Symmetric embeddings of free lattices into each other

By a 1941 result of Ph. M. Whitman, the free lattice FL(3) on three generators includes a sublattice $S$ that is isomorphic to the lattice FL($\omega$)=FL($\aleph_0$) generated freely by denumerably many elements. The first author has recently"symmetrized"this classical result by constructing a sublattice $S\cong$ FL($\omega)$ of FL(3) such that $S$ is SELFDUALLY POSITIONED in FL(3) in the sense that it is invariant under the natural dual automorphism of Fl(3) that keeps each of the three free generators fixed. Now we move to the furthest in terms of symmetry by constructing a selfdually positioned sublattice $S\cong$ FL$(\omega)$ of FL(3) such that every element of $S$ is fixed by all automorphisms of FL(3). That is, in our terminology, we embed FL$(\omega)$ into FL(3) in a TOTALLY SYMMETRIC way. Our main result determines all pairs $(\kappa,\lambda)$ of cardinals greater than 2 such that FL$(\kappa)$ is embeddable into FL$(\lambda)$ in a totally symmetric way. Also, we relax the stipulations on $S\cong$FL$\kappa$ by requiring only that $S$ is closed with respect to the automorphisms of FL$(\lambda)$, or $S$ is selfdually positioned and closed with respect to the automorphisms; we determine the corresponding pairs $(\kappa,\lambda)$ even in these two cases. We reaffirm some of our calculations with a computer program developed by the first author. This program is for the word problem of free lattices, it runs under Windows, and it is freely available.


Introduction and our results
There are many nice and deep results on free lattices of the variety of all lattices. A large part of these results were achieved by Ralph Freese and J. B. Nation, to whom this paper is dedicated. Some of these results are included in [7,8,10,11,14,15] and in the monograph Freese, Ježek, and Nation [9], but this list is far from being complete. The monograph just mentioned serves as the reference book for the present paper.
By a classical result of Whitman [20], the free lattice FL(ω) = FL(ℵ 0 ) on denumerably many free generators is isomorphic to a sublattice of the free lattice FL (3) with three free generators. In fact, we know from a deep result of Tschantz [19] that there are many copies of FL(ω) in FL(3); namely, every infinite interval of FL(3) includes a sublattice isomorphic to FL(ω). For more about free lattices, the reader is referred to Freese, Ježek and Nation [9]. In this paper, we embed free lattices into each other symmetrically. For a free lattice F , δ = δ F will denote the natural dual automorphism of F that keeps the free generators fixed; (1.1) it is uniquely determined. A subset (or a sublattice) S of F is selfdually positioned if δ(S) = S. Selfduality is a sort of symmetry, and a selfdually positioned sublattice is necessarily selfdual. As the main result of [2], the first author proved that FL(3) has a selfdually positioned sublattice that is isomorphic to FL(ω). (1.2) Besides selfduality, there is a more general concept of symmetry, which is used even outside algebra; it is based on automorphisms. For a lattice L, let Aut(L) denote the automorphism group of L. We call a subset (or a sublattice) S of L symmetric if π(S) = π for every π ∈ Aut(F ). Also, an element u ∈ L is a symmetric element of L if {u} is a symmetric subset of L. Note that there is no symmetric element in FL(ω). If S contains only symmetric elements of L, then S is element-wise symmetric in L. Our key concept is the following. Definition 1.1. A lattice embedding ϕ : L → F of a lattice L into a free lattice F is totally symmetric if its range ϕ(L) = {ϕ(u) : u ∈ L} is a selfdually positioned and element-wise symmetric sublattice of F .
To expand our notation for all cardinal numbers κ ≥ 3, we denote by FL(κ) the free lattice with κ many free generators. If κ = n is a natural number, then we often write FL(n). Following the tradition, we often denote FL(ℵ 0 ) by FL(ω) even if ω is an ordinal number. In order to avoid ambiguity about natural numbers, we adhere to the notations N + := {1, 2, 3, . . . } and N 0 := {0} ∪ N + . The elements of N 0 are also cardinals; namely, the finite cardinal numbers. Our main result is the following. For later reference, we mention the following corollary even if it trivially follows from Theorem 1.2. In addition to Theorem 1.2 on total symmetry, we have some progress in studying selfdually positioned free sublattices, which is stated as follows.
Prerequisites. The reader is expected to have some basic familiarity only with the rudiments of lattice theory. That is, only some preliminary sections of the monographs, say, Grätzer [12] or Nation [16] are assumed. The results on free lattices that we need from the literature, mainly from Freese, Ježek, and Nation [9], are known for most lattice theorists and will be quoted with sufficient details since the paper is intended to be self-contained. In Section 5, we quote some recent achievements from [2]; when reading this section, the reader does not have to but may want to look into Czédli [2] 1 to verify how we quote from it.
