Eighty-three sublattices and planarity

Let $L$ be a finite $n$-element lattice. We prove that if $L$ has at least $83\cdot 2^{n-8}$ sublattices, then $L$ is planar. For $n>8$, this result is sharp since there is a non-planar lattice with exactly $83\cdot 2^{n-8}-1$ sublattices.

Furthermore, we say that a finite lattice L has σ-many sublattices or, in other words, it has σ-many subuniverses if σ(L) > 83. This concept and notation will play a crucial role in the rest of the paper. Since | Sub(L)| is larger than the number of sublattices by 1, we can reformulate Theorem 1.1 and a part of Remark 1.3 as follows.
Theorem 2.2. If L is a finite lattice such that σ(L) > 83, then L is planar. In other words, finite lattices with σ-many sublattices are planar. Furthermore, for every natural number n ≥ 9, there exists an n-element lattice L such that σ(L) = 83 and L is not planar.
The importance of the concepts introduced in this section so far is well explained by the following easy lemma. Proof. Let m := |B| and n := |A|. Then k := n − m = |A \ B| ≥ 0. Define an equivalence relation ∼ on Sub(A) by letting X ∼ Y mean that X ∩B = Y ∩B. Since X∩B ∈ Sub(B) for every X ∈ Sub(A), this equivalence has at most | Sub(B)| blocks. Every block of ∼ is a subset of {U ∪ X : X ⊆ A \ B} for some U ∈ Sub(B). Since A \ B has 2 k subsets, every block of ∼ consists of at most 2 k elements of Sub(A). Therefore, | Sub(A)| ≤ | Sub(B)| · 2 k . Dividing this inequality by 2 n−8 = 2 m−8 · 2 k , we obtain the validity of the lemma. 2.2. The Kelly-Rival list. For a poset P , its dual will be denoted by P δ . With reference to Kelly and Rival [27] or, equivalently, to , the Kelly-Rival list of lattices is defined as follows.
L KR := {A n , E n , E δ n , F n , G n , H n : n ≥ 0} ∪ {B, B δ , C, C δ , D, D δ }. Note that A n , F n , G n , and H n are selfdual lattices. The key tool we need is the following deep result. Theorem 2.4 (Kelly and Rival [27]). A finite lattice is planar if and only if it does not contain any lattice in L KR as a subposet. Lemma 2.3 and Theorem 2.4 raise the following problem; B, C, . . . , H 0 still denote lattices in L KR . Note that being a subposet is a weaker assumption than being a sublattice.
Problem 2.5. Let X and L be finite lattices such that X is a subposet of L.
(i) Is σ(L) ≤ σ(X) necessarily true in this case? If we could answer at least least part (ii) of Problem 2.5 affirmatively, then the proof of Theorem 2.2 would only require the lemmas of (the present) Section 2 and an easy application of a straightforward computer program. Later, Remark 2.20 and Example 2.21 will point out why Problem 2.5 is not as easy as it may look at first sight.
Remark 2.6. There are a lot of finite lattices X such that for every finite lattice L, σ(L) ≤ σ(X) if X is a subposet of L. For example, X = F 0 has this property.
Proof of Remark 2.6. Every finite chain obviously has the property above, whence there are "a lot of" such lattices. Although the proof of Lemma 3.6 will, in effect, establish the above property of X = F 0 , we can present a short argument for this fact right now. (But this short argument is not independent from Lemma 3.6.) For the sake of contradiction, suppose that X = F 0 is a subposet of L but σ(L) > σ(F 0 ). For a computer, it is straightforward to show that σ(F 0 ) = 83; see Lemma 2.8 later. So σ(L) > 83, and we obtain from Theorem 2.2 that L is planar. Hence, by Theorem 2.4, F 0 cannot be a subposet of L, which is a contradiction.  Since it would be a very tedious task to compute σ(X) manually even for the smallest lattice X ∈ L KR , we have developed a straightforward computer program for Windows 10 to do it. This program, called subsize, is downloadable from the authors website. The input of the program is an unformatted text file describing a finite binary partial algebra A = (A; F ); there are several word processors that can produce such a file. In particular, the description of A includes a list of strings x * y = z of length four where * is an operation symbol in F , (x, y) ∈ Dom(f A ) and f A (x, y) = z; these strings are called constraints in the input file. The output, σ(A), is displayed on the screen and saved into a text file. The algorithm is trivial: the program lists all the 2 |A| subsets of A and counts those that are closed with respect to all constraints. Remark 2.7. There are two kinds of difficulties we have to face. First, we could not solve Problem 2.5; see the paragraph following it. Second, the running time of our program depends exponentially on the input size |A|. Hence, a lot of theoretical considerations is necessary before resorting to the program and what is even worse, many cases have to be input into the program. Because of its exponential time, it is not clear (and it is note hopeful) whether the appropriate cases could be found by a much more involved (and so less reliable) computer program without a lot of human work. So the program is simple, we believe it is reliable, and it is not to hard to write another program to test our input files. On the other hand, the exceptionally tedious work to find the appropriate cases and to create the input files needed several weeks.
