Canonical Relativized Cylindric Set Algebras and Weak Associativity

TL;DR

This paper refines the representation theorem for weakly associative relation algebras using canonical relativized cylindric set algebras, ensuring atoms are orbits under permutation groups, building on a 1989 result.

Contribution

It introduces a sharper version of the relative representation theorem for weakly associative relation algebras using canonical relativized cylindric set algebras.

Findings

Atoms are orbits of sequences under permutation groups.

The theorem was previously proved for the Resek-Thompson case in 1989.

Provides a more precise algebraic characterization.

Abstract

Canonical relativized cylindric set algebras are used to sharpen the relative representation theorem for weakly associative relation algebras, that every complete atomic weakly associative relation algebra is isomorphic with the relativization of a set relation algebra to a symmetric and reflexive binary relation, by insuring that the atoms of the set relation algebra and its relativization are orbits of single sequences under a group of permutations of the underlying set. This sharpening of the relative representation theorem was first proved for the Resek-Thompson Theorem in 1989.

Equations

$$\mathfrak{A}\cong \mathfrak{Ra}\mathfrak{Rl}{}_{V(\mathfrak{B}(\mathfrak{A}))}\mathfrak{C}\cong \mathfrak{Rl}{}_{{\mathsf{C}}_{2}V(\mathfrak{B}(\mathfrak{A}))}\mathfrak{Ra}\mathfrak{C}\cong \mathfrak{Rl}{}_E{\mathfrak{A}}^{\prime }.$$

$$x\cdot y=\overline{\overline{x}+\overline{y}},0\text{raise}5.59721pt,=\overline{1\text{raise}5.59721pt,},1=1\text{raise}5.59721pt,+0\text{raise}5.59721pt,,0=\overline{1\text{raise}5.59721pt,+0\text{raise}5.59721pt,}.$$

$$x+y$$

$$x+(y+z)$$

$$\overline{\overline{x}+\overline{y}}+\overline{\overline{x}+y}$$

$$((x\cdot 1\text{raise}5.59721pt,);1);1$$

$$(x+y);z$$

$$x;1\text{raise}5.59721pt,$$

$$\breve{\stackrel{˘}{x}}$$

$$\left(x+y\right)\stackrel{˘}{}$$

$$\left(x;y\right)\stackrel{˘}{}$$

$$\breve{x};\overline{x;y}+\overline{y}$$

$$I{d}_U=\{⟨u,u⟩:u\in U\}.$$

$$\mathfrak{Re}(U)=⟨\mathcal{P}(U\times U),\cup ,{}_U\setminus ,|,{}^{-1},I{d}_U⟩$$

$$R\cup S$$

$${}_U\setminus R$$

$$R|S$$

$$R^{-1}$$

$$\displaystyle Fd({{E}})=\{{{u}}:\exists{{v}}\big{(}\langle{{u}},{{v}}\rangle\in{{E}}\text{ or }\langle{{v}},{{u}}\rangle\in{{E}}\big{)}\}.$$

$$I{d}_{Fd(E)}\cap (E|E^{-1}\cup E^{-1}|E)\subseteq E,$$

$$\displaystyle\mathfrak{Rl}\mskip 1.5mu_{{E}}\left(\mathfrak{{{A}}}\right)=\big{\langle}\{{{R}}\cap{{E}}:{{R}}\in{{A}}\},+,\overline{\phantom{x}},{\mskip 1.5mu;\mskip 1.5mu},\breve{\phantom{x}},{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\big{\rangle}$$

$$R+S$$

$$\overline{R}$$

$$R;S$$

$$\breve{R}$$

$$1\text{raise}5.59721pt,$$

$$\mathfrak{{{A}}}\cong\mathfrak{{{A}}}^{\prime}\subseteq\mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)}.$$

$$F(x)=\{⟨u,v⟩:u,v\in U,⟨u,a,v⟩\in T\text{ for some a∈AtA such that a≤x}\},$$

$${{F}}\in\text{Ism}\big{(}\mathfrak{{{A}}},\mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)}\big{)}.$$

$${{F}}\in\text{Hom}\big{(}Bl\,\mathfrak{{{A}}},Bl\,\mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)}\big{)}.$$

$$a\le x;y=\sum _{x\ge b\in \mathsf{At}\mathfrak{A},y\ge c\in \mathsf{At}\mathfrak{A}}b;c,$$

$$⟨I^{(\mathfrak{A})},\{⟨a,a,a⟩:a\in I^{(\mathfrak{A})}\}⟩$$

$$T^{\prime }=T\cup \{⟨u,b,w⟩,⟨w,c,v⟩,⟨w,\breve{b},u⟩,⟨v,\breve{c},w⟩,⟨w,b^{\mathsf{r}},w⟩\},$$

$$⟨u_0,x_1,u_2⟩,⟨u_2,x_0,u_1⟩,⟨u_0,x_2,u_1⟩\in T^{\prime }.$$

$$\begin{array}[]{|r|c|c|c|c|c|c|c|}\hline\cr&\,\,\,1&\,\,\,2&\,\,\,3&\,\,\,4&\,\,\,5&\,\,\,6&7\\ \hline\cr&{{u}}_{0}&{{u}}_{1}&{{u}}_{2}&{{x}}_{2}&{{x}}_{1}&{{x}}_{0}&\,{{x}}_{2}\leq{{x}}_{1}{\mskip 1.5mu;\mskip 1.5mu}{{x}}_{0}\\ \hline\cr 1&{{u}}&{{v}}&{{w}}&{{a}}&{{b}}&{{c}}&{{a}}\leq{{b}}{\mskip 1.5mu;\mskip 1.5mu}{{c}}\\ 2&{{u}}&{{w}}&{{v}}&{{b}}&{{a}}&\breve{{{c}}}&{{b}}\leq{{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}\\ 3&{{v}}&{{w}}&{{u}}&\breve{{{c}}}&\breve{{{a}}}&{{b}}&\breve{{{c}}}\leq\breve{{{a}}}{\mskip 1.5mu;\mskip 1.5mu}{{b}}\\ 4&{{v}}&{{u}}&{{w}}&\breve{{{a}}}&\breve{{{c}}}&\breve{{{b}}}&\breve{{{a}}}\leq\breve{{{c}}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{b}}}\\ 5&{{w}}&{{u}}&{{v}}&\breve{{{b}}}&{{c}}&\breve{{{a}}}&\breve{{{b}}}\leq{{c}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{a}}}\\ 6&{{w}}&{{v}}&{{u}}&{{c}}&\breve{{{b}}}&{{a}}&{{c}}\leq\breve{{{b}}}{\mskip 1.5mu;\mskip 1.5mu}{{a}}\\ 7&{{u}}&{{w}}&{{w}}&{{b}}&{{b}}&{{{b}}}^{\mathsf{r}}&{{b}}\leq{{b}}{\mskip 1.5mu;\mskip 1.5mu}{{{b}}}^{\mathsf{r}}\\ 8&{{w}}&{{u}}&{{w}}&\breve{{{b}}}&{{{b}}}^{\mathsf{r}}&\breve{{{b}}}&\breve{{{b}}}\leq{{{b}}}^{\mathsf{r}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{b}}}\\ 9&{{w}}&{{w}}&{{u}}&{{{b}}}^{\mathsf{r}}&\breve{{{b}}}&{{b}}&{{{b}}}^{\mathsf{r}}\leq\breve{{{b}}}{\mskip 1.5mu;\mskip 1.5mu}b\\ 10&{{v}}&{{w}}&{{w}}&\breve{{{c}}}&\breve{{{c}}}&{{{b}}}^{\mathsf{r}}&\breve{{{c}}}\leq\breve{{{c}}}{\mskip 1.5mu;\mskip 1.5mu}{{{b}}}^{\mathsf{r}}\\ 11&{{w}}&{{v}}&{{w}}&{{c}}&{{{b}}}^{\mathsf{r}}&{{c}}&{{c}}\leq{{{b}}}^{\mathsf{r}}{\mskip 1.5mu;\mskip 1.5mu}{{c}}\\ 12&{{w}}&{{w}}&{{v}}&{{{b}}}^{\mathsf{r}}&{{c}}&\breve{{{c}}}&{{{b}}}^{\mathsf{r}}\leq{{c}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}\\ 13&{{u}}&{{u}}&{{w}}&{{{a}}}^{\mathsf{d}}&{{b}}&\breve{{{b}}}&{{{a}}}^{\mathsf{d}}\leq{{b}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{b}}}\\ 14&{{u}}&{{w}}&{{u}}&{{b}}&{{{a}}}^{\mathsf{d}}&{{b}}&{{b}}\leq{{{a}}}^{\mathsf{d}}{\mskip 1.5mu;\mskip 1.5mu}{{b}}\\ 15&{{w}}&{{u}}&{{u}}&\breve{{{b}}}&\breve{{{b}}}&{{{a}}}^{\mathsf{d}}&\breve{{{b}}}\leq\breve{{{b}}}{\mskip 1.5mu;\mskip 1.5mu}{{{a}}}^{\mathsf{d}}\\ 16&{{v}}&{{v}}&{{w}}&{{{a}}}^{\mathsf{r}}&\breve{{{c}}}&{{c}}&{{{a}}}^{\mathsf{r}}\leq\breve{{{c}}}{\mskip 1.5mu;\mskip 1.5mu}c\\ 17&{{v}}&{{w}}&{{v}}&\breve{{{c}}}&{{{a}}}^{\mathsf{r}}&\breve{{{c}}}&\breve{{{c}}}\leq{{{a}}}^{\mathsf{r}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}\\ 18&{{w}}&{{v}}&{{v}}&{{c}}&{{c}}&{{{a}}}^{\mathsf{r}}&{{c}}\leq{{c}}{\mskip 1.5mu;\mskip 1.5mu}{{{a}}}^{\mathsf{r}}\\ 19&{{w}}&{{w}}&{{w}}&{{{b}}}^{\mathsf{r}}&{{{b}}}^{\mathsf{r}}&{{{b}}}^{\mathsf{r}}&{{{b}}}^{\mathsf{r}}\leq{{{b}}}^{\mathsf{r}}{\mskip 1.5mu;\mskip 1.5mu}{{{b}}}^{\mathsf{r}}\\ \hline\cr\end{array}$$

$$\langle{{U}}_{\lambda},T_{\lambda}\rangle=\Big{<}\bigcup_{\kappa<\lambda}{{U}}_{\kappa},\bigcup_{\kappa<\lambda}T_{\kappa}\Big{>}.$$

$$\langle{{U}}^{\prime},{{T}}^{\prime}\rangle=\Big{<}\bigcup_{n<\omega}{{U}}_{n},\bigcup_{n<\omega}{{T}}_{n}\Big{>}$$

$$L_p(q)=\{\begin{array}{ll}p\odot q& \text{ if }{t}_n=s_0\\ \breve{p}\odot q& \text{ if }{t}_n\ne s_0=t_0\\ q& \text{ if }{t}_n\ne s_0\ne t_0,\end{array}$$

