Proof.

1. (1) This follows from a general result; see [Reference Sapir30, Proposition 3.1].

2. (2) See [Reference Jackson and Lee16, Lemma 2.1].

3. (3) See [Reference Lee, Rhodes and Steinberg27, Proposition 4.1].

Proof.

1. (1) This is routinely verified.

2. (2) See [Reference Gusev and Sapir11, Fact 2.1].

Proof. By Lemma 2.2(2), the variety $\mathbf {V}$ satisfies the identities for some $k \geq 2$ .

(1) Suppose that $\mathbf {V}$ is completely regular and so satisfies the identity $\sigma : x^{n+1} \approx x$ for some $n \geq 1$ . Then, $\mathbf {V}$ satisfies ⓐ $_1$ because

and $\mathbf {V}$ satisfies ⓒ $_1$ because

Therefore, the inclusion $\mathbf {V} \subseteq \mathbf {J}_1$ holds. Conversely, every idempotent monoid, and so every subvariety of $\mathbf {J}_1$ , is completely regular.

(2) If $\mathbf {V}$ is commutative, then it satisfies $xy \approx yx$ and so $xy$ is not an isoterm for $\mathbf {V}$ . Conversely, suppose that $xy$ is not an isoterm for $\mathbf {V}$ . Then, it is easily shown that $\mathbf {V}$ is either commutative or idempotent; see, for instance, [Reference Gusev and Sapir11, Lemma 2.7]. If $\mathbf {V}$ is idempotent, then it is completely regular and so is commutative by part (1).

(3) This follows from the description of all commutative varieties [Reference Head13].

Proof. (1) By Lemma 2.2(2), the variety $\mathbf {V}$ satisfies the identities for some $n \geq 2$ . By Lemma 2.4, the assumption ${Rq}\{xhx\} \notin \mathbf {V}$ implies that $\mathbf {V}$ satisfies a nontrivial identity $\sigma : xhx \approx \mathbf {w}$ for some $\mathbf {w} \in \mathscr {A}^{\ast }$ . Let $p = |\mathbf {w}|_x$ and $q = |\mathbf {w}|_h$ . There are two cases.

Case 1: $p \leq 1$ or $q \neq 1$ or there exists $z \in \mathscr {A} \backslash \{x,h\}$ such that $|\mathbf {w}|_z = r \geq 1$ . Then, $\mathbf {V}$ satisfies $x^2 \approx x^p$ or $h \approx h^q$ or $1 \approx z^r$ , whence $\mathbf {V}$ is completely regular. Therefore, the inclusion $\mathbf {V} \subseteq \mathbf {J}_1$ holds by Lemma 2.3(1), so that $\mathbf {V}$ is a subvariety of all the varieties in (2-1).

Case 2: $p \geq 2$ and $q = 1$ with ${\textsf {con}}(\mathbf {w}) = \{ x,y\}$ . Then, $\mathbf {w} = x^{p_1}hx^{p_2}$ for some $p_1,p_2 \geq 0$ such that $p_1+p_2 = p \geq 2$ and $(p_1,p_2) \neq (1,1)$ . If $p = 2$ , then $(p_1,p_2) \in \{ (2,0),(0,2)\}$ , so that either $\mathbf {V} \subseteq \mathbf {J}_n\{xhx \approx x^2h\}$ or $\mathbf {V} \subseteq \mathbf {J}_n\{xhx \approx hx^2\}$ . Therefore, suppose that ${p_1+p_2 = p \geq 3}$ , so that either $p_1 \geq 2$ or $p_2 \geq 2$ . Then, $\mathbf {V}$ satisfies the identity $\tau : x^2 \approx x^{2+k}$ with $k = p-2$ and so also the identities because

Now, if $p_1 \geq 2$ , then

similarly, if $p_2 \geq 2$ , then $\mathbf {V} \models (\blacktriangleright )$ . Consequently, either $\mathbf {V} \subseteq \mathbf {J}_2\{(\blacktriangleleft )\}$ or $\mathbf {V} \subseteq \mathbf {J}_2\{(\blacktriangleright )\}$ .

(2) This is [Reference Lee25, Corollary 3.6].

Proof.

1. (1) This part can be extracted from [Reference Gusev and Sapir11, Proof of Proposition 3.1].

2. (2) This part follows from Lemma 2.3(2) and part (1).

Proof. It follows from [Reference Gusev9, Lemma 6.7] that the equivalence

holds for some possibly empty set $\Sigma $ that consists of some of the following identities:

1. (a) $xhx^2 \approx hx^2$ , $x^2hx \approx x^2h$ , $x^2hxtx \approx x^2htx$ ;

2. (b) $(\prod _{i=1}^{n-1} (xh_i))xh_nx^2 \approx (\prod _{i=1}^{n-1} (xh_i))xh_nx$ for some $n \geq 1$ ;

3. (c) $(\prod _{i=1}^{n-1} (xh_i))x^2h_nx \approx (\prod _{i=1}^{n-1} (xh_i))xh_nx^2$ for some $n \geq 1$ .

(Note that if $n=1$ , then the identities (b) and (c) are $xh_1x^2 \approx xh_1x$ and $x^2h_1x \approx xh_1x^2$ , respectively.) It is easily seen that the deduction (▸) holds, so the equivalence (▸) $\cup\ \Sigma _1 \sim $ (▸) $\cup\ \Sigma $ follows. Since it is routinely checked that the equivalence {(▸), $\text {(a)}, \text {(b)}, \text {(c)} \} \sim $ {(▸), (2-5) } holds, the identities in $\Sigma $ can be chosen from (2-5).

Proof. Let $\mathbf {V}$ be any noncommutative subvariety of $\mathbf {O} \cap \mathbf {J}_2$ . Then, by Lemma 2.6(2), the variety $\mathbf {V}$ is defined by {(2-2), for some $\Sigma _1 \subseteq $ (2-3) and some $\Sigma _2 \subseteq $ (2-4). Since the deduction (2-2) $\vdash $ (▸) holds, it follows from Lemma 2.7 that the identities in $\Sigma _1$ can be chosen from (2-5). Hence, it remains to show that the identities in $\Sigma _2$ can be chosen from (2-6).

