5 :=\{T\in L_4^d\mid (\phi(1),3)\in T,\,(\phi(0),3)\not\in T,\,(\phi(2),2)\in T\}\)

• \(R_6 :=\{T\in L_4^d\mid (\phi(0),3)\in T\}\)

These sets form a partition of all the torclosed sets of \(L_4^d\), thus \(|L_4^d|=\sum_{i=1}^6 |R_i|\). It is immediate that \(|R_1|=|R_6|=|L_3^d|=d+3+5\times 2^d\).

A torclosed set of \(R_2\) is a torclosed set of \(\mathop{\mathrm{Tam}}(os_4^d,\phi_{|\{0,1\}})\) together with an order filter of \(\big(\mathcal{C}_3\setminus\{(\phi(3),3)\}\big) \setminus I_{\mathcal{C}_3}(\phi(2),3)\). The elements in this latter poset are those elements of \(os_4^{d+1}\) that end with two letters \(3\), but are not \((3,3,\dots,3)\). It follows that \(|R_2|=(d+4)\, (|J(os_4^{d-1})|-1)\). Thus, by 40, \(|R_2|=(d+4) (a_d-1)\).

Let \(Q=(\mathcal{C}_2 \cup \mathcal{C}_3,\leq_{prod})\). A torclosed set of \(R_3\) may or may not contain \((\phi(0),0)\), does not contain elements of \(\mathcal{C}_1\), and on the other elements it should be an order filter of \(Q\setminus \big(I_Q(\phi(1),3) \cup F_Q(\phi(2),3)\big)\). The elements in this latter poset are those of \(os_4^{d+1}\) that are in \([(0,\dots,0,2,2),(1,3,\dots,3)] \cup \{(2,\dots,2)\}=K_1\cup \{(2,\dots,2)\}\). Then by 15, \(|R_3|=2\times (|J(K_1)|+|J(I)|)\) where \(I\) is the subposet of \(K_1\cup \{(2,\dots,2)\}\) on the incomparable elements of \((2,\dots,2)\). We have \(I=[(0,\dots,0,2,3),(1,3,\dots,3)]=K_2\). Thus \(|R_3|=2\,(|J(K_1)|+|J(K_2)|)\).

The torclosed sets of \(R_4\) are the order filters of the subposet of \(os_4^{d+1}\) on the elements \([(0,\dots,0,1,1),(1,\dots,1)] \cup [(0,\dots,0,1,3),(0,3,\dots,3)]\). Denote by \(V\) this subposet. We have \(|R_4|=|J(V)|\). It will be helpful to add \((0,\dots,0,3)\) to \(V\) (the adding forms a new subposet of \(os_4^{d+1}\)). To do this we use 15, which gives \(|R_4|=|J(V\cup \{(0,\dots,0,3)\})|-|J(I')|\) where \(I'\) is the subposet of \(V\cup \{(0,\dots,0,3)\}\) on the elements that are incomparable to \((0,\dots,0,3)\). Then \(I'=[(0,\dots,0,1,1),(1,\dots,1)]\), thus \(|R_4|=|J(V\cup \{(0,\dots,0,3)\})|-(d+1)\). In \(V\cup \{(0,\dots,0,3)\}\) we give the subposet formed by the incomparable elements to each of the element of the chain \([(0,\dots,0,1,1),(1,\dots,1)]\); there is only one element incomparable to \((0,\dots,0,1,1)\) which is \((0,\dots,0,3)\), for \((1,\dots,1)\) the incomparable elements form the subposet \([(0,\dots,0,3),(0,3,\dots,3)] \cong os_4^{d-1}\), and for any \((0,\dots,0,1,\dots,1)\) with the number of zeros equal to \(i\in [d-2]\) the incomparable elements are the elements starting with \(i+1\) zeros and ending with a \(3\), which form a subposet isomorphic to \(os_4^{d-i-1}\). Remember that \(|J(os_4^{j})|=a_{j+1}\) (40). Then applying 15 successively for all the elements of the chain \([(0,\dots,0,1,1),(1,\dots,1)]\) gives at the end \[|J(V\cup \{(0,\dots,0,3)\})| = |J([(0,\dots,0,3),(0,3,\dots,3)])|+2+a_d+\sum_{i=1}^{d-2} a_{d-i}=2+2\,a_d+\sum_{i=1}^{d-2} a_{d-i} .\] Thus \(|R_4|=2+2\,a_d+\sum_{i=1}^{d-2} a_{d-i} -(d+1)= 1+a_d-d+\sum_{i=2}^d a_i\).

