determined as follows: %

INSTITUTE OF MATHEMATICS POLISH ACADEMY OF SCIENCES

WARSZAWA 1993

ON ALGEBRAS OF RELATIONS

D. A. B R E D I K H I N

Lermontova 7-22, 410002 Saratov, Russia

Throughout, by relation we mean a binary relation. Let Rel(X) be the set of all binary relations on the set X. An algebra of relations is a pair (Φ, Ω) where Ω is a set of operations on relations and Φ ⊂ Rel(X) is a set of relations closed under the operations of Ω. Each algebra of relations can be considered as ordered by the set-theoretic inclusion ⊂. Denote by M {Ω} the class of all algebras isomorphic to ones whose elements are relations and whose operations are members of Ω. The class M {Ω, ⊂} is determined in the same way.

We will consider the following operations on relations: relation product ◦, relation inverse

, intersection ∩, diagonal relation ∆, and the unary operation

determined as follows: %

= % ∩ ∆.

The class M {◦,

, ∩, ∆} was introduced and characterized in [6]. It is not finitely axiomatizable [5]. The classes M {◦,

, ∆} and M {◦,

, ∆, ⊂} were char- acterized in [1, 8]. The class M {◦,

, ∆} is not finitely axiomatizable [2].

In this paper we find a system of axioms for the class M {◦,

,

, ∆, ⊂} and use it to obtain some results about the class M {◦,

, ∩, ∆}.

Theorem 1. An algebra (A, ·,

,

, 1, ≤) belongs to M {◦,

,

, ∆, ⊂} iff it satisfies the following conditions:

(1) (A, ·,

, 1) is an involuted monoid , i.e. (xy)z = x(yz), 1x = x1 = x, (x

)

= x, (xy)

= y

x

.

(2) ≤ is an order relation and all operations are monotonic, i.e. x ≤ y implies xz ≤ yz, zx ≤ zy, x

≤ y

, x

≤ y

.

(3) The following identities are satisfied :

(3.1) (x

)

= x

,

1991 Mathematics Subject Classification: 03G15.

Key words and phrases: agebras of relations, representations, involuted semigroups.

The paper is in final form and no version of it will be published elsewhere.

[191]

(3.2) x

x

= x

,

(3.3) x

y

= y

x

,

(3.4) (x

y

)

= x

y

,

(3.5) (xx

)

x = x ,

(3.6) (xyy

x

)

= (x(yy

)

x

)

,

(3.7) x

≤ 1 ,

(3.8) x

≤ x .

Suppose that an algebra A = (A, ·,

, ∧, 1) ∈ M {◦,

, ∩, ∆}. Then A is a semilattice ordered involuted monoid, i.e. (A, ·,

, 1) is an involuted monoid, (A, ∧,

) is an involuted semilattice and the identity x(y ∧ z) ≤ xy ∧ xz holds (≤ is the natural order of the semilattice) [6]. It is known [3] that A also satisfies

(4) xy ∧ z ≤ (x ∧ yz

)y .

We say that a semilattice ordered involuted monoid A is weakly representable if there exists a mapping F : A → Rel(X) for some X such that F (ab) = F (a) ◦ F (b), F (a

) = F (a)

, F (1) = ∆, F (a ∧ 1) = F (a) ∩ ∆ and a ≤ b iff F (a) ⊂ F (b) for all a, b ∈ A.

Theorem 2. Suppose that a semilattice ordered involuted monoid satisfies (4).

Then it is weakly representable.

The following condition plays an important role in applications of algebras of relations to logic [4]:

(5) (∃u, v)(∀a) u

u ≤ 1 & v

v ≤ 1 & a ≤ u

v . Consider the condition

(6) (∀a)(∃u, v) u

u ≤ 1 & v

v ≤ 1 & a ≤ u

v . Obviously, (5) implies (6).

Theorem 3. Suppose that a semilattice ordered involuted monoid A satis- fies (6). Then A belongs to M {◦,

, ∩, ∆} iff it satisfies (4).

P r o o f o f T h e o r e m 1 . Necessity. Consider the ordered algebra of re- lations of the form (Φ, ◦,

,

, ∆, ⊂). It is well known that (Φ, ◦,

, ∆) is an involuted monoid and the operations ◦ and

are monotonic [1, 8]. Suppose that

%, π ∈ Φ. Since %

, π

⊂ ∆, we have %

◦ π

= %

∩ π

. It follows that (3.1)–(3.4) hold. If % ⊂ π then %

= % ∩ ∆ ⊂ π ∩ ∆ = π

, i.e. the operation

is monotonic.