Main ideas of the paper. In this subsection, we deal mainly with the totally symmetric embeddability of FL(ω) into FL (3), that is, with Part (B) of Theorem 1.2; the rest of the results are derived from or proved like Theorem 1.2(B), or they are easier.
First, we define symmetric elements m 1 < · · · < m 4 in FL (3), see (3.2), and m i will be the dual of m i for i ∈ {1, . . . , 4}. With some computation based on Whitman's condition, we can prove that (plan1) P = {m 1 , . . . , m 4 } ∪ {m 1 , . . . , m 4 } is the cardinal sum of two 4-element chains; see Figure 1 for an illustration, and (plan2) we prove some properties of P implying that the sublattice [P ] FL(3) generated by P in FL (3) is isomorphic to the completely free lattice CF(P ; ≤) generated by the ordered set (P ; ≤). Combining the isomorphism [P ] FL(3) ∼ = CF(P ; ≤) with the main result of Rival and Wille [17], we could immediately obtain an element-wise symmetric sublattice S of FL(3) such that S ∼ = FL(ω). However, we want more. Hence, (plan3) we define a, b ∈ [P ] FL(3) in (3.2), see also Figure 1, such that some computation based on Whitman's condition yields that FL(4) is isomorphic to [a, b, a, b] FL(3) , where a and b are the duals of a and b, respectively. Note that the restriction of δ FL(3) to FL(4) is not the natural dual automorphism of FL(4) since it swaps the free generators of FL(4). At this stage, Corollary 1.3 is already proved. Next, let δ sw 4 denote the unique dual automorphism of FL(4) = FL(y 0 , y 1 , y 2 , y 3 ) for which we have that δ sw 4 (y 0 ) = y 1 , δ sw 4 (y 1 ) = y 0 , δ sw 4 (y 3 ) = y 4 , δ sw 4 (y 4 ) = y 3 ; we call it the swapping dual automorphism of FL(4).
(plan4) By the "diagonal method" of Czédli [2], FL(ω) is a sublattice of FL(4) closed with respect to δ sw 4 . Finally, a straightforward computation will show that if we combine (plan3) and (plan4), then their "swapping" features neutralize each other and we obtain a totally symmetric embedding FL(ω) → FL(3), as required.
If we embed FL(ω) or FL(κ) into FL(λ), rather than into FL(3), then some of the above-mentioned computations, most of which can be done by a computer, become longer. Fortunately, we can often rely on the following fact, which deserves separate mentioning here: with two trivial exceptions, every symmetric element of FL(λ) is given by a near-unanimity term.
Let us note that the isomorphism [P ] FL(3) ∼ = CF(P ; ≤) and the above-mentioned result of Rival and Wille [17] are only motivating facts and will not be used in the detailed proof. Note also that this subsection will not be used in the rest of the paper; due to elaborated details and many internal references, the proofs are readable without keeping the main ideas in mind. Finally, since we also need to prove (our second) Theorem 1.4, we will prove more on {a, b, a, b} than what is required by (plan3).
Outline. Our main results, Theorems 1.2 and 1.4, and our main ideas have already been presented; the rest of the paper is structured as follows. We add some comments and two corollaries to the main result in Section 2. These corollaries characterize the pairs (κ, λ) of cardinals having the property that there is an embedding FL(κ) → FL(λ) with symmetric range or with selfdually positioned and symmetric range. The lion's share of our construction and (the Key) Lemma 3.1 stating that this construction works are given in Section 3. The Key Lemma is proved in Section 4. Section 5 combines the construct given in Section 3 with that given in Czédli [2]. Section 6 completes the proofs of our theorems and proves the corollaries. Finally, Section 7 describes our computer program for the word problem of free lattices; note that this program and its source file are freely available and the program proves Corollary 1.3 in less than a millisecond.

Remarks and corollaries
We will often use the convenient notation FL(κ) = FL(x i : i < κ) in order the specify the free generating set {x i : i < κ} of size κ; in this case, i ∈ N 0 denotes an ordinal number and i < κ is understood as |i| < κ. (Equivalently, we could say that a cardinal number is the smallest ordinal with a given cardinality; then i < κ would have its usual meaning for ordinals.) Mostly, κ is not larger than ℵ 0 and then i ∈ N 0 will denote a nonnegative integer. Of course, we can write FL(x 0 , . . . , x κ−1 ) or FL(y i : i < κ) if κ is finite or we need to denote the free generators in a different way, respectively. Also, we write FL(X) if we do not want to specify the size |X| of the free generating set X. An element u of a lattice L is doubly irreducible if L \ {u} is closed with respect to both joins and meets, that is, if u is both join irreducible and meet irreducible. The set of doubly irreducible elements of L will be denoted by D ir (L). We know from Freese, Ježek, and Nation [9], see Corollary 1.9 and the first sentence of the proof of Corollary 1.12, that Since a dual automorphism maps join-irreducible elements to meet irreducible ones and vice versa, (2.1) and (2.2) imply that for every dual automorphism ψ of FL(κ), we have that Remark 2.1. The concept of totally symmetric embeddings might raise the question whether we could consider even stronger embeddings whose ranges are elementwise symmetric and are in element-wise selfdual positions. We obtain from (2.3) that the answer is negative, since at most the free generators are in element-wise selfdual positions and they form an antichain. This justifies our terminology to call the embeddings in Theorem 1.2 totally symmetric.
