On the Equational Base of SMB Algebras

Petar Đapić; Sara Bačević; Fedor Kovačević

doi:10.3390/math11051206

,

and

Faculty of Sciences, University of Novi Sad, Trg Dositeja Obradovića 3, 21000 Novi Sad, Serbia

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Author to whom correspondence should be addressed.

Mathematics2023, 11(5), 1206;https://doi.org/10.3390/math11051206

This article belongs to the Section A: Algebra and Logic

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Abstract

The “semilattices of Mal’cev blocks”, for short SMB algebras, were defined by A. Bulatov. In a recently accepted paper by P. Đapić, P. Marković, R. McKenzie, and A. Prokić, the class of all SMB algebras and its subclass of regular SMB algebras were proved to be varieties of algebras. In this paper, we find an equational base of the first variety and simplify the previously known equational base of the second variety.

Keywords:

finite axiomatizability; variety

MSC:

08B05; 08A05; 08A45

1. Introduction

The variety of semilattices of Mal’cev blocks, for short SMB algebras, first appeared in 2009–2010 in some unpublished notes by P. Marković and R. McKenzie and by M. Maróti. SMB algebras generalize the class of semilattices of groups (from semigroup theory), equivalently semigroups that are simultaneously inverse semigroups and unions of groups. This is one of the best-behaved and best-understood types of semigroups (cf. [1], Theorem 4.11).

The real motivation for studying SMB algebras comes from the structural theory of finite algebras (the so-called tame congruence theory, exposed in the monograph [2]) and its application to the complexity theory. We will not use the tame congruence theory in this article beyond this motivation part of the Introduction, so we will not define the terminology we are about to use in the next few lines.

SMB algebras and regular SMB algebras arise naturally when one considers a specific situation: the finite algebra

A

is Taylor (so no tame congruence theory type

1

occurs in the variety that algebra generates), and it has a cover-chain of three congruences

α ≺ β ≺ γ

, such that the type of the cover

α ≺ β

is

2

, while the type of

β ≺ γ

is

5

. In that situation,

A

has a term-reduct whose subalgebra is a nontrivial SMB algebra. For more details, see the discussion before Proposition 3.11 in [3].

SMB algebras are a “bad case” of Taylor algebras since they satisfy no other well-known Mal’cev condition, except for having a Taylor term. In addition to having both type

2

and

5

(thus they are neither congruence modular nor congruence meet-semidistributive), they also have no difference term (see [3], Example 7.8). Hence, if one wants to prove a result about all Taylor algebras, it is reasonable to first prove it for SMB algebras and then generalize the proof to all Taylor algebras.

This is precisely what A. Bulatov did when he was proving the dichotomy conjecture for the complexity of the constraint satisfaction problem (CSP): he first proved that SMB algebras have a tractable CSP in [4] and then went on to generalize this tractability to all Taylor algebras, thus resolving the dichotomy conjecture in [5]. The second result won him a share of Best Paper award at the conference FOCS 2017, among other prizes. It is reasonable to hope that other results about Taylor algebras can be first proved for SMB algebras and then generalized, so our paper sets out to increase our basic understanding of SMB algebras, hoping this understanding can be useful in future research. More precisely, we will clean up the axiomatization of SMB algebras in this paper.

A. Bulatov first gave the definition of SMB algebras in [4]. It was proved that they are a finitely axiomatizable variety with various interesting properties in [3]. However, no finite equational base was explicitly found in that paper, and in a recent lecture, P. Marković mentioned that it may be useful to find such a finite base. Moreover, the subvariety of regular SMB algebras was defined and a finite equational base for regular SMB algebras was given in [3]. In this paper, we find a small equational base for the variety of SMB algebras, simplify the equational base for regular SMB algebras, and prove that one of our bases is minimal in the sense of inclusion order, while the other base might have one superfluous equation.

2. Preliminaries

We assume that the reader is familiar with the basics of universal algebra. The readers who need these facts and definitions are referred to classic textbooks [6,7]. We mention that, as usual, the algebras are denoted by boldface Roman letters

A, B, C, \dots

, and their respective universes are denoted by the unbolded versions of the same letters. Moreover, when p and q are two terms of some language, the identity

p \approx q

stands for the sentence stating that for all valuations of variables, the two terms are equal. In this paper, we try to deduce all identities that hold throughout some classes of algebras from a small finite subset of those identities, i.e., find the equational bases of those classes.

An operation f of an algebra

A

is said to be idempotent if the identity

f (x, x, \dots, x) \approx x

holds in

A

. An algebra is idempotent if all of its fundamental operations are idempotent.