However, to obtain the following statement with the help of the program is quite easy. Except for its equality σ(A 0 ) = 74, this lemma will not be used in the proof of Theorem 2.2. However, a part of this lemma will be used in the proof of Remark 1.3 below, and it is this lemma that tells us how the theorem was conjectured. Even the proof of (part (i) of) this lemma requires more computation than a human is willing to carry out or check.
Proof of Remark 1.3. For n = 9, the equality σ(F 0 ) = 83 from Lemma 2.8 proves the validity of Remark 1.3 since F 0 is not planar by Theorem [27]. Assume that n > 9, let C be an (n − 9)-element chain, and let L be the ordinal sum of F 0 and C. That is, L is the disjoint union of its ideal F 0 and its filter C. By Theorem [27], L is not planar. Since a subset of L is a subuniverse if and only if it is of the form X ∪ Y such that X ∈ Sub(F 0 ) and Y ⊆ C, it follows that | Sub(L)| = | Sub(F 0 )| · 2 n−9 = (83 · 2 |F0]−8 ) · 2 n−9 = 83 · 2 n−8 , whereby σ(L) = 83, as required.
Proof Technique 2.9. For Lemma 2.8 and also for all other statements that refer to the program or mention σ(. . . ), the corresponding input files are available from the author's website http://www.math.u-szeged.hu/ czedli/ . The output files proving these statements are also available there and some of them are attached as appendices to the extended version of the paper 1 ; see https://arxiv.org/ .
2.4. Lattice theoretical preparations. The proof of Theorem 2.2 will be organized as follows. Due to Theorem 2.4, it suffices to show that for each lattice X ∈ L KR , whenever L is a lattice with σ-many subuniverses (that is, σ(L) > 83), then X cannot be a subposet of L. Although we present some uniform arguments for several infinite sub-families of L KR , separate arguments will be needed for most of the small lattices in L KR . The following lemma is crucial. Lemma 2.10 (Antichain Lemma). If {a 0 , a 1 , a 2 } is a three-element antichain in a finite lattice with σ-many subuniverses, then Part (ii) of this lemma is trivial; we present it here to emphasize its implicit use in our considerations and in the input files of the program.
Proof. For the sake of contradiction, suppose that (i) fails for a lattice L with σmany subuniverses. Then X := {a 0 ∨a 1 , a 0 ∨a 2 , a 1 ∨a 2 } is a three-element antichain. It is well known that such an antichain generates a sublattice isomorphic to A 0 , the eight-element boolean lattice; see, for example, Grätzer [10,Lemma 73]. Combining Lemmas 2.3 and 2.8, we obtain that σ(L) ≤ σ(A 0 ) = 74, which contradicts the assumption that σ(L) > 83. Lemma 2.11. If L is a finite lattice with σ-many subuniverses, then A 0 is not a subposet of L.
Proof. For the sake of contradiction, suppose that A 0 is a subposet of L and σ(L) > 83. Since a, b, c play symmetric roles, Lemma 2.10(i) allows us to assume that The following lemma needs a bit longer proof and the use of the program. This proof exemplifies many ideas that will be needed later. Note that K 5 , defined by Figure 2, is a sublattice of G n and H n for n ≥ 1, this is why it deserves our attention.