Let $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ be any identity in $\Sigma _2 \subseteq $ (2-4). If $\mathbf {c} \in \{ 1,\, txy\}$ , then by definition, the identity $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ is already in (2-6). Therefore, it suffices to consider the case when

$$ \begin{align*} \mathbf{c} = \prod_{i=1}^m (t_i\mathbf{c}_i) = t_1\mathbf{c}_1 t_2 \mathbf{c}_2 \cdots t_m \mathbf{c}_m, \end{align*} $$

where $\mathbf {c}_1,\mathbf {c}_2,\ldots ,\mathbf {c}_m \in \{ 1,x,y \}$ and $m \geq 1$ . In the remainder of this proof, it is shown that if either $\mathbf {c}_k = 1$ or $\mathbf {c}_k = \mathbf {c}_{k+1} \in \{ x,y \}$ for some k, then the equivalence

(2-7) $$ \begin{align} \{ (2\hbox{-}2),\, \mathbf{h}\mathbf{p}\mathbf{c} \approx \mathbf{h}\mathbf{q}\mathbf{c}\} \sim \{ (2\hbox{-}2),\, \mathbf{h}\mathbf{p}\mathbf{c}' \approx \mathbf{h}\mathbf{q}\mathbf{c}'\} \end{align} $$

holds, where $\mathbf {c}'$ is obtained from $\mathbf {c}$ by removing the factor $t_k\mathbf {c}_k$ . It follows that the identity $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ can be chosen so that $(\mathbf {c}_1,\mathbf {c}_2,\ldots ,\mathbf {c}_m) \in \{ (x,y,x,y,\ldots ),\, (y,x,y,x,\ldots ) \}$ ; in other words, $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ can be chosen from (2-6).

Case 1: $\mathbf {c}_k = 1$ with $1 \leq k \leq m$ . If $\mathbf {c}_m=1$ so that $\mathbf {c} = ( \prod _{i=1}^{m-1} (t_i\mathbf {c}_i) ) t_m$ and $\mathbf {c}' = \prod _{i=1}^{m-1} (t_i\mathbf {c}_i)$ , then it is obvious that $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c} \sim \mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}'$ . Thus, assume that $1 \leq k < m$ so that

$$ \begin{align*} \mathbf{c} = \bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) t_k \bigg( \prod_{i=k+1}^m (t_i\mathbf{c}_i) \bigg) \quad \text{ and } \quad \mathbf{c}' = \bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) \bigg( \prod_{i=k+1}^m (t_i\mathbf{c}_i) \bigg). \end{align*} $$

The deduction $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c} \vdash \mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}'$ is obvious. Conversely, since $\mathbf {c}$ is obtained by making the substitution $t_{k+1} \mapsto t_kt_{k+1}$ in $\mathbf {c}'$ , the deduction $\mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}' \vdash \mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ holds. It follows that the equivalence (2-7) holds.

Case 2: $\mathbf {c}_k = \mathbf {c}_{k+1} \in \{ x,y\}$ with $1 \leq k < k+1 \leq m$ . Then,

$$ \begin{align*} \mathbf{c} = \bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) t_k\mathbf{c}_k t_{k+1} \mathbf{c}_k \bigg( \prod_{i=k+2}^m (t_i\mathbf{c}_i) \bigg) \quad \text{and} \quad \mathbf{c}' = \bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) t_{k+1}\mathbf{c}_k \bigg( \prod_{i=k+2}^m (t_i\mathbf{c}_i) \bigg). \end{align*} $$

Deleting the variable $t_{k+1}$ from both sides of $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ followed by making the substitution $t_k \mapsto t_{k+1}$ results in the identity

$$ \begin{align*} \mathbf{h}\mathbf{p}\bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) t_{k+1}\mathbf{c}_k^2 \bigg( \prod_{i=k+2}^m (t_i\mathbf{c}_i) \bigg) \approx \mathbf{h}\mathbf{q}\bigg( \prod_{i=1}^{k-1} (t_i\mathbf{c}_i) \bigg) t_{k+1}\mathbf{c}_k^2 \bigg( \prod_{i=k+2}^m (t_i\mathbf{c}_i) \bigg); \end{align*} $$

since $\mathbf {c}_k \in \{ x,y \} = {\textsf {con}}(\mathbf {p}) = {\textsf {con}}(\mathbf {q})$ , the identity (▸) deducible from (2-2) can be used to eliminate from both sides the occurrence of $\mathbf {c}_k$ that immediately follows $t_{k+1}$ , resulting in $\mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}'$ . Thus, the deduction $\{$ (2-2), $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c} \} \vdash \mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}'$ holds. Conversely, making the substitution $t_{k+1} \mapsto t_k\mathbf {c}_kt_{k+1}$ on both sides of $\mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}'$ results in $\mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ , so the deduction $\mathbf {h}\mathbf {p}\mathbf {c}' \approx \mathbf {h}\mathbf {q}\mathbf {c}' \vdash \mathbf {h}\mathbf {p}\mathbf {c} \approx \mathbf {h}\mathbf {q}\mathbf {c}$ holds. Hence, the equivalence (2-7) holds.

Proof. Let $\sigma \in \{ x^2hxtx \approx x^2htx,\, \circledR _m \}$ . Since $\mathbf {O} \models $ (2-2) $\vdash $ (▸), the finitely based variety $\mathbf {O} \cap \mathbf {J}_2\{ \sigma \}$ is locally finite by Lemma 2.1(1). In what follows, this variety is shown to be small and so is Cross by Lemma 2.1(2).

Let $\mathbf {V}$ be any subvariety of $\mathbf {O} \cap \mathbf {J}_2\{ \sigma \}$ . By Lemma 2.3(3), only three subvarieties of $\mathbf {O} \cap \mathbf {J}_2\{ \sigma \}$ are commutative; specifically, they are , where $n \in \{ 0,1,2\}$ . Therefore, assume that $\mathbf {V}$ is not commmutative, so that by Proposition 2.8, it is defined by for some $\Sigma _1 \subseteq $ (2-5) and $\Sigma _2 \subseteq $ (2-6). Since (2-5) is finite, there are only finitely many choices for the set $\Sigma _1$ . As for the set $\Sigma _2$ , there are infinitely many choices due precisely to identities from (2-6) of the following types:

$$ \begin{align*} \lambda_m: \mathbf{h}\mathbf{p} \prod_{i=1}^m (t_i\mathbf{a}_i) \approx \mathbf{h}\mathbf{q} \prod_{i=1}^m (t_i\mathbf{a}_i) \quad \text{and} \quad \rho_m: \mathbf{h}\mathbf{p} \prod_{i=2}^{m+1} (t_i\mathbf{a}_i) \approx \mathbf{h}\mathbf{q} \prod_{i=2}^{m+1} (t_i\mathbf{a}_i), \quad m \geq 1. \end{align*} $$

However, in each of the following cases, it is shown that whenever $\lambda _m$ or $\rho _m$ is required for the definition of $\mathbf {V}$ , it is possible to choose m from a fixed finite set. It follows that there are only finitely many choices for $\Sigma _2$ . Consequently, there are only finitely many choices for $\mathbf {V}$ , whence $\mathbf {O} \cap \mathbf {J}_2\{ \sigma \}$ is small.