Some examples of idempotent algebras that we care about in this paper include the semilattice

A = ⟨ A; \land ⟩

, where ∧ is a commutative, associative, and idempotent operation. Semilattices can equivalently be viewed as partially ordered sets in which each pair of elements has an infimum.

Another example is a Mal’cev algebra

A = ⟨ A; d ⟩

, where d is a ternary idempotent operation that satisfies the identities

d (x, y, y) \approx d (y, y, x) \approx x

. Mal’cev operations are term operations in many familiar and classical algebras. For example, in groups, the term

d (x, y, z) = x y^{- 1} z

satisfies the Mal’cev equations. Quasigroups, when defined using the multiplication, the left division, and the right division, also have a Mal’cev term. Having a Mal’cev term operation is necessary and sufficient for an algebra to generate the variety in which all congruence relations permute (with respect to the composition of relations).

First, we recall from [3] what is an SMB algebra:

Definition 1.

Let

A = (A; \land, d)

be an idempotent algebra and

\sim \in Con A

. We say that

A

is a semilattice of Mal’cev blocks with respect to ∼, an SMB algebra over ∼ for short, if

1.: $(A / \sim; \land^{A / \sim})$ is a semilattice;
2.: On each ∼-class D, the operation $\land ↾_{D}$ acts as the second projection, while $d ↾_{D}$ acts as a Mal’cev operation.

We say that

A = (A; \land, d)

is a semilattice of Mal’cev blocks, an SMB algebra for short, if

A

is an idempotent algebra such that there exists a congruence

\sim \in Con A

so that

A

is an SMB algebra over ∼. We denote the class of all SMB algebras by

S

.

Theorem 1

(Theorem 15. in [3]). The class

S

of all SMB algebras is a variety.

We will also recollect what regular SMB algebras are.

Definition 2.

We say that an SMB algebra

A = ⟨ A; \land, d ⟩

over

\sim \in Con A

is regular if

1.: For all $a, b, c \in A$ , ${[d (a, b, c)]}_{\sim} = {[(a \land b) \land c]}_{\sim}$ ;
2.: For all $a, b \in A$ such that ${[a]}_{\sim} \geq {[b]}_{\sim}$ , $a \land b = b$ ,;
3.: $d (x, y, z) \approx d ((y \land z) \land x, (x \land z) \land y, (x \land y) \land z)$ ;
4.: $A ⊧ (x \land y) \land y \approx x \land y$ .

The following equational base was given in [3] for regular SMB algebras.

Proposition 1

(Proposition 23 in [3]). The following list of identities is an equational base for the variety

R

of regular SMB algebras:

Idem1): $x \land x \approx x$
Idem2): $d (x, x, x) \approx x$
Comm): $(x \land y) \land (y \land x) \approx y \land x$
Assoc1): $(x \land (y \land z)) \land ((x \land y) \land z) \approx (x \land y) \land z$
Assoc2): $((x \land y) \land z) \land (x \land (y \land z)) \approx x \land (y \land z)$
Mal): $d (x \land y, y \land x, y \land x) \approx d (y \land x, y \land x, x \land y) \approx x \land y$
Regi1): $((x \land y) \land z) \land d (x, y, z) \approx d (x, y, z)$
Regi2): $d (x, y, z) \land ((x \land y) \land z) \approx (x \land y) \land z$
Regii1): $x \land (x \land y) \approx x \land y$
Regii2): $x \land (y \land x) \approx y \land x$
Regiii): $d (x, y, z) \approx d ((y \land z) \land x, (x \land z) \land y, (x \land y) \land z)$
Regiv): $(x \land y) \land y \approx x \land y$

3. Equational Base of SMB Algebras

In the next theorem, we will find an equational base of the variety of all SMB algebras. This is a new contribution since it was proved in [3] that this is a finitely axiomatizable variety, but no explicit equational base was known.

Theorem 2.

The following list of identities is an equational base for the variety

S

of SMB algebras:

Idem1): $x \land x \approx x$
Idem2): $d (x, x, x) \approx x$
Comm): $(x \land y) \land (y \land x) \approx y \land x$
Assoc1): $(x \land (y \land z)) \land ((x \land y) \land z) \approx (x \land y) \land z$
Assoc2): $((x \land y) \land z) \land (x \land (y \land z)) \approx x \land (y \land z)$
Mal): $d (x \land y, y \land x, y \land x) \approx d (y \land x, y \land x, x \land y) \approx x \land y$
Tran): $((z \land y) \land x) \land ((x \land y) \land z) \approx (x \land y) \land z$
Comp1): $((y \land x) \land (u \land z)) \land ((x \land y) \land (z \land u)) \approx ((x \land y) \land (z \land u))$
Comp2): $d (x \land x^{'}, y \land y^{'}, z \land z^{'}) \land d (x^{'} \land x, y^{'} \land y, z^{'} \land z) \approx d (x^{'} \land x, y^{'} \land y, z^{'} \land z)$

Proof.