Full text

**Keywords ** Cylindric set algebras $\cdot$ Set relation algebras $\cdot$ Relativized cylindric algebras $\cdot$ Relativized relation algebras $\cdot$ Symmetric relation $\cdot$ Reflexive relation $\cdot$ Relativization July 15, 2018, revised October 16, 2019, corrected and revised February 16, 2021

1 Introduction

My first meeting and mathematical encounter with Hajnal and István occurred in 1984 on a beach in Charleston, South Carolina, at a meeting organized by Steve Comer. As the three of us came out of the water, Hajnal asked me, “How do you prove…”, and I answered, “Read the proof in my dissertation, not the published proof.” Her question referred to a theorem about relative representation, proved in my dissertation (M78, , Theorem 5(10)) for the class $\mathsf{SA}$ of semi-associative relation algebras, and published in (MR662049, , Theorem 5.20) for the strictly larger class $\mathsf{WA}$ of weakly associative relation algebras. The original proof in (M78, , pp. 76–83) is more direct, and can be illustrated, as I did on the beach, by drawing triangles in the sand (á la Archimedes) that share edges. On any given edge, labelled with an atom included in the product of two other atoms, make a new triangle by drawing two new edges labelled with the two atoms in the product, and continue ad infinitum. See the proof of Theorem 5.1 in §5.

Axioms (1)–(10) for $\mathsf{WA}$ are given in §2. Elementary consequences (11)–(43) of these axioms, used in §§2–11, are relegated (along with complete proofs) to §12. Weak associativity (4) is needed only for (38)–(43). The closure of $\mathsf{WA}$ under the formation of canonical extensions is discussed in §3. In §4 the algebra $\mathfrak{Re}({{U}})$ of all binary relations on a set ${{U}}$ and the relativizations of its subalgebras are defined. Corollary 1 shows relativizing $\mathfrak{Re}({{U}})$ to a symmetric and reflexive relation yields a weakly associative relation algebra. §5 presents a converse to this observation in Theorem 5.1, whose proof (not previously published and quoted here from (M78, , pp. 76–83)) I recommended to Hajnal and István: every $\mathfrak{{{A}}}\in\mathsf{WA}$ is isomorphic to a subalgebra of \mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)} for some symmetric and reflexive ${{E}}$. Theorem 5.1 is used to characterize $\mathsf{WA}$ in Corollary 2 as the class of algebras isomorphic to subalgebras of relativizations (to symmetric and reflexive relations) of the algebras of binary relations on sets. In §§6–11 we prove (in Theorem 11.1) a sharpening of Corollary 2 for complete atomic algebras in $\mathsf{WA}$, namely, that every such algebra is a relativization of a complete atomic set relation algebra whose atoms are orbits of single sequences under a group of permutations of the underlying set. The proof uses results from MR987611 , where a similar sharpening of the Resek-Thompson Theorem AT is presented.

In §6 we recall from MR987611 the concepts of suitable structure (of any dimension) and, for an arbitrary suitable structure $\mathfrak{{{B}}}$, its canonical relativized cylindric set algebra $\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}$. In §7, we construct from any atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ a 3-dimensional suitable structure $\mathfrak{{{B}}}(\mathfrak{{{A}}})$. In §8 we define the complex algebra $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ for any suitable structure $\mathfrak{{{B}}}$. We show (in Lemmas 5 and 7) that for any atomic $\mathfrak{{{A}}}\in\mathsf{WA}$, $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}(\mathfrak{{{A}}})$ is a non-commutative 3-dimensional cylindric algebra satisfying the merry-go-round identities. The main result of (MR987611, , Theorem C) applies to such algebras, yielding an isomorphism between $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}(\mathfrak{{{A}}})$ and $\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}(\mathfrak{{{A}}})$ in Theorem 8.1. In §9 we recall the definition of relation-algebraic reduct $\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}$ of a non-commutative 3-dimensional cylindric algebra $\mathfrak{{{C}}}$, and show, in Theorem 9.1, that every complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ is isomorphic to $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}(\mathfrak{{{A}}})$. The proof of Theorem 10.1 in §10 combines the isomorphisms of Theorem 8.1 and Theorem 9.1 with the definition of canonical relativized cylindric set algebra from §6 to show that every complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ is isomorphic to the relation-algebraic reduct of the relativization, to a ternary relation ${{V}}(\mathfrak{{{B}}}(\mathfrak{{{A}}}))$, of a complete atomic 3-dimensional cylindric set algebra $\mathfrak{{{C}}}$, giving us the first of the following three isomorphisms:

[TABLE]

Most of the proof of Theorem 11.1 in §11 is devoted to the second of the isomorphisms above. In this case, relativization commutes with relation-algebraic reduct. According to the link in HMT85 between cylindric set algebras and relation set algebras, the third isomorphism holds for some set relation algebra $\mathfrak{{{A}}}^{\prime}$ isomorphic to $\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}$ and some symmetric and reflexive binary relation ${{E}}$, namely the image of the ternary relation $\mathsf{C}_{2}{{V}}(\mathfrak{{{B}}}(\mathfrak{{{A}}}))$. The elements of the base set of $\mathfrak{{{A}}}^{\prime}$ are equivalence classes of certain finite sequences of triples of atoms of $\mathfrak{{{A}}}$ alternating with members of $\{0,1,2\}$.

2 Definition of $\mathsf{WA}$

$\mathsf{WA}$ is the class of weakly associative relation algebras, algebras of the form $\mathfrak{{{A}}}=\langle{{A}},+,\overline{\phantom{x}},{\mskip 1.5mu;\mskip 1.5mu},\breve{\phantom{x}},{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\rangle$ satisfying equational axioms (1)–(10) below. These equations are obtained from Tarski’s classic axiomatization of the class $\mathsf{RA}$ of relation algebras (see (zbMATH06783012, , Definition 2.1)) by replacing the associative law $({{x}}{\mskip 1.5mu;\mskip 1.5mu}{{y}}){\mskip 1.5mu;\mskip 1.5mu}{{z}}={{x}}{\mskip 1.5mu;\mskip 1.5mu}({{y}}{\mskip 1.5mu;\mskip 1.5mu}{{z}})$ with the weak associative law (4), accompanied by the definitions

[TABLE]

Axioms (1)–(3) assert that $\langle{{A}},+,\overline{\phantom{x}}\,\rangle$ is a Boolean algebra. If the weak associative law (4) were replaced by the semi-associative law $({{x}}{\mskip 1.5mu;\mskip 1.5mu}1){\mskip 1.5mu;\mskip 1.5mu}1={{x}}{\mskip 1.5mu;\mskip 1.5mu}1,$ then the axioms (1)–(10) would define the class $\mathsf{SA}$ of semi-associative relation algebras.

[TABLE]

All the consequences we need are proved in §12. Among them are equations that can form alternative axiom sets for $\mathsf{WA}$. The axiomatization based on (M78, , Theorem 1(22)) uses axioms (1)–(4), (6), (7), the definitions of $\cdot$ and the constants, plus (18), (19), (20), and (21) from §12. Replacing (4) with either the semi-associative law or the associative law gives alternative axiomatizations for $\mathsf{SA}$ and $\mathsf{RA}$ as well.

3 Canonical Extensions

A positive equation is one that involves only $+$, $\cdot$, constants, ${\mskip 1.5mu;\mskip 1.5mu}$, and $\breve{\phantom{x}}$, but not complementation. Examples include axioms (1), (2), (4)–(9), the alternative axioms listed in §2, the semi-associative law, and the associative law. Jónsson-Tarski MR0044502 ; MR0045086 proved that every Boolean algebra with completely additive operators has a canonical extension, and the extension satisfies the same positive equations as the original algebra. From this they conclude that the canonical extension of a relation algebra is a relation algebra. For details see MR1935083 , MR2269199 , or zbMATH06783011 (recommended). The Jónsson-Tarski results apply to $\mathsf{WA}$ because ${\mskip 1.5mu;\mskip 1.5mu}$ and $\breve{\phantom{x}}$ are completely additive by (30) and (31) in §12. Indeed, by (MR662049, , Theorem 4.2), every weakly associative relation algebra has a canonical extension that is in $\mathsf{WA}$. (Since the associative laws are positive, the same applies to $\mathsf{SA}$ and $\mathsf{RA}$.) The canonical extension of an algebra is complete and atomic, but all we need in the proof of Theorem 5.1 is that every $\mathsf{WA}$ is a subalgebra of an atomic $\mathsf{WA}$.

4 Algebras of Binary Relations

For any set ${{U}}$, the identity relation on ${{U}}$ is

[TABLE]

The algebra of all binary relations on ${{U}}$ is

[TABLE]

where $\mathcal{P}$ is the power set operator, and for all ${{R}},{{S}}\subseteq{{U}}\times{{U}}$,

[TABLE]

The field of a binary relation ${{E}}$ is

[TABLE]

A relation ${{E}}$ is reflexive if it contains the identity relation on its field, i.e.,

[TABLE]

and symmetric if ${{E}}^{-1}={{E}}$. If $\mathfrak{{{A}}}$ is a subalgebra of $\mathfrak{Re}({{U}})$ and ${{E}}\subseteq{{U}}\times{{U}}$ then the relativization of $\mathfrak{{{A}}}$ to ${{E}}$ is the algebra

[TABLE]

where, for all ${{R}},{{S}}\in{{A}}$,

[TABLE]

The following direct consequence of the relevant definitions is also a special case of (MR662049, , Theorem 5.8(2)), that the relativization of a $\mathsf{WA}$ (such as $\mathfrak{Re}({{U}})$) to a symmetric-reflexive element (e.g., any symmetric and reflexive relation) is in $\mathsf{WA}$.