Lemma 2.12. If L is a finite lattice with σ-many subuniverses, then K 5 is not a subposet of L.
Proof. For the sake of contradiction, suppose that σ(L) > 83 but K 5 is a subposet of L. For the notation of the elements of K 5 , see Figure 2.
Lattice theoretical preparatory part. We modify K 5 in L if necessary. The operations ∨ and ∧ will be understood in L. We can assume that e ∨ f = g, since otherwise we can replace g by e∨f . Of course, we have to show that this replacement results in an isomorphic subposet, but this is easy; analogous tasks will often be left to the reader. Namely, e ∨ f ≤ h would lead to e ≤ h, a contradiction, while e ∨ f ≥ h combined with g ≥ e ∨ f would lead to g ≥ h, another contradiction. By duality, we also assume that e∧f = c. Next, we can assume c∧d = b and, dually, g ∨h = k, because otherwise we can replace b and k by c ∧ d and g ∨ h, respectively. This is possible since, for example, a ≤ d implies that a ≤ c ∧ d while c ∧ d ≥ b and a ≥ b exclude that a ≥ c ∧ d. In the next step, we assume similarly that a ∧ b = o and j ∨ k = i. Note that the equalities assumed so far and the comparability relations among the elements imply further equalities: defines a partial algebra K (0) 5 on the set K 5 , which is a weak partial subalgebra of L. Note that the program calls the members of T constraints.
Computational part. The program proves that σ(K (0) 5 ) = 97.375, which means that we are not ready yet. Thus, a whole hierarchy of cases have to be investigated in general. (Here, there will be only two cases.) The idea is that for incomparable elements x and y, in notation, x y, such that x ∨ y or x ∧ y is not defined in the partial algebra, the argument splits into two cases: either x ∨ y (or x ∧ y) is one of the elements already present, or it is a new element of L that we add to the partial algebra. In terms of the program, we add a new constraint with or without adding a new element. Also, when we add a constraint, then we also add its consequences similarly to the previous paragraph where, say, e ∧ d = b. Note that if an element had three covers or three lower covers, then we would use Lemma 2.10 to split a case into three subcases, but this technique will be used later, not in the present proof. A case with name * will be denoted by (C * ).
(C1) We assume that c ∨ d = f and g ∧ h = f . Then e ∨ d = e ∨ c ∨ d = e ∨ f = g and, dually, e ∧ h = c. Adding these four constraints to the earlier ones, we get a new partial algebra K (1) 5 , which is a weak subalgebra of L. The program yields that σ(K Based on the argument for (C1) above, to make our style more concise, let us agree to the following terminological issue, which will usually be used implicitly in the rest of the paper.
Terminology 2.13. The cases we consider describe partial algebras, which are weak partial subalgebras of L; the σ-values of these partial algebras will be called the σ-values of the corresponding cases. If the σ-value of a case not greater than 83, then the case in question is excluded.
(C2) We assume that c ∨ d =: x < y := g ∧ h. We remove f from the weak partial algebra and add x and y. We remove the constraints of T that contain f but we add the new constraints c ∨ d = x, g ∧ h = y, e ∨ y = g, and e ∧ x = c. The last two constraints we add follow from x ≤ f ≤ y and the previous constraints containing f . Note that the rightmost oval in Figure 2 reminds us that now {f } is replaced by {x, y}. Since the σ-value of the present situation is 80.5625, (C2) is excluded.
After excluding both cases, that is, all possible cases, the proof of the lemma is complete.
Next, for later reference, we formulate a consequence, which trivially follows from Lemma 2.12.
Corollary 2.14. If L is a lattice with σ-many subuniverses and n ≥ 1, then none of G n and H n is a subposet of L.
In order to formulate the following lemma about the encapsulated 2-ladder F − 1 given in Figure 4, we need the following definition. This concept will be motivated by Corollary 2.17 later.
Definition 2.15. Let L and K be a finite lattices. A mapping ϕ : K → L will be called a (2.1)-embedding if ϕ is an order-embedding, ϕ(u) = ϕ(v) ∨ ϕ(w) holds for every triplet (u, v, w) ∈ K 3 of distinct elements such that u covers both v and w, and dually.