Let

A = (A, \land, d)

be any model of the equations from this theorem.

A

is idempotent because of identities Idem1) and Idem2).

At first, we define the relation ∼ as

a \sim b

iff

a \land b = b

and

b \land a = a

.

Now, we will prove that the ∼ relation is an equivalence relation on

A

.

Let

a \in A

. We have

a \land a = a

from Idem1), and by definition of ∼, this implies

a \sim a

.

Let

a, b \in A

.

a \sim b

if

a \land b = b

and

b \land a = a

if

b \sim a

. This is obvious from the definition of the relation ∼.

Let

a, b, c \in A

be such that

a \sim b

and

b \sim c

. Then, we have

a \land b = b

,

b \land a = a

,

b \land c = c

and

c \land b = b

. From this we obtain the following:

(a \land b) \land c = b \land c = c,

(1)

\begin{matrix} (a \land b) \land c \overset{Tran)}{=} & ((c \land b) \land a) \land ((a \land b) \land c) \\ \overset{(1)}{=} & ((c \land b) \land a) \land c \\ = & (b \land a) \land c \\ = & a \land c . \end{matrix}

(2)

From identities (1) and (2) follows

a \land c = c

. Analogously, we obtain

c \land a = a

, so

a \sim c

. We can see that the relation ∼ satisfies reflectivity (R), symmetricity (S) and transitivity (T), so the relation ∼ is an equivalence relation.

Next, we show that the relation ∼ is a congruence on algebra

A

. First, we are showing compatibility with the operation ∧. Therefore, we need to show that, if

a \sim b

, then

a \land c \sim b \land c

and

c \land a \sim c \land b

for every

c \in A

.

Let

a, b \in A

such that

a \sim b

. Then, we have

\begin{matrix} a \land c = (b \land a) \land c \overset{Idem 1)}{=} (b \land a) \land (c \land c) \overset{Comp 1)}{=} \\ ((a \land b) \land (c \land c)) \land ((b \land a) \land (c \land c)) \overset{Idem 1)}{=} \\ ((a \land b) \land c) \land ((b \land a) \land c) = \\ (b \land c) \land (a \land c), \end{matrix}

The proof that

b \land c = (a \land c) \land (b \land c)

is completely analogous; we simply flip letters a and b, so we prove that

a \land c \sim b \land c

. From Comm) follows that

c \land a \sim a \land c

and similarly that

b \land c \sim c \land b

. Thus, from the transitivity of the relation ∼, we obtain

c \land a \sim c \land b

. With this, we show the compatibility of the relation ∼ with the operation ∧.

The next step is showing compatibility of ∼ with the operation d. We need to show that, if

a \sim a^{'}

,

b \sim b^{'}

,

c \sim c^{'}

, then

d (a, b, c) \sim d (a^{'}, b^{'}, c^{'})

.

Let

a, b, c, a^{'}, b^{'}, c^{'} \in A

be such that

a \sim a^{'}

,

b \sim b^{'}

and

c \sim c^{'}

. Then, we have

\begin{matrix} d (a^{'}, b^{'}, c^{'}) = \\ d (a \land a^{'}, b \land b^{'}, c \land c^{'}) \overset{Comp 2)}{=} \\ d (a^{'} \land a, b^{'} \land b, c^{'} \land c) \land d (a \land a^{'}, b \land b^{'}, c \land c^{'}) = \\ d (a, b, c) \land d (a^{'}, b^{'}, c^{'}), \end{matrix}

Once again, the proof that

d (a, b, c) = d (a^{'}, b^{'}, c^{'}) \land d (a, b, c)

is completely analogous; we simply need to flip

a \leftrightarrow a^{'}

,

b \leftrightarrow b^{'}

and

c \leftrightarrow c^{'}

.

Hence, we have seen that

d (a, b, c) \land d (a^{'}, b^{'}, c^{'}) = d (a^{'}, b^{'}, c^{'})

and

d (a^{'}, b^{'}, c^{'}) \land d (a, b, c) = d (a, b, c)

, that is

d (a, b, c) \sim d (a^{'}, b^{'}, c^{'})

, which gives us the compatibility of ∼ with d.

Now the identities Idem1), Comm), Assoc1), and Assoc2) imply that

A /

∼ is a semilattice. From the definition of the relation ∼ follows that the operation ∧ is the second projection on each ∼ class. Taken together with that fact, Mal) implies that d is Mal’cev on each ∼ class. Therefore,

A

is an SMB algebra. On the other hand, the base identities are easily verifiable in each SMB algebra. □

In the following three lemmas, we trim down the equational base we found in Theorem 2.