Since (% ◦ %

)

⊂ ∆, we have (% ◦ %

)

◦ % ⊂ %. Conversely, if (x, y) ∈ % then (x, x) ∈ % ◦ %

∩ ∆ = (% ◦ %

)

and (x, y) ∈ (% ◦ %

)

◦ %, i.e. % ⊂ (% ◦ %

)

◦ %.

Since (π◦π

)

⊂ π◦π

, we have (%◦(π◦π

)

◦%

)

⊂ (%◦π◦π

◦%

)

. Con-

versely, if (x, x) ∈ (% ◦ π ◦ π

◦ %

)

then there exist y, z such that (x, y) ∈ % and

(y, z) ∈ π, hence (y, y) ∈ π◦π

∩∆ = (π◦π

)

and (x, y) ∈ (%◦(π◦π

)

◦%

)

.

Therefore, (% ◦ π ◦ π

◦ %

)

⊂ (% ◦ (π ◦ π

)

◦ %

)

.

Sufficiency. Suppose that (A, ·,

,

, 1, ≤) satisfies the conditions of Theo- rem 1. Put E(A) = {a ∈ A : a

= a}.

Lemma 1. (E(A), ·) is a semilattice; 1 ∈ E(A); (a

)

= a

; (a

)

= a

; a

≤ b

iff a

b

= a

; if a ≤ b

then a ∈ E(A).

It follows from (3.1)–(3.4) that (E(A), ·) is a semilattice. Note that 1

= 1 · 1

= (1 · 1

)

= 1. Since 1 = (1 · 1

)

· 1 = 1

· 1 = 1

, we have 1 ∈ E(A).

Since (a

)

= ((a

)

((a

)

)

)

(a

)

≤ (((a

)

)

)

= (a

)

= a

and a

= ((a

)

)

≤ (a

)

, we have (a

)

= a

. Since (a

)

≤ a

, we have (a

)

= ((a

)

)

≤ (a

)

= a, hence (a

)

≤ a

. Analogously, a

≤ (a

)

, i.e. (a

)

= a

.

If a

b

= a

then a

= a

b

≤ b

. Conversely, if a

≤ b

then a

= a

a

≤ a

b

. Since a

b

≤ a

, we have a

b

= a

.

Suppose that a ≤ b

. Then a

≤ (b

)

= b

and a = (aa

)

a ≤ (aa

)

b

≤ (a(b

)

)

b

≤ a

b

≤ a

. Since a

≤ a, we have a

= a, i.e. a ∈ E(A). This completes the proof of Lemma 1.

Define the unary operations R and L as follows: Ra = (aa

)

; La = (a

a)

. Then R(a

) = La and Ra = a, La = a for each a ∈ E(A). It follows from (3.5) that Raa = a and aLa = a. It follows from (3.6) that R(ab) = R(aRb) and L(ab) = R((ab)

) = R(b

a

) = R(b

R(a

)) = R(b

La) = R((Lab)

) = L(Lab).

Lemma 2. If La = Rb then R(ab) = Ra and L(ab) = Lb.

Indeed, if La = Rb then R(ab) = R(aRb) = R(aLa) = Ra and L(ab) = L(Lab) = L(Rbb) = Lb.

For each B, C ⊂ A we define BC = {bc : b ∈ B & c ∈ C} and B ≤ C iff (∀c ∈ C)(∃b ∈ B) b ≤ c. Note that if B ⊂ C then C ≤ B.

Let N = {0, 1, . . . , n, . . .} and let f be a one-to-one mapping from N onto N

(f (k) = (n

, n

, . . . , n

)). Define functions ϕ, ψ, α, β : N → N as follows:

ϕ(k) = n

if n

≤ k, and ϕ(k) = k otherwise; ψ(k) = n

if n

≤ k, and ψ(k) = k otherwise; α(k) = n

; β(k) = n

. Clearly, for each p ∈ N and (i, j, m, n) ∈ N

there exists k ∈ N such that k ≥ p and ϕ(k) = i, ψ(k) = j, α(k) = m, β(k) = n.

Suppose that b

, . . . , b

∈ A. Define subsets B

⊆ A for k = 1, . . . , n + 1 and i, j ≤ k as follows:

(D1) B

= {b

} , B

= {b

} , B

= {Rb

} , B

= {Lb

} ;

(D2) B

=

k

[

p=0

B

B

;

(D3)

B

= B

{b

} ∪ B

{b

} ;

B

= {b

}B

∪ {b

}B

;

(D4) B

= {Lb

Rb

} .