For a lattice L, let DAut(L) be the set of all automorphisms and dual automorphisms of L. As a consequence of (2.1) and (2.2), note that for κ ≤ ω, D ir (FL(κ)) is closed with respect to every π ∈ DAut(FL(κ)). Furthermore, each π ∈ DAut(FL(κ)) is determined by its restriction to D ir (FL(κ)) if we know whether it is an automorphism or a dual automorphism. Let us call a subset S of L a DAutsymmetric subset if π(S) = S for all π ∈ DAut(L). A dually positioned and element-wise symmetric sublattice of FL(λ), like the range of a totally symmetric embedding, is clearly DAut-symmetric but not conversely. This might give some hope that a counterpart of Theorem 1.2 for embeddings with DAut-symmetric ranges allows the case when κ is an odd natural number. However, the following corollary of Theorem 1.2 shows that this is not so if κ = λ. This corollary as well as Corollary 2.4 will be proved in Section 6.  The situation is different if we deal with embeddings whose ranges are symmetric with respect only to Aut FL(λ).
Since our concepts are based on the automorphisms of FL(n), where n := λ is a positive integer, let us have a look at what these automorphisms and the elements of DAut(FL(n)) are. Let Sym n = Sym({0, 1, . . . , n − 1}) denote the group of all permutations of {0, 1, . . . , n − 1} with respect to composition, and let C 2 = {1, −1} be the cyclic group of order 2, considered a subgroup of the group (R \ {0}; ·) of nonzero real numbers with respect to multiplication. Using that FL(n) = FL(x i : i < n) is freely generated by the set {x i : i < n}, the following remark is straightforward and its proof is omitted.

Construction and the Key Lemma
3.1. Notation. The elements of a free lattice FL(κ) = FL(x i : i < κ) will be represented by lattice terms over the set {x i : i < κ} of variables. Although there are many terms representing the same element of FL(κ), it will not cause any confusion that we often treat and call these terms as elements of the free lattice; (3.1) see pages 10-11 in Freese, Ježek and Nation [9] for a more rigorous setting. The dual of a term t will always be denoted by t; the overline reminds us that dualizing at visual level means to reflect Hasse diagrams across horizontal axes. The Symmetric part of the Free Lattice of FL(κ) will be denoted as follows; capitalization explains the acronym: Clearly, SFL(κ) is a sublattice of FL(κ); this fact will often be used implicitly.

3.2.
Constructing some important terms. In this subsection, we give a construction for the particular case (κ, λ) = (4, n) of Theorem 1.2(A). Let us agree to the following conventions. The set {0, 1, . . . , n − 1} will also be denoted by [n).
The inequality i < n is equivalent to i ∈ [n). Whenever x i , x j , etc. refer to a free generator of FL(n) = FL(x 0 , . . . , x n−1 ), then i, j, . . . will automatically belong to [n); this convention will often save us from indicating, say, that i < n or i, j ∈ [n) below the and operation signs. Also, we frequently abbreviate the conjunction of i ∈ [n) and j ∈ [n) \ {i} by the short form i = j, and self-explanatory similar other abbreviations may also occur. For the rest of this section, let 3 ≤ n = λ ∈ N + and FL(λ) = FL(n) = FL(x 0 , . . . , x n−1 ) = FL(x i : i < n). By induction on j, we define the following n-ary lattice terms over the set {x i : i < n} of variables; according to (3.1), they will also be considered elements of FL(n). Namely, we let For later reference, we note that the set will play an important role in the paper. We say that a subset X of a lattice freely generates if the sublattice S generated by X is a free lattice with X as the set of free generators. Next, we formulate our Key Lemma, which is stronger than asserting that the set in (3.3) freely generates. The proof of the Key Lemma will be postponed to Section 4.
Lemma 3.1 (Key Lemma). If 3 ≤ n ∈ N + , then the elements m j and m j for j ∈ N 0 , a, b, a, and b all belong to SFL(n). Furthermore, {a, a, b, b, x 0 } is a fiveelement subset of SFL(n) that freely generates.
For later reference, based on Remark 2.5, note the following trivial lemma.
Proof of Lemma 3.2. The first equality above follows from the fact that in (3.2), each stipulation of the form i 1 < i 2 can be replaced by i 1 = i 2 . The second equality follows from the first one.