Lemma 1.

The following list of identities follows from identitiesIdem1),Comm),Tran), andComp1):

1.: $(x \land (y \land z)) \land (x \land (z \land y)) \approx (x \land (z \land y))$ ;
2.: $(x \land (y \land z)) \land ((z \land y) \land x) \approx ((z \land y) \land x)$ ;
3.: $((z \land y) \land x) \land (x \land (y \land z)) \approx (x \land (y \land z))$ ;
4.: $(x \land (y \land z)) \land ((x \land y) \land z) \approx ((x \land y) \land z)$ , i.e., Assoc1);
5.: $((x \land y) \land z) \land (x \land (y \land z)) \approx (x \land (y \land z))$ , i.e., Assoc2).

Proof.

(1)

(x \land (y \land z)) \land (x \land (z \land y)) \approx (x \land (z \land y))

. From the assumptions Idem1) and Comp1), we obtain

\begin{matrix} (x \land (y \land z)) \land (x \land (z \land y)) \approx ((x \land x) \land (y \land z)) \land ((x \land x) \land (z \land y)) \approx \\ ((x \land x) \land (z \land y)) \approx (x \land (z \land y)) . \end{matrix}

(2)

(x \land (y \land z)) \land ((z \land y) \land x) \approx ((z \land y) \land x)

. From the assumptions (1), Comm), and Tran), we obtain

\begin{matrix} (x \land (y \land z)) \land ((z \land y) \land x) \overset{(1), Comm)}{\approx} \\ ((x \land (z \land y)) \land (x \land (y \land z))) \land ((x \land (z \land y)) \land ((z \land y) \land x)) \overset{Comm), (1)}{\approx} \\ ((((z \land y) \land x) \land (x \land (z \land y))) \land (x \land (y \land z))) \land \\ (((x \land (y \land z)) \land (x \land (z \land y))) \land ((z \land y) \land x)) \overset{Tran)}{\approx} \\ (((x \land (y \land z)) \land (x \land (z \land y))) \land ((z \land y) \land x)) \overset{(1)}{\approx} \\ ((x \land (z \land y)) \land ((z \land y) \land x)) \overset{Comm)}{\approx} \\ ((z \land y) \land x) \end{matrix}

(3)

((z \land y) \land x) \land (x \land (y \land z)) \approx (x \land (y \land z))

. From the assumptions (1), Comm), and Tran), we obtain

\begin{matrix} ((z \land y) \land x) \land (x \land (y \land z)) \overset{Comm), (1)}{\approx} \\ ((x \land (z \land y)) \land ((z \land y) \land x)) \land ((x \land (z \land y)) \land (x \land (y \land z))) \overset{(1), Comm)}{\approx} \\ (((x \land (y \land z)) \land (x \land (z \land y))) \land ((z \land y) \land x)) \land \\ ((((z \land y) \land x) \land (x \land (z \land y))) \land (x \land (y \land z))) \overset{Tran)}{\approx} \\ ((((z \land y) \land x) \land (x \land (z \land y))) \land (x \land (y \land z))) \overset{Comm)}{\approx} \\ (x \land (z \land y)) \land (x \land (y \land z)) \overset{()}{\approx} \\ (x \land (y \land z)) \end{matrix}

(4)

(x \land (y \land z)) \land ((x \land y) \land z) \approx ((x \land y) \land z)

. From the assumptions (2), (3), and Tran), we obtain

\begin{matrix} (x \land (y \land z)) \land ((x \land y) \land z) \overset{(3), Tran)}{\approx} \\ (((z \land y) \land x) \land (x \land (y \land z))] \land \\ (((z \land y) \land x) \land ((x \land y) \land z)] \overset{Tran), (2)}{\approx} \\ ((((x \land y) \land z) \land ((z \land y) \land x)) \land (x \land (y \land z))) \land \\ ((x \land (y \land z)) \land ((z \land y) \land x)) \land ((x \land y) \land z) \overset{Tran)}{\approx} \\ ((x \land (y \land z)) \land ((z \land y) \land x)) \land ((x \land y) \land z) \overset{(2)}{\approx} \\ ((z \land y) \land x) \land ((x \land y) \land z) \overset{Tran)}{\approx} \\ ((x \land y) \land z) \end{matrix}

(5)

((x \land y) \land z) \land (x \land (y \land z)) \approx (x \land (y \land z))