The torclosed sets of \(R_5\) are the order filters of the subposet of \(os_4^{d+1}\setminus \{(2,\dots,2)\}\) on the elements that are neither bigger than \((1,\dots,1,3)\) nor smaller than \((0,\dots,0,3)\). Denote by \(W\) this subposet. We have \(K_3=W\cup \{(0,\dots,0,1),(0,\dots,0),(0,\dots,0,2),(0,\dots,0,3),(2,\dots,2)\}\). We now use five times 15 to add the five above elements to \(W\) to obtain \(K_3\). As \((0,\dots,0,1)\) is smaller than all the elements of \(W\cup \{(0,\dots,0,1)\}\), we have \(|J(W\cup\{(0,\dots,0,1)\})|=|J(W)|+1\). Then as \((0,\dots,0,0)\) is smaller than all the elements of \(W\cup\{(0,\dots,0,1),(0,\dots,0,0)\}\), we have \(|J(W\cup\{(0,\dots,0,1),(0,\dots,0)\})|=|J(W)|+2\). The subposet of \(W\cup\{(0,\dots,0,1),(0,\dots,0),(0,\dots,0,2)\}\) on the incomparable elements of \((0,\dots,0,2)\) is \([(0,\dots,0,1,1),(1,\dots,1)]\), thus \(|J(W\cup\{(0,\dots,0,1),(0,\dots,0),(0,\dots,0,2)\})|=|J(W)|+2+(d+1)=|J(W)|+3+d\). The subposet of \(W\cup\{(0,\dots,0,1),(0,\dots,0),(0,\dots,0,2),(0,\dots,0,3)\}\) on the incomparable elements of \((0,\dots,0,3)\) is \([(0,\dots,0,1,1),(2,\dots,2)[\). Since \[[(0,\dots,0,1,1),(2,\dots,2)[ \,\cup \,\{(2,\dots,2)\}\cup [(0,\dots,0),(0,\dots,0,2)]=[(0,\dots,0),(2,\dots,2)]\cong os_3^{d+1} ,\] by using 15 we can obtain \(|J([(0,\dots,0,1,1),(2,\dots,2)[)|=|J(os_3^{d+1})|-d-4=2^{d+2}-d-4\). Thus \[|J(W\cup\{(0,\dots,0,1),(0,\dots,0),(0,\dots,0,2),(0,\dots,0,3)\})|=|J(W)|+2^{d+2}-1 .\]

The incomparable elements of \((2,\dots,2)\) in \(K_3\) are the ones that start with a \(0\) and end with a \(3\), thus they form a subposet of \(K_3\) isomorphic to \(os_4^{d-1}\). Then \[|J(K_3)|=(|J(W)|+2^{d+2}-1)+|J(os_4^{d-1})|=|J(W)|+2^{d+2}-1+a_d .\] Thus \(|R_5|=|J(W)|=|J(K_3)|-2^{d+2}-a_d+1\).

Putting all together using \(|L_4^d|=\sum_{i=1}^6 |R_i|\) finishes the proof of the proposition. ◻

Proof. Let \((y,j)\in \mathcal{C}_j\cap os_{\underline{l}}^{d+1}\) such that \((y,j)\geq (x,j)\). Let us prove that \((y,j)\in R\). We have \((y,i)\geq (x,i) \in R\), thus \((y,i)\in R\) since \(R\) is an order filter of \(\mathcal{C}_{os_n^d}^{\phi}\). For \(k \in \{0,1,\dots ,i\}\), let \[\begin{align} X_k &:=(\max(y_1,i-k),\max(y_2,i-k), \dots , \max(y_d,i-k), i), \\ L_k &:=(\max(y_2,i-k)+1, \max(y_3,i-k)+1, \dots , \max(y_d,i-k)+1, i+1, j), \\ Y_k &:=(\max(y_1,i-k),\max(y_2,i-k), \dots , \max(y_d,i-k), j). \end{align}\]