The proof of the Key Lemma
From the theory of free lattices, we only use three basic facts, which we recall below as lemmas; all of them can be found in Freese, Ježek and Nation [9]. An element u of a lattice L is join prime if for all k ∈ N + and x 0 , . . . , x k−1 ∈ L, the inequality u ≤ x 0 ∨ · · · ∨ x k−1 implies that u ≤ x i for some i ∈ [k). Meet prime elements are defined dually. An element is doubly prime if it is join prime and meet prime. The following statement says that free lattices satisfy Whitman's condition (W), see Whitman [20].
Next, we describe whether a subset of a free lattice generates freely or not.
Since the lattice operations are idempotent, it is obvious that the join and the meet of two n-ary NU-terms are NU-terms. (4.1) If t 1 and t 2 are n-ary lattice terms such that t 1 = t 2 in FL(n), see (3.1), then Proof. First, we are going to show that if g ∈ SFL(n) and , implying (4.2). Next, assume that t satisfies the assumptions of the lemma. As a binary lattice term, t(x, . . . , x, y) equals one of x, y, x ∧ y, and x ∨ y in FL(x, y). If we had that, in FL(x, y), t(x, . . . , x, y) = x ∧ y, then Hence, t(x, . . . , x, y) is distinct from x ∧ y, and it is distinct also from x ∨ y by duality. The case t(x, x, . . . , x, y) = y is impossible, because it would imply that 1 = t(0, . . . , 0, 1) ≤ t(0, 1, . . . , 1) = t(1, . . . , 1, 0) = 0 holds in the two-element lattice 2, which is a contradiction. Hence, t(x, x, . . . , x, y) = x, which means that t is an NU-term since it is symmetric.
The only coatom of SFL(n) is s. Except from its bottom 0 SFL(n) = i<n x i and top The statement of this lemma for 3 < n ∈ N + is illustrated by Figure 1, where only the two thick edges stand for coverings in SFL(n); the thin lines indicate comparabilities that need not be coverings. (These comparabilities will be proved later; see Lemma 4.9.) The reflection across the symmetry center point, which is not indicated in the figure, represents the restriction of δ FL(n) to SFL(n). We could obtain a similar figure for n = 3 by removing the vertices m 0 and m 0 and decreasing the subscripts of the remaining vertices by 1; see Lemma 4.9.
Proof. By Lemma 4.4 and the duality principle, we need to show only that t ≥ s holds for every near-unanimity term t ∈ SFL(n). This follows from the fact that In order to get some preliminary insight into SFL(n), note the following. As usual, M n denotes (n + 2)-element lattice given in Figure 2. For a permutation π : [n) → [n), let π X and π A denote the permutation of {x i : i < n} and that of {a i : i < n} defined by π X (x i ) = x π(i) and π A (a i ) = a π(i) for all i < n, respectively. These permutations uniquely extend to automorphisms π aut X ∈ Aut(SFL(n)) and π aut A ∈ Aut(M n ), respectively. Consider the natural homomorphism η : FL(n) → M n defined by η(t(x 0 , . . . , x n−1 )) = t(a 0 , . . . , a n−1 ).
Clearly, π aut A • η = η • π aut X . This implies easily that the η-image of a symmetric element of FL(n) is a symmetric element of M n . But the symmetric elements of M n form 2 (as a sublattice), and we obtain that η SFL(n) : SFL(n) → 2 is a surjective homomorphism. So the kernel of η SFL(n) cuts SFL(n) into a prime ideal, the dark-grey lower part together with the bottom element in Figure 1 Proof. First, we show by induction on j that x k for all j ∈ N + and k, r, s ∈ [n). (4.4) Suppose, for a contradiction, that (4.4) fails, and let j ∈ N + be the smallest number violating it. Pick k, r, s ∈ [n) such that p j ≤ x k . Since x k is meet prime by Lemma 4.1, we can assume that p Since this inequality does not violate (4.4) by the choice of j, it follows that j − 1 / ∈ N + . Hence, j = 1 and (3.2) turns p A straightforward induction yields that the sequence is increasing, that is, Armed with the preparations above, it suffices to prove the strict inequalities in the lemma only for i = 0, since then the case i > 0 will follow by symmetry. For the sake of contradiction, suppose that there exists a j ∈ N 0 such that (4.8) Let j be minimal with this property. By symmetry, the superscript 0 does not play a distinguished role here. That is, until the end of the proof, We are going to derive a contradiction from (4.8) by infinite descent. Since p 0 would lead to x 1 ∧ x 2 ≤ x 0 , which fails even in 2, we obtain that j > 0. So, as the first step of the descent, we conclude that j − 1 ∈ N 0 . Thus, (4.8) and j − 1 ∈ N 0 imply that (4.10) By (4.6), this inequality would fail if one of the two meetands on the left was omitted. Hence, (W) and (4.4) yield that p j−1 . We formulate this inequality together with j − 1 ∈ N 0 also in the following way: the condition j − m ∈ N 0 and there is a u m ∈ [n) such that p holds for m = 1. In order to continue the descent in infinitely many steps, we assert that for every m ∈ N + , if (4.11) holds for m, then it holds also for m + 1. In order to prove (4.12), assume (4.11) for m. The case j − m = 0 is ruled out by (4.4), whence j − (m + 1) ∈ N 0 . Hence, the inequality in (4.11) gives that (4.13) As before, we are going to apply (W) to (4.13); however, the argument for the meetands on the left is a bit longer. If u m / ∈ {1, 2}, then we obtain from (4.6) that none of the meetands on the left of (4.13) can be omitted from the inequality. If u m = 1, then p (1) j still cannot be omitted by (4.6), but we need the same fact for the other meetand, p (2) j . Observe that if p (2) j was omitted and u m equaled 1, then we would have by (4.7) that p j−1 , contradicting (4.9). So none of the two meetands in question can be omitted if u m = 1, and the same is true for u m = 2 since 1 and 2 play symmetric roles. This shows that no matter what is u m , none of the two meetands on the left of (4.13) can be omitted. Therefore, (4.13), (W) and (4.4) imply that p j−(m+1) . Thus, we conclude that (4.11) holds for m + 1, completing the proof of (4.12). Consequently, (4.11) holds for all m ∈ N + , which contradicts the finiteness of j ∈ N 0 and completes the proof of Lemma 4.6.