Note that if v and w are distinct elements covered by u in K, then u = v ∨ K w, and the dual of this observation also holds. Hence, every lattice embedding is a (2.1)-embedding but, clearly, not conversely. Proof. We can assume that F − 1 ⊆ L. The notation of the elements of F − 1 is given in Figure 4. We are going to modify these elements in L if necessary in order to obtain a (2.1)-embedding. The operations ∨ and ∧ will be understood in L. First, we let That is, f is incomparable with g; in notation, f g. We obtain similarly that f x for all x ∈ F − 1 such that x f . This allows us to replace f by f . To ease the notation, we will write f instead of f . So, F − 1 is still a subposet of L but now f = b∨c. Next, we replace c by c := f ∧g ≥ c; then it is straightforward to see that we still have a poset embedding.
Thus, after writing c instead of c , the notation still gives a poset embedding of F − 1 into L with the progress that now b ∨ c = f and f ∧ g = c. We continue in the same way step by step, always defining a new poset embedding such that the already established equalities remain true; note that the order of adjusting the elements is not at all arbitrary. In the next step, we replace b by b := e ∧ f ≥ b and g by g := c ∨ d ≤ g to add b = e ∧ f and g = c ∨ d to the list of valid equalities. We continue with setting a = b ∧ c and j = f ∨ g similarly. Finally, redefining i and o as e ∨ j and a ∧ d completes the proof.
Armed with Lemma 2.16, we can give an easy proof of the following statement.
Corollary 2.17. If L is a lattice with σ-many subuniverses, then F 1 is not a subposet of L.
Proof. Suppose the contrary. Then F − 1 , which is a sublattice of F 1 , is also a subposet of L. By Lemma 2.16, we can assume that F − 1 is a subposet of L such that the inclusion map is a (2.1)-embedding. Hence, we know that The σ-value of the situation described by (2.2)-(2.5) is 81.75.
The eight-element fence is the eight-element poset formed by the empty-filled elements on the right of Figure 2.
Lemma 2.18. If L is a finite lattice with σ-many subuniverses, then the eightelement fence is not a subposet of L.
Proof. For the sake of contradiction, suppose that σ(L) > 83 but the eight-element fence is a subposet of L. For the notation, see Figure 2.
Lattice theoretical preparatory part. We claim that we can assume that It suffices to show that if one of these equalities fails, then we can redefine the element in its right-hand side so that the valid equalities remain valid and the new eight-element poset is still (isomorphic to) a copy of the eight-element fence. By duality, it suffices to deal with the joins, which are understood in L, of course. We give the details only for c ∨ e = d, since the other cases need almost the same argument. So assume that d := c ∨ e < d.
since the first two of these equalities are due to definitions and the rest are easy consequences; for example, b ∧ e = b ∧ d ∧ e = c ∧ e = o while the rest follow by duality or symmetry.
Computational part. For the elements a, b, . . . , h, o, i subject to (2.6) and (2.7), the σ-value is 84.5; see Terminology 2.13. In other words, we obtain with our usual technique (that is, using the program and Lemma 2.3) that σ(L) ≤ 84.5. Since this estimate is too week to derive a contradiction, we distinguish two cases.
Adding these two equalities to (2.6) and (2.7), the σ-value is 79, which excludes this case.
(C2) We assume that x : have eleven elements and, in addition to the two equalities just mentioned, (2.6), and (2.7). Since the σ-value is 72, this case is also excluded.
Both cases have been excluded, which proves Lemma 2.18.
The lemma we have just proved trivially implies the following statement.
Corollary 2.19. If L is a lattice with σ-many subuniverses and n ≥ 1, then none of A n , E n+1 , and F n+1 is a subposet of L.
In the rest of the paper, due to Corollaries 2.14 and 2.19, we need to exclude only finitely many members of the infinite list L KR as subposets of a finite lattice L with σ-many subuniverses. After the proofs of Lemmas 2.12 and 2.18, our plan to exclude that a given member X of L KR occurs as a subposet of a lattice L with σ(L) > 83 is the following. After assuming that X is a subposet of L, first we need some lattice theoretical preparation to ensure a feasible computational time.