Note that these elements were already introduced in the proof of 37 and were illustrated in 10 (where \(x_1,\dots,x_d\) are replaced by \(y_1,\dots,y_d\)). Since \(X_k \geq X_i=(y,i) \in R\) for all \(k\leq i\), then \(X_k\in R\) for all \(k\leq i\) since \(R\) is an order filter of \(\mathcal{C}_{os_n^d}^{\phi}\). Since \(Y_k\geq Y_i=(y,j)\in os_{\underline{l}}^{d+1}\), then \(Y_k\in os_{\underline{l}}^{d+1}\) for all \(k\leq i\). Let us prove by induction that \(Y_k\in R\) for all \(k\leq i\). This will finish the proof of the lemma since \(Y_i=(y,j)\). We know that \(X_0=(i,\dots,i)\in R\), and \(L_0=(i+1,\dots,i+1,j)=(\phi(i+1),j)\in R\) by hyphothesis. Moreover \(X_0\rightsquigarrow \tau_d(L_0)\). Thus by condition \(2\) of 8, since \((i,\dots,i,j)=Y_0 \in os_{\underline{l}}^{d+1}\), we have \(Y_0\in R\). Let \(k\in [i]\). Suppose that for all \(p<k\), \(Y_p\in R\). Let us prove that \(Y_k\in R\). We know that \(X_k\in R\). Since \(L_k\geq Y_{k-1}\) and \(Y_{k-1}\in R\) by hypothesis, then \(L_k\in R\). Moreover \(X_k\rightsquigarrow \tau_d(L_k)\). Thus by condition \(2\) of 8, since \(Y_k \in os_{\underline{l}}^{d+1}\) and the \(l\)-th coordinate of this element is either the \(l\)-th coordinate of \(X_k\) or \(L_k\) for all \(l\in [d+1]\), we have \(Y_k\in R\). This finishes the proof of the induction, thus of the lemma. ◻

Proof. If \(T\in L_n^d\), then \(T\cap os_{\underline{l}}^{d+1} \in L_{\underline{l}}^d\). Conversely, let \(R\in L_{\underline{l}}^d\). Then \(R\subseteq \mathcal{C}_{os_n^d}^{\phi}\). Let \(T\) be the smallest torclosed set of \(L_n^d\) containing \(R\). Let us prove that \(R=T\cap os_{\underline{l}}^{d+1}\). We prove by induction that \(T_k\cap os_{\underline{l}}^{d+1} = R_k\) for all \(k<n\). It is immediate for \(k=0\). Let \(k\in [n-1]\). We suppose that \(T_i\cap\, os_{\underline{l}}^{d+1} = R_i\) for all \(i<k\). We want to prove that \(T_k \cap \,os_{\underline{l}}^{d+1} = R_k\).

By definition of \(T\), we know that \(R_k\subseteq T_k\cap os_{\underline{l}}^{d+1}\). Thus we need to prove that \(T_k\cap os_{\underline{l}}^{d+1} \subseteq R_k\). If \(R_k=\emptyset\) this is immediate, thus we suppose that \(R_k\neq \emptyset\). By 7, we can compute \(T_k\) by first looking at the completion of \(T_{k-1}\cup R_k\), which in \(\mathcal{C}_k\) forms the order filter that we denote \(F_1\). Then at the completion of \(T_{k-2}\cup F_1\), which in \(\mathcal{C}_k\) forms the order filter that we denote \(F_2\), and so on. At the end we will have \(F_k=T_k\). Let us prove by induction that \(F_j\cap os_{\underline{l}}^{d+1}\subseteq R_k\) for all \(j\in [k]\). This will finish the proof of the proposition since for \(j=k\) this will give \(T_k\cap os_{\underline{l}}^{d+1}\subseteq R_k\).

Let us prove that \(F_1\cap os_{\underline{l}}^{d+1}\subseteq R_k\). Let \((y,k)\in F_1\cap os_{\underline{l}}^{d+1}\). If \((y,k)\) is not obtained through the completion of \(T_{k-1}\cup R_k\), then \((y,k)\in R_k\) and there is nothing to prove. Thus we assume that there exists \((z,k-1)\in T_{k-1}\) such that \((y,k)\geq (z,k)\). In particular \((\phi(k),k)\in R_k\). Let \(t=(\min(y_1,k-1),\min(y_2,k-1),\dots,\min(y_d,k-1))\in os_{n}^{d}\). We have \((y,k)\geq (t,k)\). Let us prove that \((t,k-1)\in R_{k-1}\).