. From the assumptions (2), (3), and Tran), we obtain

\begin{matrix} ((x \land y) \land z) \land (x \land (y \land z)) \overset{Tran), (3)}{\approx} \\ (((z \land y) \land x) \land ((x \land y) \land z)) \land \\ (((z \land y) \land x) \land (x \land (y \land z))) \overset{(2), Tran)}{\approx} \\ (((x \land (y \land z)) \land ((z \land y) \land x)) \land ((x \land y) \land z)) \land \\ ((((x \land y) \land z) \land ((z \land y) \land x)) \land (x \land (y \land z))) \overset{Tran)}{\approx} \\ ((((x \land y) \land z) \land ((z \land y) \land x)) \land (x \land (y \land z))) \overset{Tran)}{\approx} \\ (((z \land y) \land x) \land (x \land (y \land z))) \overset{(3)}{\approx} \\ (x \land (y \land z)) \end{matrix}

□

Lemma 2.

The identityIdem2)follows from identitiesIdem1)andMal).

Proof.

\begin{matrix} d (x, x, x) \overset{Idem 1)}{\approx} \\ d (x \land x, x \land x, x \land x) \overset{Mal)}{\approx} \\ x \land x \overset{Idem 1)}{\approx} x \end{matrix}

□

Lemma 3.

The identityComm)follows from the identitiesMal)andComp2).

Proof.

\begin{matrix} (x \land y) \land (y \land x) \overset{Mal)}{\approx} \\ d (x \land y, y \land x, y \land x) \land d (y \land x, x \land y, x \land y) \overset{Comp 2)}{\approx} \\ d (y \land x, x \land y, x \land y) \overset{Mal)}{\approx} y \land x \end{matrix}

□

We summarize the results we proved so far.

Theorem 3.

The following list of identities is an equational base for the variety

S

of SMB algebras:

Idem1): $x \land x \approx x$
Mal): $d (x \land y, y \land x, y \land x) \approx d (y \land x, y \land x, x \land y) \approx x \land y$
Tran): $((z \land y) \land x) \land ((x \land y) \land z) \approx (x \land y) \land z$
Comp1): $((y \land x) \land (u \land z)) \land ((x \land y) \land (z \land u)) \approx (x \land y) \land (z \land u)$
Comp2): $d (x \land x^{'}, y \land y^{'}, z \land z^{'}) \land d (x^{'} \land x, y^{'} \land y, z^{'} \land z) \approx d (x^{'} \land x, y^{'} \land y, z^{'} \land z)$

Proof.

Follows from Theorem 2, Lemmas 1–3. □

With this theorem, we have significantly reduced the equational base, but can we remove any other identity? Maybe the identity Comp1), but we can not eliminate any others. The following algebras shown in Cayley tables satisfy all but one identity.

Theorem 4.

All identities from Theorem 3, except perhapsComp1), are independent from the others.

Proof.

We will list off algebras

(A, \land, d)

with their tables of operations

a \land b

and

d (a, b, c)

, which satisfy all but one of the identities listed in Theorem 3. Note: Here, we will break identity Mal) into 2 identities:

Mal1): $d (x \land y, y \land x, y \land x) \approx x \land y$ ;
Mal2): $d (y \land x, y \land x, x \land y) \approx x \land y$ .

which should have been done from the beginning, but we left it as is in [3].

Idem1)
Algebra $({1, 2}, \land_{1}, d_{1})$ given with Table 1 and Table 2 satisfies all except Idem1), which fails for elements $x = 2$ .
Mal1)
Algebra $({1, 2}, \land_{2}, d_{2})$ given with Table 3 and Table 4 satisfies all except Mal1), which fails for elements $x = 1, y = 2$ .
Mal2)
Algebra $({1, 2}, \land_{3}, d_{3})$ given with Table 5 and Table 6 satisfies all except Mal2), which fails for elements $x = 1, y = 2$ .
Tran)
Algebra $({1, 2, 3}, \land_{4}, d_{4})$ given with Table 7 and Table 8 satisfies all except Tran), which fails for elements $x = 1, y = 1, z = 3$ .
Comp2)
Algebra $({1, 2, 3}, \land_{5}, d_{5})$ given with Table 9 and Table 10 satisfies all except Comp2), which fails for elements $x = 2, x^{'} = 2, y = 2, y^{'} = 3, z = 3, z^{'} = 2$ .

□

Table 1. Idem1) Meet.

$\land_{1}$	1	2
1	1	1
2	1	1

Table 2. Idem1) d.

$d_{1}$	a	1		2
$d_{1}$	b	1	2	1	2
c	1	1	1	1	1
c	2	1	1	1	1

Table 3. Mal1) Meet.

$\land_{2}$	1	2
1	1	2
2	1	2

Table 4. Mal1) d.