Corollary 1

If ${{U}}$ is a set, ${{E}}\subseteq{{U}}\times{{U}}$, ${{E}}$ is symmetric, and ${{E}}$ is reflexive, then the relativization of $\mathfrak{Re}({{U}})$ to ${{E}}$ is a complete atomic weakly associative relation algebra: \mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)}\in\mathsf{WA}.

5 Characterizing $\mathsf{WA}$

In the proof of the following converse to Corollary 1, a relation ${{E}}$ is constructed as a graph by adding a new vertex at each step, along with two new edges connecting that vertex to the ends of a previously selected edge. In the proof, Lemma 1 asserts that each step is possible, while Lemmas 2 and 3 say the process can be completed to give the desired relation ${{E}}$. The whole construction is mediated by a labelling of the edges with atoms of an atomic $\mathsf{WA}$, called a “labelling system”. (The theorem was originally stated and proved for $\mathsf{SA}$ but it applies to $\mathsf{WA}$ with no changes.)

Theorem 5.1** ((M78, , Theorem 5(10))**)

Let $\mathfrak{{{A}}}\in\mathsf{WA}$. Then there is a set ${{U}}$ and a symmetric and reflexive binary relation ${{E}}\subseteq{{U}}\times{{U}}$ such that $\mathfrak{{{A}}}$ is isomorphic to a subalgebra $\mathfrak{{{A}}}^{\prime}$ of the relativization to ${{E}}$ of the algebra of all binary relations on ${{U}}$:

[TABLE]

Proof

(M78, , pp. 76–83) We shall prove the theorem under the additional assumption that $\mathfrak{{{A}}}$ is atomic. The original theorem will then follow from the fact that every $\mathsf{WA}$ is a subalgebra of an atomic $\mathsf{WA}$ (namely, its canonical extension). Suppose a set ${{U}}$ and a set $T\subseteq{{U}}\times\mathsf{At\,}{\mathfrak{{{A}}}}\times{{U}}$, where $\mathsf{At\,}{\mathfrak{{{A}}}}$ is the set of atoms of $\mathfrak{{{A}}}$, have been constructed so that the following six properties hold for all $a,b,c\in\mathsf{At\,}{\mathfrak{{{A}}}}$ and for all $u,v,w\in{{U}}$:

1. (i)

if $\langle u,a,v\rangle,\langle u,b,v\rangle\in T$, then $a=b$. 2. (ii)

if $a\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ then there exists some $u\in{{U}}$ such that $\langle u,a,u\rangle\in T$, 3. (iii)

$u=v$ iff there exists some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$ such that $a\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ and $\langle u,a,v\rangle\in T$. 4. (iv)

if $\langle u,a,v\rangle\in T$ then $\langle v,\breve{a},u\rangle\in T$, 5. (v)

if $\langle u,a,v\rangle,\langle v,b,w\rangle,\langle u,c,w\rangle\in T$, then $a{\mskip 1.5mu;\mskip 1.5mu}b\geq c$, 6. (vi)

if $\langle u,a,v\rangle\in T$ and $a\leq b{\mskip 1.5mu;\mskip 1.5mu}c$, then there exists some $w\in{{U}}$ such that $\langle u,b,w\rangle,\langle w,c,v\rangle\in T$.

For all $x\in A$, set

[TABLE]

and let ${{E}}={{F}}(1)$. We can then use properties (i)–(vi) to prove that ${{F}}$ is an isomorphism from $\mathfrak{{{A}}}$ into the algebra of all binary relations contained in ${{E}}$:

[TABLE]

To prove the above, it is enough to show that conditions (vii)–(xii) below hold for all $x,y\in A$.

(vii) ${{F}}(x+y)={{F}}(x)\cup{{F}}(y)$.

Proof

Suppose $\langle u,v\rangle\in{{F}}(x+y)$. Hence $\langle u,a,v\rangle\in T$ for some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$ with $a\leq x+y$. Then we get $a\leq x$ or $a\leq y$ since $a$ is an atom, so either $\langle u,v\rangle\in{{F}}(x)$ or $\langle u,v\rangle\in{{F}}(y)$. This shows ${{F}}(x+y)\subseteq{{F}}(x)\cup{{F}}(y)$. On the other hand, if $\langle u,v\rangle\in{{F}}(x)$, then for some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$ we have $\langle u,a,v\rangle\in T$ and $a\leq x$, hence $a\leq x+y$, so $\langle u,v\rangle\in{{F}}(x+y)$. Thus ${{F}}(x)\subseteq{{F}}(x+y)$, and similarly, ${{F}}(y)\subseteq{{F}}(x+y)$.

(viii) ${{F}}(\overline{{{x}}})={{E}}\setminus{{F}}(x)$.

Proof

Let $\langle u,v\rangle\in{{F}}(\overline{{{x}}})$. Then $\langle u,a,v\rangle\in T$ and $a\leq\overline{{{x}}}$ for some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$. Suppose $\langle u,v\rangle\in{{F}}(x)$, i.e., $\langle u,b,v\rangle\in T$ for some $b\in\mathsf{At\,}{\mathfrak{{{A}}}}$. Then $a=b$ by (i), and the conditions $a\leq\overline{{{x}}}$ and $b\leq x$ are contradictory. Thus $\langle u,v\rangle\notin{{F}}(x).$ From (vii) we get $\langle u,v\rangle\in{{F}}(\overline{{{x}}})\subseteq{{F}}(\overline{{{x}}}+x)={{F}}(1)={{E}}$. Thus $\langle u,v\rangle\in{{E}}\setminus{{F}}(x)$. Now let $\langle u,v\rangle\in{{E}}\setminus{{F}}(x)$. Then there is some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$ such that $\langle u,a,v\rangle\in T$ and for no $b\in\mathsf{At\,}{\mathfrak{{{A}}}}$ is it the case that $\langle u,b,v\rangle\in T$ and $b\leq x$. In particular, since $\langle u,a,v\rangle\in T$, it is not the case that $a\leq x$. This implies that $a\leq\overline{{{x}}}$ since $a$ is an atom, and hence $\langle u,v\rangle\in{{F}}(\overline{{{x}}})$.

(ix) ${{F}}$ is one-to-one.

Proof

From (vii) and (viii) we conclude that ${{F}}$ is a homomorphism from the Boolean part of $\mathfrak{{{A}}}$ into the Boolean algebra of all subrelations of ${{E}}$:

[TABLE]

We therefore need only show that ${{F}}(x)\neq\emptyset$ whenever $0\neq x\in A$. Suppose $0\neq x\in A$. Since $\mathfrak{{{A}}}$ is atomic, there is some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$ with $a\leq x$. Then $a{\mskip 1.5mu;\mskip 1.5mu}\breve{a}\cdot{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\in\mathsf{At\,}{\mathfrak{{{A}}}}$ by (34) and (41), so by (ii) there is some $u\in{{U}}$ such that $\langle u,a{\mskip 1.5mu;\mskip 1.5mu}\breve{{{a}}}\cdot{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt},u\rangle\in T$. Now ${{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{a}}}\cdot{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\leq{{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{a}}}$, so by (vi) there is some $w\in{{U}}$ such that $\langle u,a,w\rangle,\langle w,\breve{a},u\rangle\in T$. But then $\langle u,w\rangle\in{{F}}(x)$, so ${{F}}(x)\neq\emptyset$.

(x) ${{F}}({1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt})=Id_{{U}}$.

Proof

Note that (x) is just a restatement of (iii).

(xi) ${{F}}(\breve{x})={{F}}(x)^{-1}$.

Proof

Let $\langle u,v\rangle\in{{F}}(\breve{x})$, so that $\langle u,a,v\rangle\in T$ and $a\leq\breve{x}$ for some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$. Then $\langle v,\breve{a},u\rangle\in T$ by (iv), $\breve{a}\leq\breve{\breve{x}}=x$ by (7), and $\breve{a}\in\mathsf{At\,}{\mathfrak{{{A}}}}$ by (32), so $\langle v,u\rangle\in{{F}}(x)$, and hence $\langle u,v\rangle\in{{F}}(x)^{-1}$. This shows that ${{F}}(\breve{x})\subseteq{{F}}(x)^{-1}$ and hence also ${{F}}(\breve{x})^{-1}\subseteq{{F}}(x)$. Since the latter formula holds for all $x\in A$, we can substitute $\breve{x}$ for $x$ and obtain ${{F}}(x)^{-1}={{F}}(\breve{\breve{x}})^{-1}\subseteq{{F}}(\breve{x})$, which completes the proof of (xi).

(xii) ${{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)=({{F}}(x)|{{F}}(y))\cap{{E}}$.

Proof

Suppose $\langle u,v\rangle\in{{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)$, i.e., $\langle u,a,v\rangle\in T$ and $a\leq x{\mskip 1.5mu;\mskip 1.5mu}y$ for some $a\in\mathsf{At\,}{\mathfrak{{{A}}}}$. Since $\mathfrak{{{A}}}$ is atomic and ${\mskip 1.5mu;\mskip 1.5mu}$ is completely additive by (31),

[TABLE]

so there must be some $b,c\in\mathsf{At\,}{\mathfrak{{{A}}}}$ with $a\leq b{\mskip 1.5mu;\mskip 1.5mu}c$, $b\leq x$, and $c\leq y$. By (vi) there is some $w\in{{U}}$ such that $\langle u,b,w\rangle,\langle w,c,v\rangle\in T$. From these facts we get $\langle u,w\rangle\in{{F}}(x)$ and $\langle w,v\rangle\in{{F}}(y)$, and hence $\langle u,v\rangle\in{{F}}(x)|{{F}}(y)$. We also have $\langle u,v\rangle\in{{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)\subseteq{{F}}(1)={{E}}$, so ${{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)\subseteq({{F}}(x)|{{F}}(y))\cap{{E}}$.