Case 1: $\sigma $ is $x^2hxtx \approx x^2htx$ . Consider the identity $\lambda _m$ with $m \geq 7$ . Removing the variables $t_3$ and $t_4$ from both sides of $\lambda _m$ results in

(2-8) $$ \begin{align} \mathbf{h}\mathbf{p} \cdot t_1x \cdot t_2y \cdot x \cdot y \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i) \approx \mathbf{h}\mathbf{q} \cdot t_1x \cdot t_2y \cdot x \cdot y \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i); \end{align} $$

in other words, $\lambda _m \vdash $ (2-8). Recall that (2-2) $\vdash $ (▸) and $x,y \in {\textsf {con}}(\mathbf {p}) = {\textsf {con}}(\mathbf {q})$ ; hence, the deduction $\{$ (2-2), $\sigma ,\lambda _m\} \vdash \lambda _{m-2}$ holds because

$$ \begin{align*} \{(2\text{-}2),\sigma,\lambda_m\} \vdash\! \mathbf{h}\mathbf{p} \cdot t_1x \cdot t_2y \cdot t_5x \cdot t_6y \hspace{-1pt}\prod_{i=7}^m (t_i\mathbf{a}_i) & \stackrel{(\blacktriangleright)}{\approx} \mathbf{h}\mathbf{p} \cdot t_1x^2 \cdot t_2y^2 \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i) \\ & \stackrel{\sigma}{\approx} \mathbf{h}\mathbf{p} \cdot t_1x^2 \cdot t_2y^2 \cdot x \cdot y \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i) \\ & \stackrel{(2\text{-}8)}{\approx} \hspace{-1pt}\mathbf{h}\mathbf{q} \cdot t_1x^2 \hspace{-1.5pt}\cdot t_2y^2 \hspace{-1.5pt}\cdot x \cdot y \cdot t_5x \cdot t_6y \hspace{-1pt}\prod_{i=7}^m (t_i\mathbf{a}_i) \\ & \stackrel{\sigma}{\approx} \mathbf{h}\mathbf{q} \cdot t_1x^2 \cdot t_2y^2 \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i) \\ & \stackrel{(\blacktriangleright)}{\approx} \mathbf{h}\mathbf{q} \cdot t_1x \cdot t_2y \cdot t_5x \cdot t_6y \prod_{i=7}^m (t_i\mathbf{a}_i) \vdash \lambda_{m-2}. \end{align*} $$

The deduction $\lambda _{m-2} \vdash \lambda _m$ is obvious, so that $\{(2\text{-}2),\sigma ,\lambda _m\} \sim \{(2\text{-}2),\sigma ,\lambda _{m-2}\}$ . If ${m - 2 \geq 7}$ , then the same argument gives $\{$ (2-2), $\sigma ,\lambda _{m-2}\} \sim \{$ (2-2), $\sigma ,\lambda _{m-4}\}$ . This can be repeated to obtain

$$ \begin{align*} \{(2\text{-}2),\sigma,\lambda_m\} \sim \{(2\text{-}2),\sigma,\lambda_{m-2}\} \sim \cdots \sim \{(2\text{-}2),\sigma,\lambda_k\} \end{align*} $$

for some $k \leq 6$ . Similarly, $\{(2\text{-}2),\sigma ,\rho _m\} \sim \{(2\text{-}2),\sigma ,\rho _k\}$ for some $k \leq 6$ . Therefore, whenever any of the identities $\lambda _m$ and $\rho _m$ are required for the definition of $\mathbf {V}$ , it is possible to choose it from $\{\lambda _k,\rho _k \,|\, 1 \leq k \leq 6\}$ .

Case 2: $\sigma $ is Ⓡ $_m$ . Since $\mathbf {p},\mathbf {q} \in \{x,y\}^+$ are such that $1 \leq |\mathbf {p}|_x = |\mathbf {q}|_x, \, |\mathbf {p}|_y = |\mathbf {q}|_y \leq 2$ , it is easily seen that the deduction Ⓡ $_m \vdash \{ \lambda _k, \rho _k\}$ holds for all $k \geq m$ . Therefore, whenever any identity from $\{\lambda _k,\rho _k \,|\, k \geq 1\}$ is required for the definition of $\mathbf {V}$ , it is possible to choose it from $\{\lambda _k,\rho _k \,|\, 1 \leq k \leq m-1\}$ .

Proof.

1. (1) This follows from [Reference Lee, Rhodes and Steinberg27, Theorem 5.23].

2. (2) If $\mathbf {V}$ is completely regular, then $\mathbf {V} \subseteq \mathbf {J}_1$ by Lemma 2.3(1), so that both ${Q} \notin \mathbf {V}$ and $\mathbf {V} \models x^nhxtx^n \approx x^nhtx^n$ hold. If $\mathbf {V}$ is not completely regular, then the equivalence holds by [Reference Gusev8, Lemma 2.12].

3. (3) This follows from [Reference Almeida1, Proposition 11.10.2].

Proof. By assumption, the variety $\mathbf {V}$ satisfies the identities and so is locally finite by Lemma 2.1(1).

(1) Suppose that ${A_0} \notin \mathbf {V}$ . Then, $\mathbf {V}$ satisfies $\sigma : (x^2y^2)^2 \approx x^2y^2$ by Lemma 2.11(1) so that

Therefore, the inclusion $\mathbf {V} \subseteq \mathbf {Q}$ holds by Lemma 2.10(4). Conversely, if $\mathbf {V} \subseteq \mathbf {Q}$ , then $\mathbf {V}$ satisfies the identity $x^2y^2 \approx y^2x^2$ ; but since ${A_0}$ does not satisfy this identity, ${A_0} \notin \mathbf {V}$ .