$d_{2}$	a	1		2
$d_{2}$	b	1	2	1	2
c	1	1	1	1	1
c	2	2	1	1	2

Table 5. Mal2) Meet.

$\land_{3}$	1	2
1	1	2
2	1	2

Table 6. Mal2) d.

$d_{3}$	a	1		2
$d_{3}$	b	1	2	1	2
c	1	1	1	2	1
c	2	1	1	1	2

Table 7. Tran) Meet.

$\land_{4}$	1	2	3
1	1	1	2
2	1	2	1
3	2	1	3

Table 8. Tran) d.

$d_{4}$	a	1			2			3
$d_{4}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	1	1	1
	2	1	1	1	1	2	1	1	1	1
	3	1	1	1	1	1	1	1	1	3

Table 9. Comp2) Meet.

$\land_{5}$	1	2	3
1	1	1	1
2	1	2	3
3	1	2	3

Table 10. Comp2) d.

$d_{5}$	a	1			2			3
$d_{5}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	1	1	1
	2	1	1	1	1	2	1	1	3	2
	3	1	1	1	1	3	2	1	1	3

4. Equational Base of Regular SMB Algebras

We will first prove a lemma that will help us remove an identity from the equational base for regular SMB algebras (from Proposition 1).

Lemma 4.

Let

G = (G; \land)

be a groupoid such that it satisfies the following identities:

Idem1): $x \land x \approx x$ ;
Assoc1): $(x \land (y \land z)) \land ((x \land y) \land z) \approx (x \land y) \land z$ ;
Comm): $(x \land y) \land (y \land x) \approx y \land x$ ;
Regiv): $(x \land y) \land y \approx x \land y$ .

Then,

G

also satisfies the identityAssoc2):

((x \land y) \land z) \land (x \land (y \land z)) \approx x \land (y \land z) .

Proof.

Let

G

be any model of the above equations. Firstly, we will use following claim from [3].

Claim 2. Let $a, b, c \in G$ such that $a \land b = b$ and $b \land c = c$ ; then, $a \land c = c$ .

To prove this, we will use Assoc1) and Regiv).

c = b \land c = (a \land b) \land c = (a \land (b \land c)) \land ((a \land b) \land c) =

(a \land c) \land (b \land c) = (a \land c) \land c = a \land c .

In addition to this claim, we will need to prove an additional claim, which is as follows.

Claim 3. $x \land (y \land z) \approx (y \land (z \land x)) \land (x \land (y \land z)) .$

By using Comm), Asocc1), and Regiv), we obtain the following:

\begin{matrix} x \land (y \land z) \overset{Comm)}{\approx} \\ ((y \land z) \land x) \land (x \land (y \land z)) \overset{Asocc 1)}{\approx} \\ ((y \land (z \land x)) \land ((y \land z) \land x)) \land (x \land (y \land z)) \overset{Asocc 1)}{\approx} \\ ((y \land (z \land x)) \land (((y \land z) \land x) \land (x \land (y \land z)))) \overset{}{\land} \\ (((y \land (z \land x)) \land ((y \land z) \land x)) \land (x \land (y \land z))) \overset{Asocc 1)}{\approx} \\ ((y \land (z \land x)) \land (((y \land z) \land x) \land (x \land (y \land z)))) \overset{}{\land} \\ (((y \land z) \land x) \land (x \land (y \land z))) \overset{Comm)}{\approx} \\ ((y \land (z \land x)) \land (((y \land z) \land x) \land (x \land (y \land z)))) \land (x \land (y \land z)) \overset{Comm)}{\approx} \\ ((y \land (z \land x)) \land (x \land (y \land z))) \land (x \land (y \land z)) \overset{Regiv)}{\approx} \\ (y \land (z \land x)) \land (x \land (y \land z)), \end{matrix}

which is what we needed. If we rearrange letters in Claim 3, we obtain the following:

\begin{matrix} (z \land (x \land y)) \land (y \land (z \land x)) \approx y \land (z \land x), \\ (y \land (z \land x)) \land (x \land (y \land z)) \approx x \land (y \land z), \end{matrix}

and by using Claim 2, we have that

(z \land (x \land y)) \land (x \land (y \land z)) \approx x \land (y \land z)

, and with Comm) we obtain:

\begin{matrix} ((x \land y) \land z) \land (z \land (x \land y)) \approx z \land (x \land y), \\ (z \land (x \land y)) \land (x \land (y \land z)) \approx x \land (y \land z) . \end{matrix}

By again using Claim 2, we proved what we needed. □

Now, we prove that several other identities from Proposition 1 can also be removed.

Theorem 5.