Lemma 3. (B

)

= B

and B

⊂ B

, B

≤ B

for k ≤ m.

According to (D1), (B

)

= B

. Suppose that (B

)

= B

. Then

(B

)

=  [

p=0

B

B



=

k

[

p=0

(B

B

)

=

k

[

p=0

(B

)

(B

)

=

k

[

p=0

B

B

= B

,

(B

)

= (B

{b

} ∪ B

{b

})

= {b

}B

∪ {b

}B

= B

, (B

)

= B

.

Suppose that B

⊂ B

. Then B

= B

B

⊂ B

B

⊂ B

. If B

≤ B

then B

≤ B

B

≤ B

B

≤ {1}B

= B

.

Since M {◦,

,

, ∆, ⊂} is a quasivariety [7], without loss of generality we may suppose that A is countable, i.e. A = {a

, a

, . . . , a

, . . .}.

Using induction for each d ∈ A we define sequents b

, . . . , b

, b

, . . . and r

, . . . , r

, . . . of elements of A and E(A) respectively such that b

= d and for all n the following conditions hold:

(a) B

≤ {a

a

};

(b) Rb = r

and Lb = r

for each b ∈ B

and B

= {r

}.

Base of induction. Put b

= d and r

= Rb

, r

= Lb

.

Inductive step. Suppose that b

, . . . , b

, b

and r

, . . . , r

have already been defined and (a), (b) are satisfied for n = 1, . . . , m and i, j ≤ m. Put

b

= r

a

R(a

r

) ,

b

= L(r

a

)a

r

, r

= Lb

if B

≤ {a

a

}, and b

= b for some b ∈ B

, b

= r

, r

= r

, otherwise.

If b

= b for some b ∈ B

, b

= r

, r

= r

then b = bLb = br

= b

b

, i.e. B

≤ b

b

and Lb

= Rb

= r

, hence B

= {Lb

Rb

} = {r

}.

If B

≤ {a

a

}, i.e. b ≤ a

a

for some b ∈ B

, then

b = r

br

≤ r

a

a

r

= r

a

L(r

a

)R(a

r

)a

r

= r

a

R(a

r

)L(r

a

)a

r

= b

b

,

i.e. B

≤ b

b

.

Since b ≤ b

b

for some b ∈ B

, we have

r

= Rb ≤ R(b

b

) ≤ R(b

Rb

) ≤ R(b

) . On the other hand,

Rb

= R(r

a

R(a

r

))

= R(r

R(a

R(a

r

))) ≤ R(r

) = r

,

hence Rb

= r

. Analogously, Rb

= r

. Since r

= Lb

, we have

r

= Lb

= L(r

a

R(a

r

))

= L(L(r

a

)R(a

r

)) = R(L(r

a

)R(a

r

))

= R(L(r

a

)a

r

) = Rb

.

Therefore, using Lemma 2 and the definition (D1)–(D4) we conclude that (b) is satisfied for i, j ≤ m + 1.

Put B

= S{B

: n ∈ N}. Then B

B

≤ B

. Lemma 4. {b

} ≤ B

and {r

} ≤ B

.

Note that if b ∈ B

then b can be represented as a product of elements b

, . . . , b

, . . . and b

, . . . , b

, . . . This product constructed according to (D1)–

(D4) will be called the canonical form of b. Let Q

(b) be the number of occurrences of elements a

, a

in the canonical form of b and Q(b) = max{k : Q

(b) > 0}.

Suppose that b ∈ B

. If Q(b) = 0 then according to (D1)–(D4), b = b

(b

b

)

for some m and we have b

= b

Lb

= b

(b

b

)

≤ b

b

b

≤ . . . ≤ b

(b

b

)

= b. Assume that for every b ∈ B

if Q(b) ≤ k and Q

(b) = p then b

≤ b. Suppose that Q

(b) = p + 1. According to (D1)–(D4) the following cases are possible:

1) b = c

b

b

c

where c

∈ B

and c

∈ B

. Since B

≤ {b

b

}, i.e. c ≤ {b

b

} for some c ∈ B

, using the inductive as- sumption we have b

≤ c

cc

≤ c

b

b

c

= b.

2) b = c

b

b

c

where c

∈ B

and c

∈ B

. This case is analo- gous to Case 1.

3) b = c

b

b

c

where c

∈ B

and c

∈ B

. Since Lc

= Rb

= r

, using the inductive assumption we have b

≤ c

c

= c

Lc

c

= c

Rb

c

= c

(b

b

)

c

≤ c

b

b

c

= b.