(2) Suppose that ${Q} \notin \mathbf {V}$ . Then, $\mathbf {V}$ satisfies $\tau : x^2hxtx^2 \approx x^2htx^2$ by Lemma 2.11(2) so that

$$ \begin{align*} \mathbf{V} \models xhxtx \stackrel{(\blacktriangleleft),(\blacktriangleright)}{\approx} x^2hxtx^2 \stackrel{\tau}{\approx} x^2htx^2 \stackrel{(\blacktriangleleft),(\blacktriangleright)}{\approx} xhtx \vdash xhxtx \approx xhtx. \end{align*} $$

Therefore, the inclusion $\mathbf {V} \subseteq \mathbf {{A_0}}$ holds by Lemma 2.10(1). Conversely, if $\mathbf {V} \subseteq \mathbf {{A_0}}$ , then $\mathbf {V}$ satisfies the identity $xhxtx \approx xhtx$ ; but since ${Q}$ does not satisfy this identity, ${Q} \notin \mathbf {V}$ .

(3) Suppose that $\mathbf {{A_0}} \vee \mathbf {Q} \subseteq \mathbf {V}$ . Let $\mathbf {U}$ be any variety in $\mathfrak {L}(\mathbf {V})$ that is not in $\mathfrak {L}(\mathbf {{A_0}} \vee \mathbf {Q})$ . Then, $\mathbf {U} \nsubseteq \mathbf {Q}$ and $\mathbf {U} \nsubseteq \mathbf {{A_0}}$ so that ${A_0},{Q} \in \mathbf {U}$ by parts (1) and (2). Hence, ${\mathbf {U} \in [\mathbf {{A_0}} \vee \mathbf {Q},\mathbf {V}]}$ .

Proof. (1) It is easily checked, either directly or by referring to Lemma 2.10, that the monoids ${A_0}$ and ${Q}$ satisfy the identities (2-9). It follows from [Reference Sapir32, Theorem 4.5] that the variety $\mathbf {{A_0}} \vee \mathbf {Q}$ can be defined by (2-9) $\cup\ \Sigma $ for some set $\Sigma $ of identities from (2-3). Consider any identity

$$ \begin{align*} \sigma: x^{e_0} \prod_{i=1}^m (h_ix^{e_i}) \approx x^{f_0} \prod_{i=1}^m (h_ix^{f_i}) \end{align*} $$

from $\Sigma $ where $e_0,f_0, e_1,f_1,\ldots ,e_m,f_m \geq 0$ ; $\sum _{i=0}^me_i, \sum _{i=0}^mf_i \geq 2$ ; and $m \geq 0$ . It follows from Lemma 2.13 that for each $i \in \{ 0,1,\ldots ,m \}$ , either $e_i = f_i = 0$ or $e_i,f_i \geq 1$ . If ${m \geq 1}$ , then the identity $\sigma $ is clearly deducible from $\{$ (◂), (▸) $\}$ ; if $m = 0$ , then the identity $\sigma $ is $x^{e_0} \approx x^{f_0}$ with $e_0,f_0 \geq 2$ and so is deducible from the consequence $x^3 \approx x^2$ of (◂). Therefore, the identities in $\Sigma $ are not required in the definition of the variety $\mathbf {{A_0}} \vee \mathbf {Q}$ , whence $\mathbf {{A_0}} \vee \mathbf {Q} = {\textsf {var}}\{ $ (2-9) $\}$ .

(2) If $\mathbf {V}$ is any proper subvariety of $\mathbf {{A_0}} \vee \mathbf {Q}$ , then either ${A_0} \notin \mathbf {V}$ or ${Q} \notin \mathbf {V}$ , whence by Corollary 2.12, either $\mathbf {V} \subseteq \mathbf {Q}$ or $\mathbf {V} \subseteq \mathbf {{A_0}}$ . Therefore, $\mathbf {{A_0}}$ and $\mathbf {Q}$ are the only maximal subvarieties of $\mathbf {{A_0}} \vee \mathbf {Q}$ . Now, the variety $\mathbf {{B_0}}$ is the unique maximal subvariety of both $\mathbf {{A_0}}$ and $\mathbf {Q}$ , and the lattice $\mathfrak {L}(\mathbf {{B_0}})$ is as given in the proposition [Reference Lee20, Reference Lee25].

Proof. (1) If $\mathbf {V}$ is completely regular, then $\mathbf {V} \subseteq \mathbf {J}_1$ by Lemma 2.3(1), so that both ${F}_1 \notin \mathbf {V}$ and $\mathbf {V} \models $ (◂) hold. If $\mathbf {V}$ is not completely regular, then the result follows from [Reference Gusev8, Lemma 2.10].

(2) Suppose that $\mathbf {F} \nsubseteq \mathbf {V}$ and $\overleftarrow {{E}} \notin \mathbf {V}$ . Then, since $\overleftarrow {\mathbf {E}} \subseteq \mathbf {{B_0}}$ by Proposition 2.14(2), it follows from the assumption $\overleftarrow {{E}} \notin \mathbf {V}$ that ${B_0} \notin \mathbf {V}$ . Therefore, by Lemma 2.11(3), one has $\mathbf {V} \models x^2 \approx x$ or $\mathbf {V} \models x^2hx^2 \approx x^2h$ or $\mathbf {V} \models x^2hx^2 \approx hx^2$ .

Case 1: $\mathbf {V} \models x^2 \approx x$ . Then, $\mathbf {V}$ is completely regular and so is Cross by Lemma 2.3.

Case 2: $\mathbf {V}$ satisfies $\sigma : x^2hx^2 \approx x^2h$ . Then, since

$$ \begin{align*} \mathbf{V} \models x^2h \stackrel{\sigma}{\approx} x^2hx^2 \stackrel{(\blacktriangleright)}{\approx} x^2hx \vdash x^2hx \approx x^2h, \end{align*} $$

the inclusion $\mathbf {V} \subseteq \mathbf {J}_2 \{$ (▸) $,\, x^2hx \approx x^2h \} = \mathbf {F}$ holds. However, this inclusion is proper due to the assumption $\mathbf {V} \neq \mathbf {F}$ . It follows that $\mathbf {V}$ is Cross since $\mathbf {F}$ is almost Cross by Proposition 3.1(1).