In the second phase, we reduce the estimate on σ(L) by assuming equations and introducing new elements in a systematic way as long as we obtain that σ(L) ≤ 83. In other words, we keep branching cases as long as all "leaves of our parsing tree" have σ-values at most 83. Unfortunately, this plan requires quite a lot of work; see Table (3.2) later. In the rest of the paper, we present some of the details in order the give a better impression how our plan works. The rest of the details are given by the output files of our program and some of them in the extended version of the paper; see Proof Technique 2.9 for their coordinates.
Remark 2.20. One may think of the following possibility: if X ∈ L KR is a subposet of L with σ(L) > 83, c, e ∈ X, and c ∨ X e = g, then either c ∨ L e = g in L, or x := c ∨ L e < g ∈ L \ X. If we could show that the second alternative (with x) always yields a better (that is, smaller) estimate of σ(L),  Figure 5. Note that X is a lattice but not a sublattice of H 0 . If X is a subposet of a finite lattice L such that c ∨ L e = g and g ∧ L f = d,

The rest of the lemmas and some proofs
In order to complete the proof of Theorem, we still need the following eight lemmas, in which L denotes a finite lattice.    Proof of Lemma 3.6. For the sake of contradiction, suppose that σ(L) > 83 but F 0 is a subposet of L.
Lattice theoretical preparatory part. Unless otherwise stated, the lattice operations are understood in L; in notation, x ∨ y will mean x ∨ L y and dually. Note that F 0 is a selfdual lattice and it has a unique selfdual automorphism Hence, after replacing c by c if necessary, we can assume that e ∨ g = c. In the next step, after replacing e by e := b ∧ c, we assume that b ∧ c = e; we still have a subposet (isomorphic to) F 0 .
In the next step, we can clearly assume that a ∨ b = i and f ∧ g = o. To summarize, we have assumed that the inclusion map is a (2.1)-embedding of F 0 into L, that is, Computational part. While splitting the possibilities into cases and subcases, we will benefit from the fact that both F 0 and (3.1) are selfdual. We keep splitting (sub)cases to more specific subcases only as long as their σ-values are larger than 83; this tree-like splitting structure will have thirteen leaves, that is, thirteen subcases with small σ-values that cover all possibilities. Every case below is either evaluated, that is, its σ-value is computed by the program, or the case is split further. Of courses, we evaluated all cases to see which of them need further splitting, but we present the σ-values only of the non-split cases. In fact, only the thirteen evaluated cases are needed in the proof. The (sub)cases are denoted by strings. When a case (C x) is mentioned, all the "ancestor cases", that is, (C y) for all meaningful prefixes of x are automatically assumed.
Since this is the dual of the previous assumption, we are in a selfdual situation. Observe that (Note that this case describes the situation when F 0 is a sublattice of L.) Since the σ-value of this case is 83, L has few subuniverses, whereby (C1a.1a.1a) is excluded.
(C1a.1a.1b): f ∨ g =: x such that x = c. (The notation "=:" means that x is defined as f ∨ g and f ∨ g = x is a new constraint.) Clearly, x < c. Using the incomparabilities among the elements of F 0 , it is straightforward to see that x is a new element. (In the rest of the paper, we will not emphasize that an element with a new notation is distinct from the rest of elements.) Since c = e ∨ g ≤ e ∨ x <= c, we have that e ∨ x = c. Since the σ-value of this case is 74.25, (C1a.1a.1b) is excluded. Thus, the case (C1a.1a.1) is also is excluded. Since (C1a.1a) is seldfual, the dual of (C1a.1a.1) is also excluded; this will be used in the next case.