4) b = c

b

b

c

where c

∈ B

and c

∈ B

. This case is analogous to Case 3.

Suppose that b ∈ B

. If Q(b) = 0 then according to (D1)–(D4), i = 0 or i = 1. If i = 0 then b = r

or b = (b

b

)

for some m and we have r

= (r

)

= (Rb

)

= ((b

b

)

)

≤ (b

b

)

. The case i = 1 is analogous. Assume that for every b ∈ B

if Q(b) ≤ k and Q

(b) = p then r

≤ b. Suppose that Q

(b) = p + 1.

According to (D1)–(D4) the following cases are possible:

1) b = c

b

b

c

where c

∈ B

and c

∈ B

. Since B

≤ {b

b

}, i.e. c ≤ {b

b

} for some c ∈ B

, using the inductive as- sumption we have r

≤ c

cc

≤ c

b

b

c

= b.

2) b = c

b

b

c

where c

∈ B

and c

∈ B

. This case is analogous to Case 1.

3) b = c

b

b

c

where c

∈ B

and c

∈ B

. Since Lc

= Rb

= r

, using the inductive assumption we have r

≤ c

c

= c

Lc

c

= c

Rb

c

= c

(b

b

)

c

≤ c

b

b

c

= b.

4) b = c

b

b

c

where c

∈ B

and c

∈ B

. This case is analogous to Case 3.

This completes the proof of Lemma 3.

Define the mapping F

: A → Rel(N) as follows:

F

(a) = {(i, j) : B

≤ {a}} .

Since (B

)

= B

, we have F

(a

) = (F

(a))

. Obviously, a ≤ b implies F

(a) ⊂ F

(b).

We show that F

(ab) = F

(a)◦F

(b). If (i, j) ∈ F

(a)◦F

(b), i.e. (i, k) ∈ F

(a) and (k, j) ∈ F

(b) for some k, then B

≤ {a} and B

≤ {b}, hence B

≤ B

B

≤ {ab}, i.e. (i, j) ∈ F

(ab). Conversely, suppose that (i, j) ∈ F

(ab), i.e. B

≤ {ab}. Then B

≤ {ab} for some p and there exists k ≥ p such that ϕ(k) = i, ψ(k) = j, a = a

, b = a

. Since B

≤ B

= B

≤ {ab} = {a

a

}, we have b

= r

a

R(a

r

), b

= L(r

a

)b

r

, r

= Lb

, hence b

≤ a and b

≤ b. Since b

= r

b

∈ B

b

⊂ B

, we have (i, k + 1) ∈ F

(b

) ⊂ F

(a).

Analogously, (k + 1, j) ∈ F

(b

) ⊂ F

(b). Thus, (i, j) ∈ F

(a) ◦ F

(b).

We show that F

(a

) = F

(a) ∩ F

(1). Since a

≤ a and a

≤ 1, we have F

(a

) ⊂ F

(a) and F

(a

) ⊂ F

(1). Conversely, suppose that (i, j) ∈ F

(a) ∩ F

(1); then B

≤ {a} and B

≤ {1}, hence B

≤ B

(B

)

≤ {a1

} = {a}.

Since {r

} ≤ B

, we have r

≤ a, hence r

= r

≤ a

. Since r

∈ B

, we have (i, i) ∈ F

(r

) ⊂ F

(a

). It now follows from (i, i) ∈ F

(a

) and (i, j) ∈ F

(1) that (i, j) ∈ F

(a

) ◦ F

(1) = F

(a

1) = F

(a

).

Put X

= X × {d} and X = S{X

: d ∈ A}, F

(a) = {((i, d), (j, d)) : (i, j) ∈ F

(a)} and F (a) = S{F

(a) : d ∈ A}. Obviously F (ab) = F (a) ◦ F (b), F (a)

= F (a)

, F (a

) = F (a) ∩ F (1), and a ≤ b implies F (a) ⊂ F (b). Suppose that F (a) ⊂ F (b); then F

(a) ⊂ F

(b). Since (0, 1) ∈ F

(a), we have (0, 1) ∈ F

(b), i.e. B

≤ {b}. Then using Lemma 4, we obtain {a} ≤ B

≤ {b}, i.e.

a ≤ b. Therefore, F is an isomorphism of (A, ·,

, ≤) into (Rel(X), ◦,

, ⊂) and F (a

) = F (a) ∩ F (1).