Case 3: $\mathbf {V}$ satisfies $\tau : x^2hx^2 \approx hx^2$ . Then, since

it follows from Lemma 2.5(2) that $\mathbf {V}$ is Cross.

Proof. It is easy to check that ${A_0}$ and ${Q}$ satisfy the identities $\{$ ⓒ $_2$ , (◂), (▸), (3-1), Ⓡ $_2 \}$ that define $\mathbf {H}_2$ , so the inclusion $\mathbf {{A_0}} \vee \mathbf {Q} \subseteq \mathbf {H}_2$ holds. To establish the reverse inclusion, it suffices to show that $\mathbf {H}_2$ satisfies the four identities from (2-9) that define ${\mathbf {{A_0}} \vee \mathbf {Q}}$ ; see Proposition 2.14(1). The first three identities {(◂), (▸), Ⓡ $_2 \}$ from (2-9) are clearly satisfied by $\mathbf {H}_2$ . The fourth identity $xhytxy \approx xhytyx$ from (2-9) is also satisfied by $\mathbf {H}_2$ because

Proof. The variety $\mathbf {H}$ satisfies the identities (3-4)–(3-6) because

The deduction (3-4) $\vdash $ (3-3) holds vacuously, so that $\mathbf {H} \models $ (3-3). Hence,

Proof. It is easily seen that $\mathbf {H}_m \in [\mathbf {{A_0}} \vee \mathbf {Q},\, \mathbf {H}]$ for all $m \geq 2$ . Conversely, let $\mathbf {V}$ be any variety in $[\mathbf {{A_0}} \vee \mathbf {Q},\, \mathbf {H}]$ . By Lemma 3.4, the variety $\mathbf{H}$ is a subvariety of $\mathbf{O} \cap \mathbf{J}_2$ because it satisfies the identities $\{$ (3-5), (3-6) $\}$ = $\{$ (2-2) $\}$ . Hence, it follows from Proposition 2.8 that $\mathbf {V} = \mathbf {H}(\Sigma _1 \cup \Sigma _2)$ for some $\Sigma _1 \subseteq $ (2-5) and some $\Sigma _2 \subseteq $ (2-6). Since $\mathbf {{A_0}} \vee \mathbf {Q} \subseteq \mathbf {V} = \mathbf {H}(\Sigma _1 \cup \Sigma _2)$ :

1. (a) ${A_0}$ and ${Q}$ satisfy every identity in $\Sigma _1 \cup \Sigma _2$ .

However, any identity in $\Sigma _1 \cup \Sigma _2$ that is satisfied by $\mathbf {H}$ is redundant in the definition of $\mathbf {V} = \mathbf {H}(\Sigma _1 \cup \Sigma _2)$ . Hence, it can further be assumed that:

1. (b) $\mathbf {H}$ does not satisfy any identity in $\Sigma _1 \cup \Sigma _2$ .

It is easily checked that each identity from (2-5) is either not satisfied by ${Q}$ or satisfied by $\mathbf {H}$ . Therefore, by (a) and (b), none of the identities from (2-5) belong to $\Sigma _1$ . Hence, $\Sigma _1$ is empty and so $\mathbf {V} = \mathbf {H}\Sigma _2$ . Consider any identity from ${\Sigma _2 \subseteq }$ (2-6), say

$$ \begin{align*} \sigma: \mathbf{h}\mathbf{p}\mathbf{c} \approx \mathbf{h}\mathbf{q}\mathbf{c}, \end{align*} $$

where $\mathbf {h} \in \{ 1,\, yh\}$ , $\mathbf {p},\mathbf {q} \in \{x,y\}^+$ , and $\mathbf {c} \in \{ 1,\, txy,\, \prod _{i=1}^m (t_i\mathbf {a}_i), \,\prod _{i=2}^{m+1} (t_i\mathbf {a}_i) \,|\, m \geq 1 \}$ with

$$ \begin{align*} 1 \leq |\mathbf{p}|_x = |\mathbf{q}|_x,\, |\mathbf{p}|_y = |\mathbf{q}|_y \leq 2; \quad |\mathbf{h}\mathbf{p}\mathbf{c}|_x = |\mathbf{h}\mathbf{q}\mathbf{c}|_x,\, |\mathbf{h}\mathbf{p}\mathbf{c}|_y = |\mathbf{h}\mathbf{q}\mathbf{c}|_y \geq 2; \end{align*} $$

and $(\mathbf {a}_1,\mathbf {a}_2,\mathbf {a}_3,\mathbf {a}_4,\ldots ) = (x,y,x,y,\ldots )$ . In the following, it is shown that $\mathbf {H}\{\sigma \} = \mathbf {H}_m$ for some $m \geq 2$ . The proof is then complete since the inclusions $\mathbf {H}_2 \subseteq \mathbf {H}_3 \subseteq \mathbf {H}_4 \subseteq \cdots $ are easily checked.

Now, according to the above conditions, the words $\mathbf {p}$ and $\mathbf {q}$ can be any of

$$ \begin{align*} x^ry^s, \quad y^rx^s, \quad xy^rx, \quad yx^ry, \quad xyxy, \quad yxyx, \quad r,s \in \{1,2\}. \end{align*} $$

Since x and y are nonsimple variables of both $\mathbf {h}\mathbf {p}\mathbf {c}$ and $\mathbf {h}\mathbf {q}\mathbf {c}$ , the identities $\{$ (◂), (▸) $\}$ can be used to convert the identity $\sigma $ into $\mathbf {h}\mathbf {p}'\mathbf {c} \approx \mathbf {h}\mathbf {q}'\mathbf {c}$ , where $\mathbf {p}'$ and $\mathbf {q}'$ can be any of

$$ \begin{align*} x^2y^2, \quad y^2x^2, \quad xy^2x, \quad yx^2y, \quad xyxy, \quad yxyx. \end{align*} $$