The following list of identities is an equational base for the variety

R

of regular SMB algebras:

Idem1): $x \land x \approx x$
Assoc1): $(x \land (y \land z)) \land ((x \land y) \land z) \approx (x \land y) \land z$
Mal): $d (x \land y, y \land x, y \land x) \approx d (y \land x, y \land x, x \land y) \approx x \land y$
Regi1): $((x \land y) \land z) \land d (x, y, z) \approx d (x, y, z)$
Regi2): $d (x, y, z) \land ((x \land y) \land z) \approx (x \land y) \land z$
Regiii): $d (x, y, z) \approx d ((y \land z) \land x, (x \land z) \land y, (x \land y) \land z)$
Regiv): $(x \land y) \land y \approx x \land y$

Proof.

Let

A

be any model of the above equations. We will prove that identities:

Idem2): $d (x, x, x) \approx x$ ;
Comm): $(x \land y) \land (y \land x) \approx y \land x$ ;
Regii1): $x \land (x \land y) \approx x \land y$ ;
Regii2): $x \land (y \land x) \approx y \land x$ ;
Assoc2): $((x \land y) \land z) \land (x \land (y \land z)) \approx x \land (y \land z)$ .

also hold, and as we prove them we will start using them to prove other identities.

Idem2)

d (x, x, x) \approx x

. From the assumptions Idem1) and Mal), we obtain

\begin{matrix} d (x, x, x) \overset{Idem 1)}{\approx} d (x \land x, x \land x, x \land x) \overset{Mal)}{\approx} x \land x \overset{Idem 1)}{\approx} x . \end{matrix}

Comm)

(x \land y) \land (y \land x) \approx y \land x

. From the assumptions Idem1), Regi1), Mal), and Regiv), we obtain

\begin{matrix} (x \land y) \land (y \land x) \overset{Regiv)}{\approx} ((x \land y) \land (y \land x)) \land (y \land x) \overset{Mall)}{\approx} \\ ((x \land y) \land (y \land x)) \land d (x \land y, x \land y, y \land x) \overset{Idem 1)}{\approx} \\ (((x \land y) \land (x \land y)) \land (y \land x)) \land d (x \land y, x \land y, y \land x) \overset{Regi 1)}{\approx} \\ d (x \land y, x \land y, y \land x) \overset{Mal)}{\approx} y \land x \end{matrix}

Regii1)

x \land (x \land y) \approx x \land y

. From the assumptions Regiv), Idem1), and Assoc1), we obtain

\begin{matrix} x \land (x \land y) \overset{Regiv)}{\approx} (x \land (x \land y)) \land (x \land y) \overset{Idem 1)}{\approx} \\ (x \land (x \land y)) \land ((x \land x) \land y) \overset{Assoc 1)}{\approx} (x \land x) \land y \overset{Idem 1)}{\approx} (x \land y) \end{matrix}

Regii2)

x \land (y \land x) \approx y \land x

. From the assumptions Comm), Assoc1), Regii1), and Regiv), we obtain

\begin{matrix} (y \land x) \overset{Comm)}{\approx} (x \land y) \land (y \land x) \overset{Assoc 1)}{\approx} \\ (x \land (y \land (y \land x))) \land ((x \land y) \land (y \land x)) \overset{Regii 1)}{\approx} \\ (x \land (y \land x)) \land ((x \land y) \land (y \land x)) \overset{Comm)}{\approx} \\ (x \land (y \land x)) \land (y \land x) \overset{Regiv)}{\approx} x \land (y \land x) \end{matrix}

Assoc2)

((x \land y) \land z) \land (x \land (y \land z)) \approx x \land (y \land z)

. It is clear that

(A; \land)

satisfies all of the conditions for Lemma 4; therefore, this identity holds.

Now it follows that, by Proposition 1,

A

is a regular SMB algebra. □

With this proposition, we have almost halved the equational base, but can we remove any other identity? Unfortunately, no. The following algebras shown in Cayley tables satisfy all but one identity.

Theorem 6.

All identities in Theorem 5 are independent.

Proof.

Note: As in Theorem 4, we will break identity Mal) into two:

Mal1)

d (x \land y, y \land x, y \land x) \approx x \land y

;

Mal2)

d (y \land x, y \land x, x \land y) \approx x \land y

.