It is clear that ε = F (1) is an equivalence relation on X. Let Y = X/ε and let

η be the natural mapping of X onto Y . Put P (a) = η ◦ F (a) ◦ η

. It is easy to see

that P is an isomorphism of (A, ·,

, ≤) into (Rel(Y ), ◦,

, ⊂) and P (1) = ∆. It

follows that P (a

) = P (a) ∩ P (1) = P (a) ∩ ∆ = P (a)

. This completes the proof of Theorem 1.

P r o o f o f T h e o r e m s 2 a n d 3 . Suppose that (A, ·,

, ∧, 1) is a semilat- tice ordered involuted monoid and (4) holds. Let ≤ be the canonical order relation of the semilattice (A, ∧). Put a

= a ∧ 1. Obviously, a

≤ 1 and a

≤ a.

Lemma 5. The operations ·,

,

are monotonic.

If a ≤ b, i.e. a ∧ b = a, then a

∧ b

= (a ∧ b)

= a

, i.e. a

≤ b

, and a

= a∧1 ≤ b∧1 = b

. Also if a ≤ b, i.e. a∧b = a, then ac = (a∧b)c ≤ ac∧bc ≤ bc and c(a ∧ b) ≤ ca ∧ cb ≤ cb.

Lemma 6. xy ∧ z ≤ x(y ∧ x

z), x ≤ xx

x.

Indeed, xy ∧ z = (y

x

∧ z

)

≤ ((y

∧ z

x)x

)

= x(y ∧ xz), x = x1 ∧ x ≤ x(1 ∧ x

x) ≤ xx

x.

Lemma 7. (x

)

= x

, x

x

= x

, (x

)

= x

.

Indeed, x

= x ∧ 1 = (x ∧ 1)1 ∧ 1 ≤ (x ∧ 1)(1 ∧ (x ∧ 1)

1) ≤ (x ∧ 1)

= x

∧ 1 = (x

)

and (x

)

≤ ((x

)

)

= x

. The second assertion follows from

x

x

= (x ∧ 1)(x ∧ 1) ≤ (x ∧ 1)1 ≤ x ∧ 1 = x

and

x

= x ∧ 1 = (x ∧ 1)1 ∧ 1 ≤ (x ∧ 1)(1 ∧ (x ∧ 1)

1)

≤ (x ∧ 1)(1 ∧ x

) = x

(x

)

= x

x

.

Finally, (x

)

= (x ∧ 1)

= x

∧ 1

= x

∧ 1 = (x

)

= x

. Lemma 8. x

y

= x

∧ y

, x

y

= y

x

, (x

y

)

= x

y

.

Since x

y

≤ x

1 = x

and x

y

≤ 1y

= y

, we have x

y

≤ x

∧ y

. Conversely, x

∧ y

= x

1 ∧ y

≤ x

(1 ∧ (x

)

y

) ≤ x

x

y

= x

y

. Thus, x

y

= x

∧ y

. It follows that x

y

= x

∧ y

= y

∧ x

= y

x

and (x

y

)

= x

y

∧ 1 = x

∧ y

∧ 1 = x

∧ y

= x

y

.

Lemma 9. (xyy

x

)

≤ (xx

)

, (xx

)

x = x.

Indeed, (xyy

x

)

= xyy

x

∧ 1 = (xyy

x

∧ 1) ∧ 1 ≤ x(yy

x

∧ x

1) ∧ 1 ≤ xx

∧ 1 = (xx

)

. For the second assertion, (xx

)

x ≤ 1x = x and x = 1x ∧ x ≤ (1 ∧ xx

)x = (xx

)

x.

Lemma 10. (xyy

x

)

= (x(yy

)

x

)

. Indeed,

(xyy

x

)

= (x((yy

)

y)((yy

)

y)

x

)

= (x(yy

)

yy

((yy

)

)

x

)

≤ (x(yy

)

((yy

)

)

x

)

= (x(yy

)

(yy

)

x

)

= (x(yy

)

x

)

and

(x(yy

)

x

)

≤ (xyy

x

)

.

According to Lemmas 5–10, (A, ·,

,

, 1, ≤) satisfies the conditions of Theo- rem 1. This immediately implies the conclusion of Theorem 2.

Lemma 11. x ∧ y(z ∧ tz) ≤ y(y

xz

∧ t)z.