Now suppose $\langle u,v\rangle\in({{F}}(x)|{{F}}(y))\cap{{E}}$. It follows that there are $a,b,c\in\mathsf{At\,}{\mathfrak{{{A}}}}$ and $w\in{{U}}$ such that $a\leq x$, $b\leq y$, and $\langle u,a,w\rangle,\langle w,b,v\rangle,\langle u,c,v\rangle\in T$. Then $a{\mskip 1.5mu;\mskip 1.5mu}b\leq x{\mskip 1.5mu;\mskip 1.5mu}y$ by (15) and (17), and $a{\mskip 1.5mu;\mskip 1.5mu}b\geq c$ by (v), so $\langle u,v\rangle\in{{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)$. Thus $({{F}}(x)|{{F}}(y))\cap{{E}}\subseteq{{F}}(x{\mskip 1.5mu;\mskip 1.5mu}y)$, and the proof of (xii) is complete.

We have shown that {{F}}\in\text{Ism}\big{(}\mathfrak{{{A}}},\mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)}\big{)}. In addition, using (xi), (vii), (x), and (12) we get ${{E}}^{-1}={{F}}(1)^{-1}={{F}}(\breve{1})={{F}}(1)={{E}}$ and $Id_{{U}}={{F}}({1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt})\subseteq{{F}}(1)={{E}}$, so ${{E}}\subseteq{{U}}\times{{U}}$ is a symmetric and reflexive relation. It follows that the theorem will be proved if we can find sets ${{U}}$ and $T$ satisfying (i)–(vi). It is easy to get sets ${{U}}$ and $T$ which satisfy (i)–(v). For example, let ${{U}}={{I}}^{(\mathfrak{{{A}}})}=\{a:{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\geq{{a}}\in\mathsf{At\,}{\mathfrak{{{A}}}}\}$ and ${{T}}=\{\langle{{a}},{{a}},{{a}}\rangle:{{a}}\in{{I}}^{(\mathfrak{{{A}}})}\}.$ Then (i)–(iii) obviously hold. To verify (iv) and (v) we only need to know that ${{a}}=\breve{{{a}}}={{a}}{\mskip 1.5mu;\mskip 1.5mu}{{a}}$ whenever $a\in I^{(\mathfrak{{{A}}})}$ (use (26) and (28)).

We shall call $\langle{{U}},T\rangle$ a labelling system if $T\subseteq{{U}}\times\mathsf{At\,}{\mathfrak{{{A}}}}\times{{U}}$ and properties (i)–(v) hold. (This terminology was chosen because it is convenient to think of each triple $\langle u,a,v\rangle\in T$ as an ordered pair $\langle u,v\rangle$ which has the atom $a$ as a label.) Thus

[TABLE]

is a labelling system. If property (vi) also holds, then we say the labelling system $\langle{{U}},T\rangle$ is complete. If $\langle{{U}},T\rangle$ and $\langle{{U}}^{\prime},T^{\prime}\rangle$ are labelling systems, then $\langle{{U}}^{\prime},T^{\prime}\rangle$ extends $\langle{{U}},T\rangle$ if ${{U}}\subseteq{{U}}^{\prime}$ and $T\subseteq T^{\prime}$. We wish to show every labelling system can be extended to a complete labelling system. A flaw in a labelling system $\langle{{U}},T\rangle$ is a quintuple $\langle u,a,v,b,c\rangle$ where $a,b,c\in\mathsf{At\,}{\mathfrak{{{A}}}}$, $u,v\in{{U}}$, $\langle u,a,v\rangle\in T$, $a\leq b{\mskip 1.5mu;\mskip 1.5mu}c$, and for every $w\in{{U}}$ either $\langle u,b,w\rangle\notin T$ or $\langle w,c,v\rangle\notin T$. Thus $\langle{{U}},T\rangle$ is complete iff it has no flaws.

Lemma 1

If $\langle{{U}},T\rangle$ is a labelling system and $\langle u,a,v,b,c\rangle$ is a flaw in $\langle{{U}},T\rangle$, then there is a labelling system $\langle{{U}}^{\prime},T^{\prime}\rangle$ which extends $\langle{{U}},T\rangle$, such that $\langle u,a,v,b,c\rangle$ is not a flaw in $\langle{{U}}^{\prime},T^{\prime}\rangle$.

Proof

Choose any ${{w}}$ such that ${{w}}\notin{{U}}$. Set ${{U}}^{\prime}={{U}}\cup\{{{w}}\}$, and

[TABLE]

where ${{{b}}}^{\mathsf{r}}$ is defined in (34), and $\breve{{{b}}},\breve{{{c}}},{{{b}}}^{\mathsf{r}}\in\mathsf{At\,}{\mathfrak{{{A}}}}$ by (32) and (41). Thus $T^{\prime}\subseteq{{U}}^{\prime}\times\mathsf{At\,}{\mathfrak{{{A}}}}\times{{U}}^{\prime}$. It is easy to check that (i) and (iii) hold in $\langle{{U}}^{\prime},{{T}}^{\prime}\rangle$. (ii) holds for $\langle{{U}}^{\prime},{{T}}^{\prime}\rangle$ simply because it holds for $\langle{{U}},{{T}}\rangle$. (iv) is easily verified by using (7) and (28). It is obvious from the construction of ${{U}}^{\prime}$ and ${{T}}^{\prime}$ that $\langle{{u}},{{a}},{{v}},{{b}},{{c}}\rangle$ is not a flaw in $\langle{{U}}^{\prime},{{T}}^{\prime}\rangle$. It remains to check (v). Suppose ${{u}}_{0},{{u}}_{1},{{u}}_{2}\in{{U}}^{\prime}$, ${{x}}_{0},{{x}}_{1},{{x}}_{2}\in\mathsf{At\,}{\mathfrak{{{A}}}}$, and

[TABLE]

We wish to show ${{x}}_{1}{\mskip 1.5mu;\mskip 1.5mu}{{x}}_{0}\geq{{x}}_{2}$. If ${{w}}\notin\{{{u}}_{0},{{u}}_{1},{{u}}_{2}\}$, then $\{{{u}}_{0},{{u}}_{1},{{u}}_{2}\}\subseteq{{U}}$, so we get ${{x}}_{1}{\mskip 1.5mu;\mskip 1.5mu}{{x}}_{0}\geq{{x}}_{2}$ from the fact that (v) holds in $\langle{{U}},{{T}}\rangle$. So we may assume ${{w}}\in\{{{u}}_{0},{{u}}_{1},{{u}}_{2}\}.$ Suppose, for example, that ${{w}}={{u}}_{0}$. Then ($*$) implies that $\langle{{w}},{{x}}_{1},{{u}}_{2}\rangle,\langle{{w}},{{x}}_{2},{{u}}_{1}\rangle\in{{T}}^{\prime}.$ Now there are no triples in ${{T}}$ having ${{w}}$ as first term, so $\langle{{w}},{{x}}_{1},{{u}}_{2}\rangle,\langle{{w}},{{x}}_{2},{{u}}_{1}\rangle\in{{T}}^{\prime}\setminus{{T}},$ and hence ${{u}}_{1},{{u}}_{2}\in\{{{u}},{{v}},{{w}}\}.$ Arguing similarly in case ${{w}}={{u}}_{1}$ or ${{w}}={{u}}_{2}$, we conclude that $\{{{u}}_{0},{{u}}_{1},{{u}}_{2}\}\subseteq\{{{u}},{{v}},{{w}}\}.$ There are 19 ways of choosing ${{u}}_{0},{{u}}_{1},{{u}}_{2}$ so that ${{w}}\in\{{{u}}_{0},{{u}}_{1},{{u}}_{2}\}\subseteq\{{{u}},{{v}},{{w}}\}$, and these ways are listed in the first 3 columns of the table below.

The next 3 columns show which atoms must be assigned to ${{x}}_{0},{{x}}_{1},{{x}}_{2}$ in order that ($*$) holds. These entries are easily determined from the definition of ${{T}}^{\prime}$ when ${{w}}$ is involved. For example, when ${{u}}_{0}={{u}}$ and ${{u}}_{1}={{w}}$ in line 2, there is only one atom ${{x}}_{2}$ such that $\langle{{u}},{{x}}_{2},{{w}}\rangle\in{{T}}^{\prime}$, namely ${{x}}_{2}={{b}}$. When two of ${{u}}_{0},{{u}}_{1},{{u}}_{2}$ are either both equal to ${{u}}$, or both equal to ${{v}}$, then an additional argument is required. Consider, for example, the case ${{u}}={{u}}_{0}={{u}}_{1}$ (line 13). By ($*$), $\langle{{u}},{{x}}_{2},{{u}}\rangle=\langle{{u}}_{0},{{x}}_{2},{{u}}_{1}\rangle\in{{T}}$. From this and $\langle{{u}},{{a}},{{v}}\rangle\in T$ we have $\langle{{u}},{{x}}_{2},{{u}}\rangle,\langle{{u}},{{a}},{{v}}\rangle,\langle{{u}},{{a}},{{v}}\rangle\in{{T}}$, and hence ${{x}}_{2}{\mskip 1.5mu;\mskip 1.5mu}{{a}}\geq{{a}}$ by (v). Thus ${{x}}_{2}{\mskip 1.5mu;\mskip 1.5mu}{{a}}\neq 0$. We also have ${{x}}_{2}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ by (iii), and consequently ${{x}}_{2}={{{a}}}^{\mathsf{d}}$ by (34) and (43).

The seventh column contains the conclusions which must be proved in each of the 19 cases. Following the table are brief remarks on how each of the entries in column 7 can be derived.