Since \((y,k)\in \mathcal{C}_k\cap os_{\underline{l}}^{d+1}=\{(x,k)\in \mathcal{C}_k\mid x\geq \phi(k-l_k+1)\}\), then \(y_1\geq k-l_k+1\). Since \(k\geq 1\), we have \(l_k\geq 2\), which gives \(k-1\geq k-l_k+1\). Thus \(\min(y_1,k-1)\geq k-l_k+1\). Then \((t,k)\in os_{\underline{l}}^{d+1}\). Since \((t,k) \leq (k-1,\dots,k-1,k)\), we have \((t,k-1) \in \mathcal{C}_{k-1}\). Moreover, \((t,k-1)\in os_{\underline{l}}^{d+1}\). Indeed, since \(l_k-l_{k-1}\leq 1\), then \((k-1)-l_{k-1}+1\leq k-l_k+1\). Thus if \((t,k-1)\not\in os_{\underline{l}}^{d+1}\), which means \(t\not\geq k-1-l_{k-1}+1\), then \(t\not\geq k-l_{k}+1\), which is absurd since \((t,k)\in os_{\underline{l}}^{d+1}\). Thus \((t,k-1)\in \mathcal{C}_{k-1} \cap os_{\underline{l}}^{d+1}\). Since \((z,k-1)\in \mathcal{C}_{k-1}\), we have \(z_i\leq k-1\) for all \(i\in [d]\). Moreover \((z,k)\leq (y,k)\) gives \(z_i\leq y_i\) for all \(i\in [d]\). Then \((z,k-1)\leq (t,k-1)\). Since \((z,k-1)\in T_{k-1}\), then \((t,k-1)\in T_{k-1}\). But we proved that \((t,k-1)\in \mathcal{C}_{k-1} \cap os_{\underline{l}}^{d+1}\), thus \((t,k-1)\in T_{k-1}\cap os_{\underline{l}}^{d+1}=R_{k-1}\) by induction hypothesis.

Then we use 16 with \((t,k-1)\in R\) and \((\phi(k),k)\in R\). This gives \((y,k)\in R_k\) since \((y,k)\in \mathcal{C}_k \cap os_{\underline{l}}^{d+1}\) and \((y,k)\geq (t,k)\). This is what we wanted to prove. Now if we let \(j\in [k]\) and we suppose that \(F_i\cap os_{\underline{l}}^{d+1} \subseteq R_k\) for all \(i<j\), we need to prove that \(F_j\cap os_{\underline{l}}^{d+1} \subseteq R_k\). The proof below uses the same structure as in the proof of \(F_1\cap os_{\underline{l}}^{d+1} \subseteq R_k\) above. To help the reader follow these similar arguments, as no confusion is possible since we do not use previously introduced elements, we will reuse some variable names.

Let \((y,k)\in F_j\cap os_{\underline{l}}^{d+1}\). Let us prove that \((y,k)\in R_k\). If \((y,k)\) is not obtained through the completion of \(T_{k-j}\cup F_{j-1}\), then \((y,k)\in F_{j-1}\cap os_{\underline{l}}^{d+1}\subseteq R_k\) by induction hypothesis, and there is nothing to prove. Thus we assume that there exists \((z,k-j)\in T_{k-j}\) such that \((y,k)\geq (z,k)\). In particular \((\phi(k-j+1),k)\in F_{j-1}\). If \((\phi(k-j+1),k)\not\in os_{\underline{l}}^{d+1}\), then \((\phi(k-j+1),k)\leq (\phi(k-l_k+1),k)\). Since \((\phi(k-j+1),k)\in F_{j-1}\) then \((\phi(k-l_k+1),k) \in F_{j-1}\cap os_{\underline{l}}^{d+1} \subseteq R_k\) by induction hypothesis, and this proves that \(F_j\cap os_{\underline{l}}^{d+1} \subseteq \mathcal{C}_k\cap os_{\underline{l}}^{d+1} = R_k\) since \(\mathcal{C}_k\cap os_{\underline{l}}^{d+1}=\{(x,k)\in \mathcal{C}_k\mid x\geq \phi(k-l_k+1)\}\). Thus we assume that \((\phi(k-j+1),k)\in os_{\underline{l}}^{d+1}\), which gives us \(l_k\geq j\). Then \((\phi(k-j+1),k)\in F_{j-1}\cap os_{\underline{l}}^{d+1} \subseteq R_k\) by induction hypothesis, then \((\phi(k-j+1),k) \in R_k\). To have something to prove we thus further assume that \(\phi(k-l_k+1)<\phi(k-j+1)\), which gives \(l_k>j\). Let \(t=(\min(y_1,k-j),\dots,\min(y_d,k-j)) \in os_n^d\). We have \((y,k)\geq (t,k)\). Let us prove that \((t,k-j)\in R_{k-j}\).