The identities $\{$ ⓒ $_2$ , (3-2), (3-3) $\}$ can then be used to convert the identity $\mathbf {h}\mathbf {p}'\mathbf {c} \approx \mathbf {h}\mathbf {q}'\mathbf {c}$ into $\mathbf {h}\mathbf {p}"\mathbf {c} \approx \mathbf {h}\mathbf {q}"\mathbf {c}$ , where $\mathbf {p}",\mathbf {q}" \in \{ x^2y^2,\, y^2x^2,\, yxyx \}$ . In summary, we can use the identities $\{$ (◂), (▸), ⓒ $_2$ , (3-2), (3-3) $\}$ of $\mathbf {H}$ to convert the identity $\sigma $ into $\mathbf {h}\mathbf {p}"\mathbf {c} \approx \mathbf {h}\mathbf {q}"\mathbf {c}$ . Therefore, $\mathbf {H}\{ \sigma \} = \mathbf {H}\{\mathbf {h}\mathbf {p}"\mathbf {c} \approx \mathbf {h}\mathbf {q}"\mathbf {c}\}$ , whence one may assume that:

1. (c) $\mathbf {p},\mathbf {q} \in \{ x^2y^2,\, y^2x^2,\, yxyx \}$ .

Suppose that $\mathbf {h} = yh$ . Then, by (b) and (c), the identity $\sigma $ can be any of the following:

$$ \begin{align*} \sigma_1: yh \cdot x^2y^2\mathbf{c} \approx yh \cdot y^2x^2\mathbf{c}, \!\quad \sigma_2: yh \cdot x^2y^2\mathbf{c} \approx yh \cdot yxyx\mathbf{c}, \!\quad \sigma_3: yh \cdot y^2x^2\mathbf{c} \approx yh \cdot yxyx\mathbf{c}. \end{align*} $$

Since $\mathbf {H} \models \{$ (▸), (3-1), (3-2) $\}$ by Lemma 3.4, and that the deductions

$$ \begin{align*} yh \cdot x^2y^2\mathbf{c} \stackrel{(3\text{-}1)}{\approx} yh \cdot yx^2y^2\mathbf{c} \stackrel{(\blacktriangleright)}{\approx} yh \cdot yx^2y\mathbf{c} \stackrel{(3\text{-}2)}{\approx} yh \cdot yxyx\mathbf{c} \end{align*} $$

hold, it follows that $\mathbf {H} \models \sigma _2$ and $\mathbf {H}\{\sigma _1\} = \mathbf {H}\{\sigma _3\}$ . Hence, $\sigma $ cannot be $\sigma _2$ due to (b), and it suffices to assume that $\sigma $ is $\sigma _3$ . In fact, $\sigma $ can be assumed to be the simpler identity $\sigma _3':y^2x^2\mathbf {c} \approx yxyx\mathbf {c}$ , obtained by removing the prefix $\mathbf {h} = yh$ from both sides of $\sigma _3$ . Indeed, the inclusion $\mathbf {H}\{\sigma _3'\} \subseteq \mathbf {H}\{\sigma _3\}$ clearly holds, while the reverse inclusion also holds because

so that $\mathbf {H}\{\sigma \} = \mathbf {H}\{\sigma _3\} = \mathbf {H}\{ \sigma _3'\}$ . Consequently, one may assume that $\mathbf {h}=1$ .

The identity $\sigma $ is thus $\mathbf {p}\mathbf {c} \approx \mathbf {q}\mathbf {c}$ . Then, it follows from (b) and (c) that $\mathbf {H}\{\sigma \}$ can be any of the following varieties:

$$ \begin{align*} \mathbf{H}\{ x^2y^2\mathbf{c} \approx y^2x^2\mathbf{c}\}, \quad \mathbf{H}\{ x^2y^2\mathbf{c} \approx yxyx\mathbf{c}\}, \quad \mathbf{H}\{ y^2x^2\mathbf{c} \approx yxyx\mathbf{c}\}; \end{align*} $$

but since $\mathbf {H} \models xyxy \approx yxyx$ , the second and third varieties coincide. Therefore, it remains to assume that:

1. (d) $\sigma $ is either $x^2y^2\mathbf {c} \approx y^2x^2\mathbf {c}$ or $y^2x^2\mathbf {c} \approx yxyx\mathbf {c}$ .

It is routinely checked that if $\mathbf {c}=1$ , then ${A_0} \not \models \sigma $ ; and if $\mathbf {c} = txy$ , then the deduction (3-4) $\vdash \sigma $ holds, so that $\mathbf {H} \models \sigma $ by Lemma 3.4. However, these contradict (a) and (b). Hence, $\mathbf {c} \notin \{1,txy\}$ and so $\mathbf {c} \in \{ \prod _{i=1}^m (t_i\mathbf {a}_i), \,\prod _{i=2}^{m+1} (t_i\mathbf {a}_i) \,|\, m \geq 1 \}$ . Equivalently:

1. (e) either $\mathbf {c} = \prod _{i=1}^m (t_i\mathbf {a}_i)$ for some $m \geq 1$ or $\mathbf {c} = \prod _{i=2}^m (t_i\mathbf {a}_i)$ for some $m \geq 2$ .

There are three cases to consider.

Case 1: $\sigma $ is $x^2y^2\mathbf {c} \approx y^2x^2\mathbf {c}$ . If $y \notin {\textsf {con}}(\mathbf {c})$ , then $\mathbf {c} = t_1x$ and it is straightforwardly checked that ${A_0} \not \models \sigma $ , contradicting (a). If $x \notin {\textsf {con}}(\mathbf {c})$ , then a similar contradiction is obtained. Therefore, $x,y \in {\textsf {con}}(\mathbf {c})$ so that $\mathbf {c}$ is either $\prod _{i=1}^m (t_i\mathbf {a}_i)$ or $\prod _{i=2}^{m+1} (t_i\mathbf {a}_i)$ for some $m \geq 2$ . Then, the equivalence

$$ \begin{align*}\{(\blacktriangleleft), x^2y^2 \mathbf {c} \approx y^2x^2 \mathbf {c}\} \sim \{(\blacktriangleleft), xy \mathbf {c} \approx yx \mathbf {c}\} \end{align*} $$

holds so that

$$ \begin{align*} \mathbf{H}\{\sigma\} = \mathbf{H}\{x^2y^2\mathbf{c} \approx y^2x^2\mathbf{c}\} = \mathbf{H}\{xy\mathbf{c} \approx yx\mathbf{c}\} = \mathbf{H}\{\circledR_m \} = \mathbf{H}_m. \end{align*} $$