Proof. It suffices to deal only with {m j : j ∈ N 0 }. The sequence in question is increasing by its definition, see (3.2), and Lemma 4.6. For the sake of contradiction, suppose that m j ≤ m j−1 holds for some j ∈ N + . Then, since all joinands of m j are less than or equal to m j−1 , we have that, in particular, (4.14) It follows from (4.5) that, with a = (a 0 , . . . , a n−1 ) ∈ M n × · · · × M n , we have that m j ( a) = 0 while p (1) j ( a) = a 1 and p (2) j ( a) = a 2 . Hence, neither of the meetands on the left of (4.14) can be omitted without breaking the inequality. Thus, applying (W) to (4.14), it follows that p j−1 . Therefore, (4.11) holds for m = 1. Consequently, it follows from (4.12) that (4.11) holds for all m ∈ N + , which is a contradiction since j − m / ∈ N 0 for m > j.
The following lemma states something on SFL(n), not on FL(n). Proof. By duality, it suffices to deal only with m j . In order to show that m j is join prime, assume that m j ≤ h 1 ∨ h 2 where h 1 , h 2 ∈ SFL(n). Remember that the containment here means that h 1 and h 2 are fixed points of every automorphism of FL(n). We have to show that m j ≤ h i for some i ∈ {1, 2}. There are two cases to consider. First, assume that there exists an i ∈ {1, 2} such that p In this case, for each (u 1 , u 2 ) ∈ [n) × [n) with u 1 < u 2 , pick a permutation σ ∈ Sym([n)) such that σ(0) = u 1 and σ(1) = u 2 . By Remark 2.5 and Lemma 3.2, Forming the join of these inequalities for all meaningful pairs (u 1 , u 2 ), we obtain that m j ≤ h i , as required. Second, assume that (4.15) fails. Then (W) applied to the (side terms of the) inequality p By (3.2) and Lemma 4.6, we have that x 0 ≤ h 1 ∨h 2 . Since x 0 is join prime by Lemma 4.1, x 0 ≤ h i holds for some i ∈ {1, 2}. Applying (4.2), we obtain that h i = 1 SFL(n) . Hence, m j ≤ 1 SFL(n) = h i , as required. Now that both the validity and the failure of (4.15) have been considered, we conclude that m j is a join prime element of SFL(n).
Next, we are going to show that m j is meet prime in SFL(n). Suppose the contrary, and pick h 1 , h 2 ∈ SFL(n) such that h 1 ∧ h 2 ≤ m j but h 1 m j and h 2 m j . We are going to obtain a contradiction by infinite descent. In order to do so, it suffices to show that the condition holds for r = 0, and to show that for every r ∈ N 0 , if (4.17) holds for r, then it also holds for r + 1. (4.18) j−0 . Hence, (4.17) holds for r = 0. Next, in order to show (4.18), assume that r ∈ N 0 satisfies condition (4.17). We cannot have that h 1 ∧ h 2 ≤ x ur , because otherwise h i ≤ x ur for some i ∈ {1, 2} by the meet primeness of x ur , see Lemma 4.1, and so (4.2) would give that h i = 0 SFL(n) ≤ m j , contradicting our assumption. Combining h 1 ∧ h 2 x ur with (4.17) and (3.2), we obtain that j − r = 0. Hence, j − (r + 1) ∈ N 0 and In order to exclude that h 1 ≤ p (ur) j−r , suppose the contrary. Then, by Lemma 4.6, j−(r+1) and (4.17) holds for r + 1. We have verified (4.18). Thus, it follows that (4.17) holds for all r ∈ N 0 . This is the required contradiction proving that m j is meet prime in SFL(n), completing the proof of Lemma 4.8.