[TABLE]

From the fact that $\langle{{u}},{{a}},{{v}},{{b}},{{c}}\rangle$ is a flaw in $\langle{{U}},{{T}}\rangle$ we get ${{a}}\leq{{b}}{\mskip 1.5mu;\mskip 1.5mu}{{c}}$, which is the conclusion needed for line 1. This is equivalent to $0\neq{{b}}{\mskip 1.5mu;\mskip 1.5mu}{{c}}\cdot{{a}}$ since ${{a}}$ is an atom. Applying (25), we get $0\neq{{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}\cdot{{b}}$, which is equivalent to ${{b}}\leq{{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}$ since $b$ is an atom, thus accounting for line 2. Each of lines 3 through 6 is obtained in a similar fashion from the preceding lines. Line 7 follows immediately from (36), line 8 from (36), (9), and (28), and line 9 from (34). Next we show that lines 10 through 18 can be equivalently transformed into formulas involving only a single atom. From lines 1, 2, and 6 we get $0\neq{{b}}{\mskip 1.5mu;\mskip 1.5mu}{{c}}$, $0\neq{{a}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{c}}}$, and $0\neq\breve{{{b}}}{\mskip 1.5mu;\mskip 1.5mu}{{a}}$. We apply (42) and (35), obtaining ${{{b}}}^{\mathsf{r}}={\breve{{{c}}}}^{\mathsf{r}}$, ${{{a}}}^{\mathsf{r}}={{{c}}}^{\mathsf{r}}$, and ${{{a}}}^{\mathsf{d}}={{{b}}}^{\mathsf{d}}$. Now substitute ${\breve{{{c}}}}^{\mathsf{r}}$ for ${{{b}}}^{\mathsf{r}}$ in lines 10 to 12, ${{{b}}}^{\mathsf{d}}$ for ${{{a}}}^{\mathsf{d}}$ in 13 to 15, and ${{{c}}}^{\mathsf{r}}$ for ${{{a}}}^{\mathsf{r}}$ in 16 to 18. Then lines 10 to 18 can be proved like lines 7 to 9. Finally we note that line 19 follows from (26).

This completes the verification of (v), so the lemma is proved.

Lemma 2

Let $\langle{{U}},{{T}}\rangle$ be a labelling system. Then there is a labelling system $\langle{{U}}^{\prime},{{T}}^{\prime}\rangle$, which extends $\langle{{U}},{{T}}\rangle$, such that there is no flaw $\langle{{u}},{{a}},{{v}},{{b}},{{c}}\rangle$ in $\langle{{U}}^{\prime},{{T}}^{\prime}\rangle$ with ${{u}},{{v}}\in{{U}}$.

Proof

Let $\langle\langle u_{\kappa},a_{\kappa},v_{\kappa},b_{\kappa},c_{\kappa}\rangle:\kappa<\alpha\rangle$ be an enumeration of the flaws in $\langle{{U}},T\rangle$. Set $\langle{{U}}_{0},T_{0}\rangle=\langle{{U}},T\rangle$. Assume the labelling system $\langle{{U}}_{\kappa},T_{\kappa}\rangle$ has been constructed. If $\langle u_{\kappa},a_{\kappa},v_{\kappa},b_{\kappa},c_{\kappa}\rangle$ is a flaw in $\langle{{U}}_{\kappa},T_{\kappa}\rangle$, let $\langle{{U}}_{\kappa+1},T_{\kappa+1}\rangle$ be a labelling system extending $\langle{{U}}_{\kappa},T_{\kappa}\rangle$ in which $\langle u_{\kappa},a_{\kappa},v_{\kappa},b_{\kappa},c_{\kappa}\rangle$ is not a flaw, by Lemma 1, and otherwise let $\langle{{U}}_{\kappa+1},T_{\kappa+1}\rangle=\langle{{U}}_{\kappa},T_{\kappa}\rangle$. If $\lambda\leq\alpha$ is a limit ordinal, then let

[TABLE]

Finally, set $\langle{{U}}^{\prime},T^{\prime}\rangle=\langle{{U}}_{\alpha},T_{\alpha}\rangle$. It is easy to see that $\langle{{U}}^{\prime},T^{\prime}\rangle$ is the desired labelling system.

Lemma 3

Every labelling system $\langle{{U}},T\rangle$ can be extended to a complete labelling system.

Proof

Set $\langle{{U}}_{0},T_{0}\rangle=\langle{{U}},T\rangle$, and for each $n<\omega$ ($\omega$ is the least infinite ordinal), let $\langle{{U}}_{n+1},T_{n+1}\rangle$ be the labelling system obtained by applying Lemma 2 to the previously constructed labelling system $\langle{{U}}_{n},T_{n}\rangle$. Then

[TABLE]

is a complete labelling system extending $\langle{{U}},{{T}}\rangle$.

According to the remarks preceding Lemma 1, the proof of Theorem 5.1 is complete.

On the beach in 1984, I encouraged Hajnal and István to read the proof just presented, rather than the considerably different method used in MR662049 to prove the following corollary.

Corollary 2** ((MR662049, , Theorem 5.20)**)

Let ${{K}}$ be the class of algebras isomorphic to \mathfrak{Rl}\mskip 1.5mu_{{E}}\big{(}\mathfrak{Re}({{U}})\big{)} for some set ${{U}}$ and some symmetric and reflexive relation ${{E}}\subseteq{{U}}\times{{U}}$. Then $\mathsf{WA}$ is the closure of ${{K}}$ under the formation of subalgebras: $\mathsf{WA}=\mathbf{S}{{K}}.$

Proof

${{K}}\subseteq\mathsf{WA}$ by Corollary 1 and $\mathsf{WA}\subseteq\mathbf{S}{{K}}$ by Theorem 5.1. We have $\mathbf{S}\mathsf{WA}=\mathsf{WA}$ since $\mathsf{WA}$ has equational axiomatizations. Therefore $\mathsf{WA}\subseteq\mathbf{S}{{K}}\subseteq\mathbf{S}\mathsf{WA}=\mathsf{WA}$.

6 The Relativized Cylindric Set Algebra of a Suitable Structure

By Theorem 5.1, every complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ is a subalgebra of a relativized set relation algebra. This result can be sharpened. By Theorem 11.1 in §11, every complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ is already itself a relativized set relation algebra, not just a subalgebra of one. For the proof we will employ suitable structures and their canonical relativized cylindric set algebras from MR987611 . According to (MR987611, , Definition 1), $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<\alpha}$ is a suitable structure if ${{B}}$ is a set, $\alpha$ is a non-zero ordinal, and, for all $\kappa,\lambda,\mu<\alpha$,

1. (i)

${{T}}_{\kappa}\subseteq{{B}}\times{{B}}$, ${{E}}_{\kappa\lambda}\subseteq{{B}}$, 2. (ii)

${{T}}_{\kappa}$ is an equivalence relation on ${{B}}$, 3. (iii)

${{E}}_{\kappa\kappa}={{B}}$, 4. (iv)

${{E}}_{\kappa\lambda}={{T}}_{\mu}^{*}({{E}}_{\kappa\mu}\cap{{E}}_{\mu\lambda})$ whenever $\kappa,\lambda\neq\mu$, 5. (v)

${{T}}_{\kappa}\cap({{E}}_{\kappa\lambda}\times{{E}}_{\kappa\lambda})\subseteq Id$ whenever $\kappa\neq\lambda$.

These are exactly the conditions in (HMT71, , 2.7.40), except the requirement that ${{T}}_{\kappa}$ and ${{T}}_{\lambda}$ commute is deleted. Assume that $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<\alpha}$ is a suitable structure. Let $Tr(\mathfrak{{{B}}})$ be the set of all sequences ${{p}}=\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{n}},\kappa_{{n}}\rangle$ such that ${{n}}\in\omega$, ${{t}}_{0},\dots,{{t}}_{{n}}\in{{B}}$, $\kappa_{0},\dots,\kappa_{{n}}<\alpha$, and, for all ${{i}}<{{n}}$, ${{t}}_{{i}}\neq{{t}}_{{{i}}+1}$ and ${{t}}_{{i}}\,{{T}}_{\kappa_{i}}\,{{t}}_{{{i}}+1}$. If ${{p}}\in Tr(\mathfrak{{{B}}})$ then ${{p}}$ is called a $\mathfrak{{{B}}}$-trail, ${{p}}$ begins at ${{t}}_{0}$, ${{p}}$ ends at ${{t}}_{{n}}$, $\kappa_{{n}}$ is the pointer of ${{p}}$, ${{p}}$ has length $|{{p}}|={{n}}+1$, and ${{p}}$ is reduced if the following conditions hold:

1. (i)

if $1=|{{p}}|$ and ${{t}}_{0}\in{{E}}_{\kappa_{0}\lambda}$, then $\kappa_{0}\leq\lambda<\alpha$, 2. (ii)

if $1<|{{p}}|$ then $\kappa_{{{n}}-1}=\kappa_{{n}}$ and for all $\lambda<\alpha$, ${{t}}_{{n}}\in{{E}}_{\kappa_{n}\lambda}$ iff $\kappa_{{n}}=\lambda$, 3. (iii)

if $0\leq{{i}}<|{{p}}|-2$, then either ${{t}}_{{i}}\neq{{t}}_{{{i}}+2}$ or $\kappa_{{i}}\neq\kappa_{{{i}}+1}$.

For every $\lambda<\alpha$, let ${{p}}\lambda=\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{{n}}-1},\kappa_{{{n}}-1},{{t}}_{{n}},\lambda\rangle.$ Let $\approx$ be the smallest equivalence relation on $Tr(\mathfrak{{{B}}})$ such that

1. (i)

$\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{i},\lambda,{{s}},\lambda,{{t}}_{{i}},\kappa_{{i}},\dots,{{t}}_{{n}},\kappa_{{n}}\rangle\approx\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{i}},\kappa_{{i}},\dots,{{t}}_{{n}},\kappa_{{n}}\rangle$

where $0\leq{{i}}\leq{{n}}$, 2. (ii)

$\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{n}},\lambda,{{s}},\kappa_{{n}}\rangle\approx\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{n}},\kappa_{{n}}\rangle$ where $\lambda\neq\kappa_{{n}}$, 3. (iii)

$\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{n}},\lambda\rangle\approx\langle{{t}}_{0},\kappa_{0},\dots,{{t}}_{{n}},\kappa_{{n}}\rangle$ where ${{t}}_{{n}}\in{{E}}_{\lambda\kappa_{{n}}}$.