Case 2: $\sigma $ is $y^2x^2 \prod _{i=2}^m (t_i\mathbf {a}_i) \approx yxyx \prod _{i=2}^m (t_i\mathbf {a}_i)$ for some $m \geq 2$ . Then, the inclusion $\mathbf {H}\{\sigma \} \subseteq \mathbf {H}\{$ Ⓡ $_m\}$ holds because

while the reverse inclusion $\mathbf {H}\{\sigma \} \supseteq \mathbf {H}\{$ Ⓡ $_m\}$ holds because

$$ \begin{align*} \mathbf{H}\{\circledR_m\} \models \circledR_m \vdash xyx\prod_{i=2}^m (t_i\mathbf{a}_i) \approx yx^2\prod_{i=2}^m (t_i\mathbf{a}_i) \vdash \sigma. \end{align*} $$

Consequently, $\mathbf {H}\{\sigma \} = \mathbf {H}\{\circledR _m\} = \mathbf {H}_m$ .

Case 3: $\sigma $ is $y^2x^2 \prod _{i=1}^m (t_i\mathbf {a}_i) \approx yxyx \prod _{i=1}^m (t_i\mathbf {a}_i)$ for some $m \geq 1$ . If $m = 1$ , then $\sigma $ is $y^2x^2 t_1 x \approx yxyx t_1 x$ and it is easily checked that ${A_0} \not \models \sigma $ , which contradicts (a). Hence, $m \geq 2$ , so that $\prod _{i=1}^m (t_i\mathbf {a}_i) = t_1xt_2y \cdots $ . Consider the identity

$$ \begin{align*} \sigma' : y^2x^2 \prod_{i=2}^m (t_i\mathbf{a}_i) \approx yxyx \prod_{i=2}^m (t_i\mathbf{a}_i) \end{align*} $$

obtained by removing the factor $t_1\mathbf {a}_1$ from both sides of $\sigma $ . Then, the inclusion ${\mathbf {H}\{\sigma \} \subseteq \mathbf {H}\{\sigma '\}}$ holds because

while the reverse inclusion $\mathbf {H}\{\sigma \} \supseteq \mathbf {H}\{\sigma '\}$ holds because $\sigma $ is obtained by making the substitution $t_2 \mapsto t_1xt_2$ in $\sigma '$ . It follows that $\mathbf {H}\{\sigma \} = \mathbf {H}\{\sigma '\} = \mathbf {H}_m$ by Case 2.

Since the three cases cover all scenarios from (d) and (e), the proof is complete.

Proof. As observed earlier, the lattice $\mathfrak {L}(\mathbf {H})$ coincides with $\mathfrak {L}(\mathbf {{A_0}} \vee \mathbf {Q}) \cup [\mathbf {{A_0}} \vee \mathbf {Q},\,\mathbf {H}]$ . It is easily seen that the deductions Ⓡ $_2 \vdash $ Ⓡ $_3 \vdash $ Ⓡ $_4 \vdash \cdots $ hold, so it follows from Lemmas 3.3 and 3.5 that the interval $[\mathbf {{A_0}} \vee \mathbf {Q},\,\mathbf {H}]$ coincides with the chain

(3-7) $$ \begin{align} \mathbf{{A_0}} \vee \mathbf{Q} = \mathbf{H}_2 \subseteq \mathbf{H}_3 \subseteq \cdots \subseteq \mathbf{H}. \end{align} $$

It is easily checked that for any $k> m$ , the identities $\{$ ⓐ $_2$ , ⓒ $_2$ , (◂), (▸), (3-1), Ⓡ $_k \}$ defining $\mathbf {H}_k$ can only convert $xy \prod _{i=1}^m (t_i\mathbf {a}_i)$ into a word of the form $x^py^q \prod _{i=1}^m (t_i\mathbf {a}_i^{r_i})$ , where $p,q,r_1,r_2,\ldots ,r_m \geq 1$ . Hence, $\mathbf {H}_k \not \models $ Ⓡ $_m$ so that $\mathbf {H}_k \neq \mathbf {H}_m$ . It follows that the inclusions in (3-7) are all proper. By Proposition 2.14(2), the lattice $\mathfrak {L}(\mathbf {H})$ is shown in Figure 1.

The variety $\mathbf {H}$ satisfies $\{$ (◂), (▸) $\}$ and so is locally finite by Lemma 2.1(1). It is clear that every proper subvariety of $\mathbf {H}$ is finitely based and small, and so also finitely generated by Lemma 2.1(2). Consequently, $\mathbf {H}$ is almost Cross. Since $\mathbf {H}$ has no maximal subvarieties, it is nonfinitely generated by Lemma 2.1(3).

Proof. The assumption $\mathbf {I},\mathbf {K} \nsubseteq \mathbf {V}$ implies that $\mathbf {V}$ satisfies the identities

(4-3) $$ \begin{align} xhxyxty & \approx xhyxty, \end{align} $$

(4-4) $$ \begin{align} \qquad\!\! xhy^2x & \approx xhy(yx)^2; \end{align} $$

see [Reference Gusev8, Lemma 4.1] and [Reference Gusev and Sapir11, Lemma 4.3]. Since

$$ \begin{align*} & \mathbf{V} \models xhytxyx \stackrel{(4\text{-}3)} {\approx} xhyt(yx)^2 \stackrel{(\blacktriangleright)}{\approx} xhyty(yx)^2 \stackrel{(4\text{-}4)}{\approx} xhyty^2x \stackrel{(\blacktriangleright).}{\approx} xhytyx \vdash xhytxyx \approx xhytyx, \end{align*} $$

the variety satisfies the identities (2-2) so that $\mathbf {V} \subseteq \mathbf {O} \cap \mathbf {J}_2$ .

Proof. The varieties $\mathbf {L}, \mathbf {P}, \overleftarrow {\mathbf {P}}, \mathbf {Y}_1, \mathbf {Y}_2$ are almost Cross by Propositions 3.8, 3.9, and 4.6. Hence, if $\mathbf {V}$ is Cross, then clearly $\mathbf {L}, \mathbf {P}, \overleftarrow {\mathbf {P}}, \mathbf {Y}_1, \mathbf {Y}_2 \nsubseteq \mathbf {V}$ .