For 3 ≤ n ∈ N + , in connection with (3.2) and Lemma 4.5, let With the ordering of SFL(n) restricted to P n , P n = (P n ; ≤) is an ordered set, which is described by the following lemma; see also Figure 1 for 3 < n ∈ N + . This lemma explains why the case n = 3 differs slightly from the case n > 3. Next, to deal with (ii), we assume that n = 3. Clearly, m 0 = s and m 0 = s. Lemma 4.5 gives that s ≤ s while M 3 witnesses that s = s. We have seen that m 0 = s < s = m 0 . For the sake of contradiction, suppose that m 1 ≤ m 1 . Then each of the joinands of m 1 is less than or equal to every meetand of m 1 . In particular, we have that p 1 is less than or equal to its dual, that is, (4.20) By (W), duality, and since x 0 and x 1 play symmetric roles, we can assume that (4.20) holds after omitting its underlined meetand. Hence, x 0 is less than or equal to the right hand side of (4.20). Using that x 0 is join prime by Lemma 4.1, we obtain that either have obtained a contradiction, proving (ii). Next, we turn our attention to (iii). Suppose, for a contradiction, that m i ≤ m j for some i, j ∈ N + . We obtain from Lemma 4.7 that In particular, m 0 < m 0 and m 1 ≤ m 1 . The first of these two inequalities contradicts part (i) if n > 3, while the second one contradicts part (ii) if n = 3. So we obtain a contradiction for all n ≥ 3, whereby (iii) holds.
Finally, we obtain (iv) basically from η(m j ) = 1 and η(m i ) = 0, see (4.3), as follows. Using (4.5), its dual, and (3.2) defining our terms, we obtain that m j (a 0 , . . . , a n−1 ) = 1 Mn and m i (a 0 , . . . , a n−1 ) = 0 Mn . This implies (iv) and completes the proof of Lemma 4.9. Now, to indicate that we are progressing in the desired direction, we are going to formulate a corollary. Note, however, that neither this corollary, nor its proof, nor the concept defined in the present paragraph will be used in this paper, so the reader can skip over them. Following Dean [3] and Dilworth [5], an ordered set P = (P ; ≤ P ) completely freely generates a lattice K if P is a subset of K, ≤ P is the restriction of the lattice order ≤ K to P , and for every lattice L and every orderpreserving map ϕ : (P ; ≤ P ) → L, there exists a lattice homomorphism K → L that extends ϕ. If so, then we denote K by CF(P ; ≤). The ordered set (P n ; ≤) was defined right before (and in) Lemma 4.9; let [P n ] FL(n) = [P n ] SFL(n) denote the sublattice generated by it.
Part (iii) is a consequence of Theorem 1.2(B); the point is that we can easily conclude Corollary 4.10(iii) from known results and the previous lemmas.
Proof of Corollary 4.10. As opposed to other proofs in the paper, the present argument relies on some outer references that are not quoted with full details. Part (i) is clear. For the validity of Part (ii), we need to show that for arbitrary (k + k)-ary lattice terms t 1 and t 2 , the inequality holds in FL(n) iff it holds in the completely free lattice CF(P n ; ≤). The satisfaction of (4.21) in CF(P n ; ≤) can be tested by Dean's algorithm, which is a generalization of Whitman's algorithm; see Dean [3] or see Freese, Ježek, and Nation [9,Theorem 5.19]. This is a recursive algorithm that uses only the following three properties of CF(P n ; ≤) and (P n ; ≤): (D1) CF(P n ; ≤) satisfies (W), (D2) the elements of P n are doubly prime, and (D3) the description of the ordering of P n .
It follows from Lemmas 4.2, 4.7, 4.8, and 4.9 that these properties hold for P n as a subset of FL(n). Therefore, Dean's algorithm gives the same result in CF(P n ; ≤) as (D1)-(D3) give in FL(n). Hence, we conclude that [P n ] FL(n) is completely freely generated by (P n ; ≤), proving Part (ii). Thus, CF(P n ; ≤) can be embedded into SFL(n). Using that FL(ω) can be embedded into FL(3) by Whitman [20] and FL (3) can be embedded into CF(P n ; ≤) by the main result of Rival and Wille [17], we conclude by transitivity that SFL(n) has a sublattice isomorphic to FL(ω). This proves Part (iii). Now, we are ready to prove our Key Lemma.

Proof of Lemma 3.1. It is clear by (3.2) that
In order to apply Lemma 4.3 and complete the proof in this way, it suffices to show that none of the inequalities holds in FL(n), because then the same will be true for their duals. For example, if a ≥ x 0 ∧ a ∧ b ∧ b held, then we could apply δ = δ FL(n) from (1.1) to this inequality to obtain that (ineq4) holds. First, we consider (ineq1). Suppose, for a contradiction, that it holds. Using (4.22), the elements we are going to deal are in SFL(n). For i ∈ N + , we have that m i < m i+1 by Lemma 4.7, whereby Lemma 4.5 gives that m i ≤ s. Since m 0 < m i by Lemma 4.7, Lemma 4.5 gives that s ≤ m i , whence m i ≤ s. Since m i ≤ s and m i ≤ s for all i ∈ N + , (3.2) gives that a ∨ a ∨ b ∨ b ≤ s. This is a contradiction, because (4.2) and (ineq1) imply that a ∨ a ∨ b ∨ b = 1 SFL(n) > s.