For each ${{p}}\in Tr(\mathfrak{{{B}}})$, let ${{p}}^{\mathfrak{{{B}}}}$ be the $\approx$-equivalence class of ${{p}}$, i.e., ${{p}}^{\mathfrak{{{B}}}}=\left\{\,{{p}}^{\prime}:{{p}}\,\approx\,{{p}}^{\prime}\,\right\}.$ The conditions defining $\approx$ are called “reductions”, e.g., a reduction of type (i) consists of the replacement of any subsequence of the form $\langle{{t}},\lambda,{{s}},\lambda,{{t}}\rangle$ by $\langle{{t}}\rangle$. Equivalent trails have the same beginnings, but may have different ends and different pointers, due to reductions of types (ii) and (iii). By (MR987611, , Lemma 6), there is a function called ${{P}}_{7}$ such that ${{P}}_{7}({{p}})$ is the unique reduced trail in ${{p}}^{\mathfrak{{{B}}}}$. Suppose ${{q}}\in{{p}}^{\mathfrak{{{B}}}}$ and ${{q}}$ is reduced. Then $|{{q}}|\leq|{{p}}|$, and if $|{{p}}|=1$, either ${{p}}$ is already reduced, or else ${{q}}={{p}}\kappa$ for some ordinal $\kappa$ which is strictly smaller than the pointer of ${{p}}$. Let ${{U}}(\mathfrak{{{B}}})=\left\{\,{{p}}^{\mathfrak{{{B}}}}:{{p}}\in Tr(\mathfrak{{{B}}})\,\right\}.$ For every point ${{u}}\in{{U}}(\mathfrak{{{B}}})$, let $|{{u}}|$ be the length of the unique reduced trail in ${{u}}$. For every $\mathfrak{{{B}}}$-trail ${{p}}=\langle{{t}}_{0},\kappa_{0},{{t}}_{1},\kappa_{1},\dots,{{t}}_{{{n}}-1},\kappa_{{{n}}-1},{{t}}_{{n}},\kappa_{{n}}\rangle$ let $\breve{{{p}}}=\langle{{t}}_{n},\kappa_{n-1},{{t}}_{n-1},\kappa_{n-2},\dots,{{t}}_{1},\kappa_{0},{{t}}_{0},\kappa_{n}\rangle.$ If ${{q}}=\langle{{s}}_{0},\lambda_{0},\dots,{{s}}_{m},\lambda_{m}\rangle$ is any other $\mathfrak{{{B}}}$-trail, then ${{p}}\odot{{q}}$ is defined iff ${{p}}$ ends where ${{q}}$ begins, in which case ${{p}}\odot{{q}}=\langle{{t}}_{0},\kappa_{0},{{t}}_{1},\kappa_{1},\dots,{{t}}_{{{n}}-1},\kappa_{{{n}}-1},{{s}}_{0},\lambda_{0},\dots,{{s}}_{m},\lambda_{m}\rangle.$ Also,

[TABLE]

and, for any $X\subseteq Tr(\mathfrak{{{B}}})$, $\ell_{{p}}({{X}})=\bigcup_{{{q}}\in{{X}}}\left({{L}}_{{p}}({{q}})\right)^{\mathfrak{{{B}}}}.$ Let

[TABLE]

By (MR987611, , Lemma 11), $Pm(\mathfrak{{{B}}})$ is a group of permutations of ${{U}}(\mathfrak{{{B}}})$, and the inverse of the permutation $\ell_{{p}}\in Pm(\mathfrak{{{B}}})$ is $\ell_{\breve{{{p}}}}$. For every ${{t}}\in{{B}}$, define a set of sequences of length $\alpha$, by

[TABLE]

and set ${{V}}(\mathfrak{{{B}}})=\bigcup_{{{t}}\in{{B}}}R^{\mathfrak{{{B}}}}_{{{t}}}.$ By (HMT85, , 3.1.2(i)), $\mathfrak{Sb}\mskip 1.5mu({}^{\alpha}{{U}}(\mathfrak{{{B}}}))$ is the full $\alpha$-dimensional cylindric set algebra of all $\alpha$-ary relations on ${{U}}(\mathfrak{{{B}}})$. If ${{t}}\in{{B}}$ then, by (MR987611, , Lemma 12(i)), $Pm(\mathfrak{{{B}}})$ preserves $R^{\mathfrak{{{B}}}}_{{{t}}}$ and acts transitively on $R^{\mathfrak{{{B}}}}_{{{t}}}$, hence $R^{\mathfrak{{{B}}}}_{{{t}}}$ is the orbit, under the action of the group $Pm(\mathfrak{{{B}}})$, of any single $\alpha$-sequence in $R^{\mathfrak{{{B}}}}_{{{t}}}$. It follows, by (MR987611, , Lemma 9), that $R^{\mathfrak{{{B}}}}_{{{t}}}$ is an atom of the subalgebra $\mathfrak{{{C}}}$ of $\mathfrak{Sb}\mskip 1.5mu({}^{\alpha}{{U}}(\mathfrak{{{B}}}))$ completely generated by $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$. This complete subalgebra $\mathfrak{{{C}}}$ is called the canonical cylindric set algebra of the suitable structure $\mathfrak{{{B}}}$. Its set of atoms includes $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$, but is typically much larger. By relativizing (in the cylindric algebraic sense of (HMT71, , 2.2.1)) to the union ${{V}}(\mathfrak{{{B}}})$ of this set of atoms, one gets $\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}=\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$, the canonical relativized cylindric set algebra (MR987611, , Definition 7). Its set of atoms is exactly $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$ by (MR987611, , Lemma 12(ii)).

7 The Suitable Structure of a $\mathsf{WA}$

We show next that every complete atomic weakly associative relation algebra $\mathfrak{{{A}}}$ has a suitable structure $\mathfrak{{{B}}}$ that is 3-dimensional.

Lemma 4

Given an atomic $\mathfrak{{{A}}}\in\mathsf{WA}$, let ${{B}}=\{{{s}}:{{s}}\in{}^{3}\mathsf{At\,}{\mathfrak{{{A}}}},\,{{s}}_{2}{\mskip 1.5mu;\mskip 1.5mu}{{s}}_{0}\geq{{s}}_{1}\}$ and, assuming $\{\kappa,\lambda,\mu\}=\{0,1,2\}$, ${{T}}_{\kappa}=\{\langle{{s}},{{t}}\rangle:{{s}}_{\kappa}={{t}}_{\kappa}\}$, ${{E}}_{\kappa\kappa}={{B}}$, and ${{E}}_{\kappa\lambda}={{E}}_{\lambda\kappa}=\{{{s}}:{{s}}\in{{B}},\,{{s}}_{\mu}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\}$. Then $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<3}$ is a suitable structure.

Proof

The first three properties (i)–(iii) required of suitable structures clearly hold. We need only show (iv) and (v). First, observe that ${{B}}$ is the union of cycles, sets of the form

[TABLE]

where ${{a}},{{b}},{{c}}\in\mathsf{At\,}{\mathfrak{{{A}}}}$. To see this, note that if $\langle{{b}},{{c}},{{a}}\rangle\in{{B}}$, then $\breve{{{a}}},\breve{{{b}}},\breve{{{c}}}\in\mathsf{At\,}{\mathfrak{{{A}}}}$ by (32), and ${{a}}{\mskip 1.5mu;\mskip 1.5mu}{{b}}\geq{{c}}$, so by (33),

[TABLE]

hence ${{C}}\subseteq{{B}}$. Because of axiom (7), this shows that ${{C}}\subseteq{{B}}$ whenever any of the listed triples of ${{C}}$ is in ${{B}}$. ${{C}}$ is a diversity cycle if ${{a}},{{b}},{{c}},\breve{{{a}}},\breve{{{b}}},\breve{{{c}}}\leq{0\kern-0.39993pt\raise 5.59721pt\hbox{,}\kern 0.39993pt}$, and an identity cycle otherwise.

If $\langle{{b}},{{c}},{{a}}\rangle\in{{B}}$ and any two of $\{{{a}},{{b}},{{c}}\}$ are below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, then so is the third, and ${{a}}={{b}}={{c}}$. (Their converses are also below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, so by one of the inequalities above the third element is below the product of two elements below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, and hence is itself below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$.)

Assume $\langle{{b}},{{c}},{{a}}\rangle\in{{B}}$ and one of $\{{{a}},{{b}},{{c}}\}$ is below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$. From the inequalities above we can use (28), (41), (43), and definition (34) of ${{{x}}}^{\mathsf{d}}$ and ${{{x}}}^{\mathsf{r}}$ to deduce that

[TABLE]

hence

[TABLE]

Proof of property (iv). Assume ${{t}}\in{{T}}_{\mu}^{*}(E_{\kappa\mu}\cap E_{\mu\lambda})$ in the case where $\{\kappa,\lambda,\mu\}=\{0,1,2\}$. Then, for some ${{s}}\in{{B}}$, ${{t}}\mathrel{{{T}}_{\mu}}{{s}}$ and ${{s}}\in{{E}}_{\kappa\mu}\cap{{E}}_{\mu\lambda}$, which gives us ${{t}}_{\mu}={{s}}_{\mu}$, ${{s}}_{\lambda}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, and ${{s}}_{\kappa}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$. Since two elements of ${{s}}$ are below ${1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, the third one is as well, i.e., ${{s}}_{\mu}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, hence ${{t}}\in{{E}}_{\kappa\mu}$ since ${{t}}_{\mu}={{s}}_{\mu}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$. This proves the equality in one direction. For the other, assume ${{t}}\in{{E}}_{\kappa\mu}$, so ${{t}}_{\lambda}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$. We want some ${{s}}\in{{E}}_{\kappa\mu}\cap{{E}}_{\mu\lambda}$ such that ${{s}}\mathrel{{{T}}_{\mu}}{{t}}$, i.e., ${{t}}_{\mu}={{s}}_{\mu}$, ${{s}}_{\lambda}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, and ${{s}}_{\kappa}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$. It suffices to let ${{s}}=\langle{{t}}_{\lambda},{{t}}_{\lambda},{{t}}_{\lambda}\rangle$.