Conversely, suppose that $\mathbf {L}, \mathbf {P}, \overleftarrow {\mathbf {P}}, \mathbf {Y}_1, \mathbf {Y}_2 \nsubseteq \mathbf {V}$ . Then, since

$$ \begin{align*} \mathbf{Rq}\{xhx\} \subseteq \mathbf{V}, \,\ \mathbf{Rq}\{ xhxyty \} \nsubseteq \mathbf{V}, \quad \text{and} \quad \mathbf{Rq}\{ xhytxy \} \vee \mathbf{Rq}\{ xyhxty\} \nsubseteq \mathbf{V}, \end{align*} $$

by Lemma 2.4, one of the following holds:

1. (a) $xhx$ is an isoterm for $\mathbf {V}$ , and $xhxyty$ and $xhytxy$ are not isoterms for $\mathbf {V}$ ;

2. (b) $xhx$ is an isoterm for $\mathbf {V}$ , and $xhxyty$ and $xyhxty$ are not isoterms for $\mathbf {V}$ .

It follows that $\mathbf {V}$ is contained in the variety $\mathbf {D} = {\textsf {var}}\{ xhxyty \approx xhyxty,\; xhytxy \approx xhytyx \}$ or its dual $\overleftarrow {\mathbf {D}}$ [Reference Lee22, Lemma 2.2]. However, $\mathbf {L}$ and $\mathbf {P}$ are the only almost Cross subvarieties of $\mathbf {D}$ , while $\mathbf {L}$ and $\overleftarrow {\mathbf {P}}$ are the only almost Cross subvarieties of $\overleftarrow {\mathbf {D}}$ [Reference Gusev8, Proposition 4.1]. Consequently, $\mathbf {V}$ does not contain any almost Cross subvarieties and so is Cross.

Proof. (1) Suppose that ${H_3} \notin \mathbf {V}$ . Then, since $\mathbf {H} = \mathbf {J}_2 \{$ (◂), (▸), $xhxy^2x \approx xhy^2x \}$ , it suffices to show that $\mathbf {V}$ satisfies the identity

(4-5) $$ \begin{align} xy^2xtx \approx xy^2tx. \end{align} $$

Now, since ${H_3}$ is embeddable in the monoid $M_\gamma (\texttt {a^+b^+ta^+})$ from [Reference Sapir33, Remark 4.4(ii)], the assumption ${H_3} \notin \mathbf {V}$ implies that $M_\gamma (\texttt {a^+b^+ta^+}) \notin \mathbf {V}$ [Reference Sapir33, Corollary 3.6]; and since $\mathbf {Q} \subseteq \mathbf {V}$ by assumption, it follows from Lemma 2.13 that $\mathbf {V}$ satisfies an identity ${xy^2tx \approx \mathbf {w} tx}$ for some word $\mathbf {w} \in \{x,y\}^{\ast }$ containing $yx$ as a factor. Multiplying both sides of this identity on the left by x and making the substitution $t \mapsto yt$ result in the identity $x^2y^3tx \approx x\mathbf {w} ytx$ of $\mathbf {V}$ . Then,

(2) This is symmetrical to part (1).

Proof. The variety $\mathbf {Z}$ is nonfinitely based by [Reference Zhang and Luo36]; see also [Reference Sapir33, Corollary 5.5]. Since $\mathbf {Z}$ is generated by the monoids ${H_3}$ and $\overleftarrow {\,{H_3}}$ , it is routinely checked that:

1. (a) $\mathbf {Z} \subseteq \mathbf {J}_2\{$ (◂),(▸), Ⓡ $_3\}$ and $\mathbf {Z} \subseteq \overleftarrow {\;\mathbf {J}_2\{({\blacktriangleleft }),({\blacktriangleright }),\, \circledR _3\}}$ .

Let $\mathbf {V}$ be any proper subvariety of $\mathbf {Z}$ , so that either ${H_3} \notin \mathbf {V}$ or $\overleftarrow {\,{H_3}} \notin \mathbf {V}$ . If ${Q} \notin \mathbf {V}$ , then $\mathbf {V} \subseteq \mathbf {{A_0}}$ by Corollary 2.12(2), so that $\mathbf {V} \subseteq \mathbf {H}_3$ by Figure 1. If ${Q} \in \mathbf {V}$ , then by Lemma 4.8, either $\mathbf {V} \subseteq \overleftarrow {\mathbf {H}}$ or $\mathbf {V} \subseteq \mathbf {H}$ , so that by (a), either

$$ \begin{align*} \mathbf{V} \subseteq \overleftarrow{\mathbf{H}} \cap \overleftarrow{\;\mathbf{J}_2\{({\blacktriangleleft}),({\blacktriangleright}),\, \circledR_3\}} = \overleftarrow{\mathbf{H}}\!{}_3 \quad \text{or } \ \mathbf{V} \subseteq \mathbf{H} \cap \mathbf{J}_2\{({\blacktriangleleft}),({\blacktriangleright}),\, \circledR_3\} = \mathbf{H}_3. \end{align*} $$

Therefore:

1. (b) $\mathbf {H}_3$ and $\overleftarrow {\mathbf {H}}\!{}_3$ are the only maximal subvarieties of $\mathbf {Z}$ .

By Lemma 3.3 and Figure 1, the variety $\mathbf {H}_2 = \mathbf {{A_0}} \vee \mathbf {Q}$ is the unique maximal subvariety of $\mathbf {H}_3$ . However, since $\mathbf {H}_2$ is self-dual (see Remark 3.7), it is also the unique maximal subvariety of $\overleftarrow {\mathbf {H}}\!{}_3$ . In other words:

1. (c) $\mathbf {H}_2$ is the unique maximal subvariety of $\mathbf {H}_3$ and of $\overleftarrow {\mathbf {H}}\!{}_3$ .

The description of $\mathfrak {L}(\mathbf {Z})$ in Figure 6 now follows from (b), (c), and the description of $\mathfrak {L}(\mathbf {H}_2)$ in Figure 1. Since the varieties $\mathbf {H}_3$ and $\overleftarrow {\mathbf {H}}\!{}_3$ are Cross by Proposition 3.6, the nonfinitely based variety $\mathbf {Z}$ is almost Cross.