Idem1): Algebra $({1, 2}, \land_{1}, d_{1})$ given with Table 1 and Table 2 satisfies all except Idem1), which falls for elements $x = 2$ .
Assoc1): Algebra $({1, 2, 3}, \land_{6}, d_{6})$ given with Table 11 and Table 12 satisfies all except Assoc1), which falls for elements $x = 1, y = 3, z = 2$ .
Mal1): Algebra $({1, 2}, \land_{2}, d_{2})$ given with Table 3 and Table 4 satisfies all except Mal1), which falls for elements $x = 1, y = 2$ .
Mal2): Algebra $({1, 2}, \land_{3}, d_{3})$ given with Table 5 and Table 6 satisfies all except Mal2), which falls for elements $x = 1, y = 2$ .
Regi1): Algebra $({1, 2, 3}, \land_{7}, d_{7})$ given with Table 13 and Table 14 satisfies all except Regi1), which falls for elements $x = 3, y = 1, z = 3$ .
Regi2): Algebra $({1, 2, 3}, \land_{8}, d_{8})$ given with Table 15 and Table 16 satisfies all except Regi2), which falls for elements $x = 2, y = 3, z = 2$ .
Regiii): Algebra $({1, 2, 3}, \land_{9}, d_{9})$ given with Table 17 and Table 18 satisfies all except Regiii), which falls for elements $x = 1, y = 2, z = 3$ .
Regiv): Algebra $({1, 2, 3}, \land_{10}, d_{10})$ given with Table 19 and Table 20 satisfies all except Regiv), which falls for elements $x = 2, y = 3$ .

□

Table 11. Assoc1) Meet.

$\land_{6}$	1	2	3
1	1	1	3
2	1	2	2
3	1	2	3

Table 12. Assoc1) d.

$d_{6}$	a	1			2			3
$d_{6}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	3	3	1
	2	1	1	2	1	2	2	3	2	2
	3	3	1	1	1	2	2	1	2	3

Table 13. Regi1) Meet.

$\land_{7}$	1	2	3
1	1	1	3
2	1	2	3
3	1	1	3

Table 14. Regi1) d.

$d_{7}$	a	1			2			3
$d_{7}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	3	3	1
	2	1	1	1	1	2	1	3	3	1
	3	3	3	1	3	3	1	2	2	3

Table 15. Regi2) Meet.

$\land_{8}$	1	2	3
1	1	1	1
2	1	2	3
3	1	2	3

Table 16. Regi2) d.

$d_{8}$	a	1			2			3
$d_{8}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	1	1	1
	2	1	1	1	1	2	1	1	3	2
	3	1	1	1	1	3	2	1	1	3

Table 17. Regiii) Meet.

$\land_{9}$	1	2	3
1	1	1	3
2	1	2	3
3	1	1	3

Table 18. Regiii) d.

$d_{9}$	a	1			2			3
$d_{9}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	3	1	1
	2	1	1	1	1	2	1	1	1	1
	3	3	1	1	1	1	1	1	1	3

Table 19. Regiv) Meet.

$\land_{10}$	1	2	3
1	1	1	3
2	1	2	1
3	1	1	3

Table 20. Regiv) d.

$d_{10}$	a	1			2			3
$d_{10}$	b	1	2	3	1	2	3	1	2	3
c	1	1	1	1	1	1	1	3	3	1
	2	1	1	1	1	2	1	3	1	1
	3	3	3	1	3	1	1	1	1	3

5. Conclusions

We have almost completed what we set out to do. For the variety of regular SMB algebras we have a minimal equational base, i.e., we prove that no identity can be deleted. In the case of the variety of all SMB algebras, we have a small equational base, which has one identity for which we could not determine whether it is necessary or it can be deleted, but all other identities cannot be deleted. This is good enough for computational purposes; for instance, it can be applied in computations carried out with the Universal Algebra Calculator, or with the Prover9/Mace4 automated theorem prover.

We hope that our results can be applied in the efforts toward settling Park’s conjecture, a famous 1970s problem that states that any finite algebra

A

in a finite language and with a finite residual bound have a finite base of equations. This conjecture is proved for algebras with a difference term in [8], but, as we stated in the introduction, SMB algebras, or even regular SMB algebras, can fail to have such a term. Hence, SMB algebras are a viable target for the next attack towards Park’s conjecture, and our equational bases could help out there.

Author Contributions

Conceptualization, P.Đ.; methodology, P.Đ.; validation, S.B., F.K. and P.Đ.; writing—original draft preparation, F.K. and S.B.; writing—review and editing, P.Đ.; visualization, S.B., F.K. and P.Đ. All authors have read and agreed to the published version of the manuscript.

Funding

Petar Đapić was supported by the Ministry of Education, Science and Technological Development of the Republic of Serbia (Grant No. 451-03-47/2023-01/200125).

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

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On the Equational Base of SMB Algebras

Abstract

1. Introduction

2. Preliminaries

3. Equational Base of SMB Algebras

4. Equational Base of Regular SMB Algebras

5. Conclusions

Author Contributions

Funding

Institutional Review Board Statement

Informed Consent Statement

Data Availability Statement

Conflicts of Interest

References

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