Since ${{E}}_{\kappa\mu}={{E}}_{\mu\kappa}$ and ${{E}}_{\kappa\kappa}={{B}}$, the remaining case of (iv) is ${{B}}=T_{\mu}^{*}E_{\kappa\mu}$ whenever $\kappa\neq\mu$. The inclusion from right to left is trivially true. Let ${{t}}=\langle{{b}},{{c}},{{a}}\rangle\in{{B}}$. Depending on the values of $\kappa$ and $\mu$, the following list contains an ${{s}}\in{{E}}_{\kappa\mu}$ such that ${{s}}\mathrel{{{T}}_{\mu}}{{t}}$:

[TABLE]

Proof of property (v). Assume $\langle{{s}},{{t}}\rangle\in{{T}}_{\kappa}\cap({{E}}_{\kappa\lambda}\times{{E}}_{\kappa\lambda})$ where $\kappa\neq\lambda$. We wish to show ${{s}}={{t}}$. From the hypothesis we get ${{s}}\mathrel{{{T}}_{\kappa}}{{t}}$ and ${{s}},{{t}}\in{{E}}_{\kappa\lambda}$, i.e., ${{s}}_{\kappa}={{t}}_{\kappa}$, ${{s}}_{\mu}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, and ${{t}}_{\mu}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, where $\{0,1,2\}=\{\kappa,\lambda,\mu\}$. Since ${{s}}$ and ${{t}}$ are triples with a subidentity element in the same position $\mu$, they must have the same form according to ($**$), and since they also have the same element at a different position $\kappa$, they are the same. For example, if $\mu=2$, then ${{s}}=\langle{{s}}_{0},{{s}}_{0},{{{s}}_{0}}^{\mathsf{d}}\rangle$ and ${{t}}=\langle{{t}}_{0},{{t}}_{0},{{{t}}_{0}}^{\mathsf{d}}\rangle$. But ${{s}}_{0}={{t}}_{0}$ follows from ${{s}}\mathrel{{{T}}_{\kappa}}{{t}}$ if $\kappa$ is either 0 or 1, so ${{s}}={{t}}$. If $\mu=1$ then ${{s}}=\langle\breve{{{s}}_{2}},{{{s}}_{2}}^{\mathsf{d}},{{{s}}_{2}}\rangle$ and ${{t}}=\langle\breve{{{t}}_{2}},{{{t}}_{2}}^{\mathsf{d}},{{{t}}_{2}}\rangle$. We get ${{s}}_{2}={{t}}_{2}$ from ${{s}}\mathrel{{{T}}_{2}}{{t}}$ and $\breve{{{s}}_{2}}=\breve{{{t}}_{2}}$ from ${{s}}\mathrel{{{T}}_{0}}{{t}}$, and again ${{s}}={{t}}$. If $\mu=0$ then ${{s}}=\langle{{{s}}_{2}}^{\mathsf{r}},{{s}}_{2},{{s}}_{2}\rangle$ and ${{t}}=\langle{{{t}}_{2}}^{\mathsf{r}},{{t}}_{2},{{t}}_{2}\rangle$, so ${{s}}_{2}={{t}}_{2}$ follows from ${{s}}\mathrel{{{T}}_{\kappa}}{{t}}$ if $\kappa$ is either 1 or 2, and again ${{s}}={{t}}$.

8 The Complex Algebra of a Suitable Structure

If $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<3}$ is a suitable structure, then the complex algebra of $\mathfrak{{{B}}}$ (HMT71, , 2.7.33) is

[TABLE]

where, for every ${{X}}\subseteq{{B}}$, ${{T}}^{*}_{{i}}({{X}})=\{{{y}}:{{y}}{{T}}_{{i}}{{x}}\in{{X}}\}$. The class $\mathsf{NCA}_{\alpha}$ is defined by the axioms for $\alpha$-dimensional cylindric algebras $\mathsf{CA}_{\alpha}$ with the commutativity of cylindrifications deleted, that is, by the axioms (C0)–(C3), (C5)–(C7) of (HMT71, , 1.1.1). The class $\mathsf{NA}_{\alpha}$ of non-commutative cylindric algebras of dimension $\alpha$ is defined by the axioms of $\mathsf{CA}_{\alpha}$ with postulate (C4) of (HMT71, , 1.1.1), $\mathsf{c}_{\kappa}\mathsf{c}_{\lambda}x=\mathsf{c}_{\lambda}\mathsf{c}_{\kappa}x$, replaced by the weaker postulate (C${}_{4}^{*}$), $\mathsf{c}_{\kappa}\mathsf{c}_{\lambda}x\geq\mathsf{c}_{\lambda}\mathsf{c}_{\kappa}x\cdot\mathsf{d}_{\lambda\mu}$ with $\mu\neq\kappa,\lambda$ (see AT , N86 , and T ). The next lemma is a sharpening of one direction of (MR662049, , Lemma 15). It is about $\mathsf{NA}_{3}$ instead of the wider class $\mathsf{NCA}_{3}$.

Lemma 5

Assume $\mathfrak{{{A}}}\in\mathsf{WA}$, $\mathfrak{{{A}}}$ is atomic, and $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<3}$ is the suitable structure built from $\mathfrak{{{A}}}$ in Lemma 4. Then $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\in\mathsf{NA}_{3}$.

Proof

By (MR987611, , Lemma 15), $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\in\mathsf{NCA}_{3}$, so we need only show $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ satisfies (C${}_{4}^{*}$). In the notation of complex algebras, (C${}_{4}^{*}$) says that if ${{X}}\subseteq{{B}}$ and $\mu\neq\kappa,\lambda$, then ${{T}}^{*}_{\lambda}{{T}}^{*}_{\kappa}{{X}}\cap{{E}}_{\lambda\mu}\subseteq{{T}}^{*}_{\kappa}{{T}}^{*}_{\lambda}{{X}}.$ Since this is trivially true if $\kappa=\lambda$, we assume $\kappa\neq\lambda$, hence $\{0,1,2\}=\{\kappa,\lambda,\mu\}$. Suppose ${{t}}\in{{B}}$ and ${{t}}=\langle{{b}},{{c}},{{a}}\rangle\in{{T}}^{*}_{\lambda}{{T}}^{*}_{\kappa}{{X}}\cap{{E}}_{\lambda\mu}.$ Then ${{t}}_{\kappa}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ since ${{t}}\in{{E}}_{\lambda\mu},$ so, by ($**$),

[TABLE]

From ${{t}}\in{{T}}^{*}_{\lambda}{{T}}^{*}_{\kappa}{{X}}$ we know there are ${{s}},{{r}}\in{{B}}$ such that ${{t}}\mathrel{{{T}}_{\lambda}}{{s}}\mathrel{{{T}}_{\kappa}}{{r}}\in{{X}}.$ We wish to find, in each of the six cases, some ${{s}}^{\prime}\in{{B}}$ such that ${{t}}\mathrel{{{T}}_{\kappa}}{{s}}^{\prime}\mathrel{{{T}}_{\lambda}}{{r}}\in{{X}}.$ The following table shows how to compute ${{s}}^{\prime}$ from the values of $\kappa,\lambda,\mu$. It is immediately obvious from the definition of ${{s}}^{\prime}$ that ${{s}}^{\prime}\mathrel{{{T}}_{\lambda}}{{r}}$, while ${{t}}\mathrel{{{T}}_{\lambda}}{{s}}^{\prime}$ follows from the reason given in the last column, deduced for each case right after the table.

[TABLE]

From ${{s}},{{r}}\in{{B}}$ it follows, by (33), (35), and (42), that statements (a)–(f) hold: (a) ${{{s}}_{1}}^{\mathsf{d}}={{{s}}_{2}}^{\mathsf{d}}$, (b) ${{{s}}_{2}}^{\mathsf{r}}={{{s}}_{0}}^{\mathsf{d}}$, (c) ${{{s}}_{0}}^{\mathsf{r}}={{{s}}_{1}}^{\mathsf{r}}$, (d) ${{{r}}_{1}}^{\mathsf{d}}={{{r}}_{2}}^{\mathsf{d}}$, (e) ${{{r}}_{2}}^{\mathsf{r}}={{{r}}_{0}}^{\mathsf{d}}$, and (f) ${{{r}}_{0}}^{\mathsf{r}}={{{r}}_{1}}^{\mathsf{r}}$. Each equation in the last column of the table above may be deduced as follows, using various combinations of statements (a)–(f) with the hypotheses ${{t}}\mathrel{{{T}}_{\lambda}}{{s}}\mathrel{{{T}}_{\kappa}}{{r}}$.

[TABLE]

Let $2\leq{{n}}<\omega$. MGRn (HMT85, , 3.2.88(1)) is the set of ${{n}}$-ary merry-go-round identities, where $\lambda,\kappa_{1},\dots,\kappa_{{n}}<\alpha$ are distinct ordinals:

[TABLE]

In particular, MGR2 (HMT85, , 3.2.88(2)) is the set of $2$-ary merry-go-round identities:

[TABLE]

and MGR3 (HMT85, , 3.2.88(3)) is the set of $3$-ary merry-go-round identities:

[TABLE]

Lemma 6

If $\mathfrak{{{B}}}$ is the suitable structure of an atomic $\mathfrak{{{A}}}\in\mathsf{WA}$, and ${{t}}\in{{B}}$, then

1. (i)

${{E}}_{10}\cap{{T}}^{*}_{0}\{{{t}}\}=\{\langle{{t}}_{0},{{t}}_{0},{{{t}}_{0}}^{\mathsf{d}}\rangle\}$, 2. (ii)

${{E}}_{01}\cap{{T}}^{*}_{1}\{{{t}}\}=\{\langle{{t}}_{1},{{t}}_{1},{{{t}}_{1}}^{\mathsf{d}}\rangle\}$, 3. (iii)

${{E}}_{12}\cap{{T}}^{*}_{2}\{{{t}}\}=\{\langle{{{t}}_{2}}^{\mathsf{r}},{{t}}_{2},{{t}}_{2}\rangle\}$, 4. (iv)

${{E}}_{21}\cap{{T}}^{*}_{1}\{{{t}}\}=\{\langle{{{t}}_{1}}^{\mathsf{r}},{{t}}_{1},{{t}}_{1}\rangle\}$, 5. (v)

${{E}}_{02}\cap{{T}}^{*}_{2}\{{{t}}\}=\{\langle\breve{{{t}}_{2}},{{{t}}_{2}}^{\mathsf{d}},{{t}}_{2}\rangle\}$, 6. (vi)

${{E}}_{20}\cap{{T}}^{*}_{0}\{{{t}}\}=\{\langle{{t}}_{0},{{{t}}_{0}}^{\mathsf{r}},\breve{{{t}}_{0}}\rangle\}$.