Indeed,

y(z ∧ tz) ∧ x ≤ y(z ∧ tz ∧ y

x)

≤ y((t ∧ zz

)z ∧ y

x) ≤ y(t ∧ zz

∧ y

xz

)z ≤ y(t ∧ y

xz

)z . Lemma 12. If x

x ≤ 1 then x(y ∧ z) = xy ∧ xz and (y ∧ z)x

= yx

∧ zx

. Indeed, x(y ∧ z) ≤ xy ∧ xz and xy ∧ xz ≤ x(y ∧ x

xz) ≤ x(y ∧ 1z) = x(y ∧ z).

The proof of the second assertion is similar.

Lemma 13. If u

u ≤ 1, v

v ≤ 1 and x ∧ y ≤ u

v, then x ∧ y = u

(uxv

)

(uyv

)

v.

Since u

u ≤ 1 and v

v ≤ 1, we have u

(uxv

)

(uyv

)

v ≤ u

(uxv

)

v ≤ u

uxv

v ≤ x. Analogously, u

(uxv

)

(uyv

)

v ≤ y.

Thus, u

(uxv

)

(uyv

)

v ≤ x ∧ y. Conversely, using Lemmas 11, 12, we obtain

x ∧ y = x ∧ y ∧ u

v

= x ∧ y ∧ u

(v ∧ 1v) = u

(u(x ∧ y)v

∧ 1)v

= u

(uxv

∧ uyv

∧ 1)v = u

((uxv

∧ 1) ∧ (uyv

∧ 1))v

= u

((uxv

)

∧ (uyv

)

)v = u

((uxv

)

(uyv

)

)v .

Lemma 14. Suppose that %, π, α, β ∈ Rel(X) and α

◦ α ⊂ ∆, β

β ⊂ ∆,

% ∩ π ⊂ α

◦ β. Then

% ∩ π = α

◦ (α ◦ % ◦ β

)

◦ (α ◦ π ◦ β

)

◦ β .

It follows from α

◦ α ⊂ ∆ and β

◦ β ⊂ ∆ that α, β are functions.

Therefore, we write y = α(x) and y = β(x) instead of (x, y) ∈ α and (x, y) ∈ β.

Since % ∩ π ⊂ α

◦ β, for each pair (x, y) ∈ % ∩ π there exists z ∈ X such that α(z) = x and β(z) = y.

Suppose that (x, y) ∈ α

◦ (α ◦ % ◦ β

)

◦ (α ◦ π ◦ β

)

◦ β. Then x = α(z) and y = β(z) for some z such that (z, z) ∈ (α ◦ % ◦ β

)

◦ (α ◦ π ◦ β

)

. It follows that (z, z) ∈ α ◦ % ◦ β

and (z, z) ∈ α ◦ π ◦ β

, hence (x, y) = (α(z), β(z)) ∈ % ∩ π.

Conversely, let (x, y) ∈ % ∩ π. Since x = α(z) and y = β(z) for some z, we have (z, z) ∈ (α ◦ % ◦ β

)

and (z, z) ∈ (α ◦ π ◦ β

)

, hence (z, z) ∈ (α ◦ % ◦ β

)

◦ (α ◦ π ◦ β

)

and (x, y) = (α(z), β(z)) ∈ α

◦ (α ◦ % ◦ β

)

◦ (α ◦ π ◦ β

)

◦ β, which completes the proof of Lemma 14.

Theorem 2 and Lemmas 13, 14 immediately imply Theorem 3.

References

[1] D. A. B r e d i k h i n, Representation of ordered involuted semigroups, Izv. Vyssh. Uchebn.

Zaved. Mat. 7 (1975), 19–29 (in Russian).

[2] —, Abstract characteristics of some relation algebras, in: Algebra and Number Theory, Nalchic 1977, 3–19 (in Russian).

[3] L. H. C h i n and A. T a r s k i, Distributive and modular laws in the arithmetic of relation algebras, Univ. Calif. Publ. Math. 1 (1951), 341–383.

[4] S. A. G i v a n t and A. T a r s k i, A formalization of set theory without variables, Amer. Math.

Soc. Colloq. Publ. 41, Amer. Math. Soc., Providence, R.I., 1987.

[5] M. H a i v a n, Arguesian lattices which are not linear , Bull. Amer. Math. Soc. 16 (1987), 121–123.

[6] B. J ´o n s s o n, Representation of modular lattices and of relation algebras, Trans. Amer.

Math. Soc. 92 (1959), 449–464.

[7] B. M. S c h e i n, Relation algebras and function semigroups, Semigroup Forum 1 (1970), 1–62.

[8] —, Representation of involuted semigroups by binary relations, Fund. Math. 82 (1974), 121–141.