Clearly, there is a lot of similarity between the treatment for (ineq2) and that for (ineq3). Namely, both arguments rely on (4.2), Lemmas 4.7, 4.8, and 4.9, and some comparabilities among the subscripts. In an analogous way, the argument for (ineq4) and that for (ineq5) are also very similar; this justifies that only the first of them will be detailed.
For the sake of contradiction, suppose that (ineq4) is satisfied, that is, We are going to parse this inequality by (W), taking into account that, according to Lemmas 4.1 and 4.8, both meetands on the left and the five non-underlined joinands on the right of (4.25) are doubly prime elements. Therefore, either one of the two meetands is less than or equal to one of the five non-underlined joinands, or m 1 ∧ m 3 ≤ m 2 ∧ m 4 ; so we need to consider only these possibilities. If we had that m 1 ≤ x 0 or m 3 ≤ x 0 , then (4.2) and Lemma 4.7 would lead to m 2 < m 1 = 0 SFL(n) or m 2 < m 3 = 0 SFL(n) , which are contradictions. If one of the two meetands was less than or equal to another non-underlined joinand, then Lemma 4.7 or Lemma 4.9 would prompt give a contradiction. We are left with the case m 1 ∧ m 3 ≤ m 2 ∧ m 4 , but then m 1 ∧ m 3 ≤ m 2 , so the meet primeness of m 2 gives that m 1 ≤ m 2 , contradicting Lemma 4.7, or m 3 ≤ m 2 , contradicting Lemma 4.9(iii). Therefore, (ineq4) fails, as required. Finally, as we have already mentioned, (ineq5) fails by an analogous argument. This completes the proof of Lemma 3.1

From the Key Lemma to a stronger statement
If a subset X of a lattice freely generates, then so do the subsets of X. Thus, (the Key) Lemma 3.1 in itself implies that, for every 3 ≤ λ ∈ N + , there is a totally symmetric embedding FL(4) → FL(λ). In particular, Lemma 3.1 implies Corollary 1.3. In this section, with the extensive help of Czédli [2], we lift the rank 4 of FL(4) to all even natural numbers κ ≥ 4 and even to ℵ 0 . That is, we are going to prove the following lemma. Remember that a, a, b and b have been defined in (3.2). Proof. First, in order to make our references to Czédli [2] convenient, we need to deal with the notation. Let (y 1 , y 2 , y 3 , y 4 ) := (a, a, b, b) ∈ SFL(n) 4  For example, FL(6) = FL(x 1+i 1 , x 1+i 2 : 2i < 6) = FL(x 1 1 , x 1 2 , x 2 1 , x 2 2 , x 3 1 , x 3 2 ). It is important that 6 and, in general, κ is even or ℵ 0 , because for an odd integer κ ∈ N + , FL(x 1+i 1 , x 1+i 2 : 2i < κ) would be FL(κ + 1) rather than FL(κ). This paragraph is to tailor the second half of [2, Lemma 4.1] to the present situation; the reader may want to skip over it. It is irrelevant for us what a and b denote in [2]; they are not the same as here. It is also irrelevant that FL(4) = FL(y 1 , . . . , y 4 ) is embedded into FL(x, y, z) in [2] but into FL(n) here; these two embeddings are different even for n = 3. Using the first page, . Therefore, since |D| = 2k + 1 = κ, we need to show only that D freely generates. By duality and Lemma 4.3, it suffices to exclude that x 0 ≤ c 0 ∨ c 1 ∨ · · · ∨ c 2k−1 , or (6.1) We know from Lemma 3.1 that the sublattice S := [a, a, b, b, x 0 ] of FL(n) is freely generated by the set {a, a, b, b, x 0 }. Therefore, the self-maps and ξ 2 := a a b b x 0 a a b b 0 S extend to endomorphisms ξ 1 : S → S and ξ 2 : S → S, respectively. Using the inclusion C ⊆ [a, a, b, b], we obtain that ξ 1 ([a, a, b, b]) = {0 S }. Taking the equality ξ 1 (x 0 ) = 1 S also into account, we obtain that the endomorphism ξ 1 does not preserve inequality (6.1). Hence, (6.1) fails, as required. Next, suppose that (6.2) holds. Using C ⊆ [a, a, b, b], it follows that the restriction of ξ 2 to [a, a, b, b] is the identity map. Therefore, since ξ 2 is order-preserving, its application to (6.2) yields that c j ≤ i∈[2k)\{j} c i . But this is a contradiction since C freely generates, and we conclude that (6.2) fails, as required.