(C1a.1a.2): a ∧ b =: x > e and f ∨ g =: y < c. Since c ∧ b = e and e ∨ g = c, it follows easily that c ∧ x = e and e ∨ y = c Since the σ-value is now 68, (C1a.1a.2) is excluded. Thus, (C1a.1a) is also excluded. (C1a.1b): This case is excluded, because its σ-value is 79.375. Thus, (C1a.1b) and so (C1a.1) are also excluded. Furthermore, since (C1a) is selfdual, we conclude that dual of (C1a.1) is also excluded; this fact will be used in the next case.   .1), is selfdual, the dual of (C1) is also excluded; this fact will be used below when (C2) is analyses. ( and so σ = 82.5, excluding this case. All cases have been excluded, and the proof of Lemma 3.6. Note that the proof above required to compute an estimate for σ(L) thirteen times. Let us call these thirteen values final σ-values. However, as mentioned previously, many more values were needed to find the proof. For example, the σ-value of (C1a.1a.1) is 90.5, and the inequality 90.5 > 83 is the reason to split the case (C1a.1a.1) into subcases. Remark 3.9 (Notes on the proofs of Lemmas 3.1-3.8). First, observe that σ(F 0 ) = 83 is the largest σ-value occurring in Lemma 2.8. Thus, Lemma 3.6 devoted to F 0 is the most crucial one in the paper. Since even the "human part" of its computerassisted proof is long and threatens with unnoticed human errors, we have elaborated two separate proofs of Lemma 3.6. One of these proofs is optimized in some sense and it has already been given, and it is also available from the corresponding file F0-output.txt. The other proof is less optimized and it is described only by its output file called F0-alternative-output.txt. The seven other lemmas in this section are less critical in the sense that their proofs need not exploit all the assumptions (but this fact does not always mean that their proofs are easy). Because of space considerations, no details of their proofs are given in the paper (except for the appendices of the extended version of the paper). Since the proofs of these seven lemmas are very similar to those we have already seen; we believe that the omitted and very tedious details would not be too exciting to the reader. Even if the reader would read these details, he should trust our computer program or develop his own computer program. However, to get some rough impression, the reader may want to cast a glance at some of these output files; these files together with the comments they contain constitute full proofs. (Some of these output files occur in the appendices of the extended version).
Next, we give the number of final σ-values that our proofs, that is, the program output files, contain. The * -labeled column refers to the second proof of Lemma 3.6 given in the downloadable file F0-alternative-output.txt. In order to explain some big numbers in the third row of the table, note the following. If X ∈ L KR contains no element with more than two covers or more than two lower covers, then the proof of the corresponding lemma is quite similar to that of Lemma 3.6; of course, we can exploit duality only if X itself is selfdual. Note that symmetries with respect to automorphisms can also be exploited. However, if there are elements with more than two lower covers or dually, like in case of H 0 , then there can be cases that we split into three subcases according to Lemma 2.10 as follows. Let a 0 , a 1 , and a 2 be the three lower covers of an element b in X ∈ L KR . (The case of upper covers is analogous.) Let c := a 0 ∨ a 1 ∨ a 2 in L. Then the following three subcases are considered. First, a 0 ∨ a 1 = c. Second, a 0 ∨ a 1 =: x < c and a 0 ∨ a 2 = c. Third, a 0 ∨ a 1 =: x < c, a 0 ∨ a 2 =: y < c, x = y are new elements, and a 1 ∨ a 2 = c. It is not surprising now that this three-direction splitting leads to more final σ-values than the two-direction splittings in the proof of Lemma 3.6. With an opposite effect, there is another factor related to the numbers of final σ-values. Namely, σ(X) < σ(F 0 ) = 83 for all X ∈ L KR \ {F 0 } that occur in the lemmas of this section, whereby we do not have to be so efficient for these X as for F 0 ; simply because our lemmas for these X state less than an affirmative answer to Problem 2.5(ii). To conclude Remark 3.9, we mention that there are many ways to prove the eight lemmas with the help of our program, and not much effort has been devoted to reduce the numbers in the third row of Table 3.2; such an effort would have required too much work. Some of these numbers might decrease in the future.
Finally, armed with our lemmas and corollaries, we are ready to present the concluding proof of the paper.