Since \((y,k)\in \mathcal{C}_k\cap os_{\underline{l}}^{d+1}=\{(x,k)\in \mathcal{C}_k\mid x\geq \phi(k-l_k+1)\}\), then \(y_1\geq k-l_k+1\). We also proved that \(l_k>j\), which gives \(k-j\geq k-l_k+1\). Thus \(\min(y_1,k-j)\geq k-l_k+1\). Then \((t,k)\in os_{\underline{l}}^{d+1}\). Since \((t,k) \leq (k-j,\dots,k-j,k)\), we have \((t,k-j) \in \mathcal{C}_{k-j}\). Moreover, \((t,k-j)\in os_{\underline{l}}^{d+1}\). Indeed, since \(l_k-l_{k-j}\leq j\), then \((k-j)-l_{k-j}+1\leq k-l_k+1\). Thus if \((t,k-j)\not\in os_{\underline{l}}^{d+1}\), which means \(t\not\geq k-j-l_{k-j}+1\), then \(t\not\geq k-l_{k}+1\), which is absurd since \((t,k)\in os_{\underline{l}}^{d+1}\). Thus \((t,k-j)\in \mathcal{C}_{k-j} \cap os_{\underline{l}}^{d+1}\). Since \((z,k-j)\in \mathcal{C}_{k-j}\), we have \(z_i\leq k-j\) for all \(i\in [d]\). Moreover \((z,k)\leq (y,k)\) gives \(z_i\leq y_i\) for all \(i\in [d]\). Then \((z,k-j)\leq (t,k-j)\). Since \((z,k-j)\in T_{k-j}\), then \((t,k-j)\in T_{k-j}\). But we proved that \((t,k-j)\in \mathcal{C}_{k-j} \cap os_{\underline{l}}^{d+1}\), thus \((t,k-j)\in T_{k-j}\cap os_{\underline{l}}^{d+1}=R_{k-j}\) by induction hypothesis.

Then we use 16 with \((t,k-j)\in R\) and \((\phi(k-j+1),k)\in R\). This gives \((y,k)\in R_k\) since \((y,k)\in \mathcal{C}_k \cap os_{\underline{l}}^{d+1}\) and \((y,k)\geq (t,k)\). This proves that \(F_j\cap os_{\underline{l}}^{d+1} \subseteq R_k\). As explained, this finishes the proof of the proposition. ◻

Proof. Let \(\equiv\) be the equivalence relation on \(L_n^d\) defined by \(T\equiv T'\) if and only if \(T\cap os_{\underline{l}}^{d+1} = T'\cap\, os_{\underline{l}}^{d+1}\). By 43, the elements of \(L_{\underline{l}}^d\) are identified with the equivalence classes of the relation \(\equiv\). Let \(K=(K_0,K_1,\dots,K_{n-1})\) be the Kupisch-like series defined by \(K_i:=\max(0,i-l_{i} +1)\) for all \(i<n\). Then the associated Kupisch equivalence relation \(\equiv_K\) on \(L_n^d\) corresponds to \(\equiv\). By 38, \(L_n^d=\mathop{\mathrm{Tam}}\big((os_n^d,\leq),\phi\big)\) with \(\phi\) defined by \(\phi(i)=(i,\dots,i)\) for all \(i<n\). Thus the proposition follows from 26 since \((os_n^d,\leq)\) is a lattice. ◻