Proof of Corollary 2.3. In order to prove the implication (ii) ⇒ (i), assume that (ii) holds. We can also assume that κ = λ since otherwise the identity map FL(κ) → FL(κ) = FL(λ) does the job. As it is pointed out right after ( Thus, the embedding given by Lemma 5.1 has a DAut-symmetric range. Hence, (ii) ⇒ (i). Before proving the converse implications, we formulate and verify some observations, some of which will be useful also in the proof of Corollary 2.4 later. This is why instead of assuming DAut-symmetry, we often assume less, the usual symmetry (with respect to automorphisms). Since Aut(FL(X)) acts transitively on the set X of free generators, it follows trivially that if S is a symmetric sublattice of FL(X) such that S ∩ X = ∅, then S = FL(λ). (6.4) As a straightforward consequence of (2.1), observe that if ϕ : FL(κ) → FL(λ) is an arbitrary embedding and S := ϕ(FL(κ)), then |D ir (S)| = κ. (6.5) Since automorphisms and dual automorphisms preserve double primeness, we obtain the following observation.

(6.6)
For u ∈ FL(λ), the set {τ (u) : τ ∈ Aut(FL(λ))} will be called the orbit of u (with respect to automorphisms). We are going to prove the following property of orbits.

A computer program and its background
Historical background. There are various known algorithms to solve the word problem of free lattices and that of finitely presented lattices. They are discussed in Sections 8 and 9 of Chapter XI of the monograph Freese, Ježek, and Nation [9]; see also Dean [4], Evans [6], McKinsey [13], and Skolem [18] for the original papers. In addition to this list, there is an additional algorithm given in Czédli [1]. We know from [9] that the algorithms given by Skolem, Freese, and Herrmann run in polynomial time; so does the one given in [1]. However, it is only Whitman's algorithm with the modifications explained in [9] that is fast enough for our purposes.
A new computer program. The first author has developed a Dev-Pascal 1.9.2 (Freepascal) program for the word problem of free lattices. This problem is based on the Freese-Whitman algorithm, as it is given in Freese, Ježek, and Nation [9]. The program runs in Windows environment (tested only under Windows 10), and it can be downloaded from the author's website. The program takes its input from a text file; several sample input files are also donwloadable. We used this program on our personal computer with IntelCore i5-4440 CPU, 3.10 GHz, and 8.00 GB RAM.
Results achieved with the computer program. First, we used the program to give alternative proofs. In particular, we used it to prove the (Key) Lemma 3.1 for n = 4.
(7.1) Also, we used the program to prove that for n = 3, the Key Lemma remains valid if we replace m 1 , m 2 , m 3 and m 4 by m 5 , m 7 , m 8 , and m 9 , respectively; (7.2) this gives an alternative proof of Corollary 1.3. By the paragraph preceding (4.25), it would not be difficult to show that the stipulation n = 3 can be omitted from (7.2), but this or a similar strengthening of (7.2) is not pursued. In addition to reaffirming some results from the previous sections, we could use the program to find an entirely new construction to prove Corollary 1.3. In order to describe it, we use the notation introduced in Remark 2.5 to define a joinhomomorphism ν In order to ease the notation, we will write x, y, and z instead of x 0 , x 1 , and x 2 , respectively. Note that the program recognizes (appropriate commands for) 1st generator 2nd generator N var (generating set) ν (∨) and µ (∧) in input files. Take the following ternary terms, that is, elements of FL(3) = FL(x, y, z).
With a and b from (7.4) and (7.5) and their duals, a and b , the program proved that {a , a , b, b } freely generates a sublattice of FL(x, y, z), (7.6) which obviously implies Corollary 1.3. Note that for each of (7.1), (7.2), and (7.6), the program ran less than a millisecond on our computer.
Finally, for a lattice term t, we define the total number N var (t) of variables of t by induction as follows: N var (t) = 1 if t is a variable and N var (t 1 ∨ t 2 ) = N var (t 1 ∧ t 2 ) = N var (t 1 ) + N var (t 2 ).
Note that, say, x = x∧(x∨y) in FL(x, y, z) but N var (x) = 1 is distinct from N var (x∧ (x ∨ y)) = 3. Hence, as opposed to what (3.1) suggests, we do not define N var for the elements of FL(x, y, z). For a set {t 1 , . . . , t k } of terms, let N var ({t 1 , . . . , t k }) = N var (t 1 ) + · · · + N var (t k ). Table 7.7 shows how the function N var compares the terms describing the free generating set given in (3.3) for n = 3 and those given in (7.4) and (7.6). Another difference between (3.3) and (7.6) is that, as opposed to the set {a, a, b, b, x 0 } from (the Key) Lemma 3.1, the program shows that {a , a , b, b , x} does not generate freely.