Proof of Theorem 2.2. For the sake of contradiction, suppose that L is a finite lattice such that σ(L) > 83. By Lemmas 2.11 and 3.1-3.8 and Corollaries 2.14, 2.17, and 2.19, none of the lattices occurring as excluded subposets in these twelve statements is a subposet of L. Using σ(L δ ) = σ(L) and applying these twelve statements to L δ , we obtain that none of the duals of the excluded lattices is a subposet of L. Hence, no member of L KR is a subposet of L, and Theorem 2.4 implies that L is planar, as required. Fence_8 is the 8-element fence; its edges are (we write xy to denote that y covers x): ab cb cd ed ef gf gh and we add o and i with oc oe and di fi |A|=10, A(without commas)={abcdefghoi}. Constraints: a+c=b c+e=d e+g=f ; the joins exist only for neighboring atoms b*d=c d*f=e f*h=g ; the meets exist only for neighboring atoms (also) d+f=i c*e=o ; since we define d+f=:i and c*e=:o (so) b*e=o c*f=o ; since b*e=b*d*e=c*e=o and c*f=c*d*f=c*e=o (and) d+g=i f+c=i; since d+g=d+e+g=d+f=i and f+c=f+e+c=f+d=i Result for A=fence_8: |Sub(A)| = 338, that is, sigma(A) = 84.5000000000000000*2^{|A|-8} .
We are not ready yet.
Fence_8 is the 8-element fence; its edges are (we write xy to denote that y covers x): ab cb cd ed ef gf fh and we add o and i with oc oe and di fi |A|=10, A(without commas)={abcdefghoi}. Constraints: a+c=b c+e=d e+g=f ; the joins exist only for neighboring atoms b*d=c d*f=e f*h=g ; the meets exist only for neighboring atoms (also) d+f=i c*e=o ; since we define d+f=:i and c*e=:o (so) b*e=o c*f=o ; since b*e=b*d*e=c*e=o and c*f=c*d*f=c*e=o (and) d+g=i f+c=i; since d+g=d+e+g=d+f=i and f+c=f+e+c=f+d=i (C1) b+d=i (so) b+e=i ; since b+e=b+c+e=b+d=i Result for A=fence_8: |Sub(A)| = 316, that is, sigma(A) = 79.0000000000000000*2^{|A|-8} .
This case is excluded.
Fence_8 is the 8-element fence; its edges are (we write xy to denote that y covers x): ab cb cd ed ef gf fh and we add o and i with oc oe and di fi |A|=11, A(without commas)={abcdefghoix}. Constraints: a+c=b c+e=d e+g=f ; the joins exist only for neighboring atoms b*d=c d*f=e f*h=g ; the meets exist only for neighboring atoms (also) d+f=i c*e=o ; since we define d+f=:i and c*e=:o (so) b*e=o c*f=o ; since b*e=b*d*e=c*e=o and c*f=c*d*f=c*e=o (and) d+g=i f+c=i; since d+g=d+e+g=d+f=i and f+c=f+e+c=f+d=i (C2) b+d=x ; since b+d=:x is distinct from i (so) b+e=x ; since b+e=b+c+e=b+d=x (and) x+f=i; since x+f>=d+f=i Result for A=fence_8: |Sub(A)| = 576, that is, sigma(A) = 72.0000000000000000*2^{|A|-8} .
This case is excluded. Both cases are excluded, q.e.d. The computation took 0/1000 seconds.

Appendix 2: H 0
Version of December 30, 2018 H_0 from Kelly-Rival: "Planar lattices"; its edges are oa ob oc ad bd be bh cg df dg eg fi gi hi Suppose, for contradiction, that H_0 is a subposet of L and L has many subuniverses. We will indicate by ' where the argument splits into 3 subcases (according to the Antichain Lemma in the paper); otherwise the argument splits into two subcases.
Few subuniverses, (C1.a3.a2') is excluded. Few subuniverses, (C1.b3') is excluded. Thus, (C1.b') is excluded. Now let us summarize where we are. Together with the automatic f*g=d, (C1) yields a selfdual situation. In this selfdual situation, (C1.a') and its dual have been excluded. Since (C1.b) is also excluded, so is its dual. By (#1), two of f,g, and h join to i; but neither f+g by (C1.a), nor f+h by (C1.b). By duality, neither a*b, nor a*c is o. This fact explain the form of (C1.c') below.