Proof

The inclusions from right to left follow immediately from definitions. The opposite inclusions have very similar proofs, so we do only the first one. Assume ${{s}}\in{{E}}_{10}\cap{{T}}^{*}_{0}\{{{t}}\}$. Then ${{s}}_{2}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ since ${{s}}\in{{E}}_{10}$, so ${{s}}=\langle{{s}}_{0},{{s}}_{0},{{{s}}_{0}}^{\mathsf{d}}\rangle$ by ($**$). We also have ${{s}}\mathrel{{{T}}_{0}}{{t}}$ since ${{s}}\in{{T}}^{*}_{0}\{{{t}}\}$, hence ${{s}}_{0}={{t}}_{0}$, yielding ${{s}}=\langle{{s}}_{0},{{s}}_{0},{{{s}}_{0}}^{\mathsf{d}}\rangle=\langle{{t}}_{0},{{t}}_{0},{{{t}}_{0}}^{\mathsf{d}}\rangle$, as desired.

Lemma 7

If $\mathfrak{{{B}}}$ is the suitable structure of an atomic $\mathfrak{{{A}}}\in\mathsf{WA}$ then $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ satisfies MGRn* for all ${{n}}$.*

Proof

MGRn is satisfied in all 3-dimensional algebras whenever $3\leq{{n}}$, because the index ordinals in MGRn are required to be distinct and dimension 3 is not large enough to find four (or more) distinct ordinals. Since $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\in\mathsf{NA}_{3}$ by Lemma 5, we need only prove MGR2. In the notation of complex algebras, MGR2 says that for every ${{X}}\subseteq{{B}}$, if $\{\kappa,\lambda,\mu\}=\{0,1,2\}$ then ${{T}}^{*}_{\kappa}({{E}}_{\kappa\lambda}\cap{{T}}^{*}_{\lambda}({{E}}_{\lambda\mu}\cap{{T}}^{*}_{\mu}({{E}}_{\mu\kappa}\cap{{T}}^{*}_{\kappa}{{X}})))={{T}}^{*}_{\kappa}({{E}}_{\kappa\mu}\cap{{T}}^{*}_{\mu}({{E}}_{\mu\lambda}\cap{{T}}^{*}_{\lambda}({{E}}_{\lambda\kappa}\cap{{T}}^{*}_{\kappa}{{X}}))).$ Here we compute both sides of a single instance of MGR2, showing that they evaluate to the same thing. (The first computation will also be used in the proof of Theorem 9.1.) The remaining cases can be obtained from this one by permuting subscripts and making other appropriate changes. First we evaluate the MGR2 term that comes from converse.

[TABLE]

Next we compute the other side of the MGR2 identity connected with converse. This term can serve as an alternative definition of converse.

[TABLE]

Theorem 8.1

Suppose $\mathfrak{{{B}}}$ is the suitable structure of an atomic $\mathfrak{{{A}}}\in\mathsf{WA}$. Then $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\cong\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}},$ via the isomorphism $R^{\mathfrak{{{B}}}}$ from $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ to $\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}$ defined for ${{X}}\subseteq{{B}}$ by

[TABLE]

Proof

(MR987611, , Theorem C) says that if $2\leq\alpha$, $\mathfrak{{{C}}}\in\mathsf{NA}_{\alpha}$, $\mathfrak{{{C}}}$ is complete, atomic, and satisfies MGR2 and MGR3, then $\mathfrak{{{C}}}\cong\mathfrak{Rc}\mskip 1.5mu\mathfrak{At}\mathfrak{{{C}}}$, where $\mathfrak{At}\mathfrak{{{C}}}$ is the atom structure of $\mathfrak{{{C}}}$ (HMT71, , 2.7.32). This theorem applies to $\mathfrak{{{C}}}=\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$, because $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ is complete, atomic, in $\mathsf{NA}_{3}$ by Lemma 5, and satisfies the MGR identities by Lemma 7, so $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\cong\mathfrak{Rc}\mskip 1.5mu\mathfrak{At}\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ via the isomorphism of (MR987611, , Definition 16). But $\mathfrak{At}\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\cong\mathfrak{{{B}}}$ by (HMT71, , 2.7.35), via the isomorphism that sends $\{{{x}}\}$ to ${{x}}$. Therefore $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\cong\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}$ via the formula above for the composed isomorphisms.

9 Relation-Algebraic Reducts

Consider an arbitrary non-commutative 3-dimensional cylindric algebra

[TABLE]

Following (HMT71, , 2.6.28), restricted to the case $\alpha=3$, let

[TABLE]

$Nr_{2}\mathfrak{{{C}}}$ is the set of $2$-dimensional elements of $\mathfrak{{{C}}}$. By (HMT85, , 5.3.7) (generalized from $\mathsf{CA}_{\alpha}$ to $\mathsf{NA}_{\alpha}$), the binary operation ${\mskip 1.5mu;\mskip 1.5mu}$ is defined for all ${{x}},{{y}}\in{{C}}$ by

[TABLE]

and the unary operation $\breve{\phantom{x}}$ is defined for every ${{x}}\in{{C}}$ by

[TABLE]

We have $\mathsf{c}_{2}\mathsf{c}_{2}{{x}}=\mathsf{c}_{2}{{x}}$ for all ${{x}}\in{{C}}$ by (HMT71, , 1.2.3). This implies the set of 2-dimensional elements is closed under the operations ${\mskip 1.5mu;\mskip 1.5mu}$ and $\breve{\phantom{x}}$. $Nr_{2}\mathfrak{{{C}}}$ is closed under $+$ because $\mathsf{c}_{2}$ distributes over $+$ by (HMT71, , 1.2.6), and $\mathsf{d}_{01}$ is a 2-dimensional element because $\mathsf{c}_{2}\mathsf{d}_{01}=\mathsf{d}_{01}$ by (HMT71, , 1.3.3). The complement $\overline{{{x}}}$ of ${{x}}\in Nr_{2}\mathfrak{{{C}}}$ is also 2-dimensional by (HMT71, , 1.2.12). We therefore have the algebra (HMT85, , 5.3.7)

[TABLE]

Theorem 9.1

If $\mathfrak{{{B}}}=\langle{{B}},{{T}}_{\kappa},{{E}}_{\kappa\lambda}\rangle_{\kappa,\lambda<3}$ is the suitable structure of a complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$, then $\mathfrak{{{A}}}\cong\mathfrak{Ra}\mskip 1.5mu\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ via the isomorphism $\varphi:{{A}}\to\mathcal{P}({{B}})$ defined by $\varphi({{x}})=\{{{t}}:{{t}}\in{{B}},\,{{t}}_{2}\leq{{x}}\}$ for all ${{x}}\in{{A}}$.

Proof

Note that ${{T}}^{*}_{2}\varphi({{x}})\subseteq\varphi({{x}})$ since ${{T}}_{2}$ is an equivalence relation, and if ${{s}}\in{{T}}^{*}_{2}\varphi({{x}})$ then, by the definition of $\varphi$, there is some ${{t}}\in{{B}}$ such that ${{s}}_{2}={{t}}_{2}$ and ${{t}}_{2}\leq{{x}}$, hence ${{s}}_{2}\leq{{x}}$, i.e., ${{s}}\in\varphi({{x}})$. This proves that ${{T}}^{*}_{2}\varphi({{x}})=\varphi({{x}})$ for every ${{x}}\in{{A}}$, so, in fact, $\varphi:{{A}}\to Nr_{2}\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}},$ and $\varphi$ is a Boolean homomorphism because

[TABLE]

Converse was handled earlier—the first computation in the proof of Lemma 7 happens to also show ${\left(\{{{t}}\}\right)}\breve{\ }=\varphi(\breve{{{t}}_{2}})$. If ${{X}},{{Y}}\in Nr_{2}\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ then

[TABLE]

so

[TABLE]

For the inclusion in the other direction, suppose ${{t}}\in\varphi({{x}}{\mskip 1.5mu;\mskip 1.5mu}{{y}})$, i.e., ${{t}}\in{{B}}$ and ${{x}}{\mskip 1.5mu;\mskip 1.5mu}{{y}}\geq{{t}}_{2}$. By complete additivity (31) there are atoms ${{a}},{{b}}\in\mathsf{At\,}{\mathfrak{{{A}}}}$ such that ${{t}}_{2}\leq{{a}}{\mskip 1.5mu;\mskip 1.5mu}{{b}}$, ${{a}}\leq{{x}}$, and ${{b}}\leq{{y}}$. Let ${{r}}=\langle{{{a}}}^{\mathsf{r}},{{a}},{{a}}\rangle$ and ${{q}}=\langle{{{b}}}^{\mathsf{r}},{{b}},{{b}}\rangle$ and note that ${{r}},{{q}}\in{{B}}$. Then ${{r}}_{2}={{a}}\leq{{x}}$ and ${{q}}_{2}\leq{{b}}\leq{{y}}$, so ${{t}}\in\varphi({{x}}){\mskip 1.5mu;\mskip 1.5mu}\varphi({{y}})$ by the first part of the computation.

10 Cylindric-Relativized Representation

Theorem 10.1

Assume $\mathfrak{{{A}}}$ is a complete atomic $\mathsf{WA}$, $\mathfrak{{{B}}}$ is the suitable structure of $\mathfrak{{{A}}}$, and for every ${{t}}\in{{B}}$,

[TABLE]

Assume $\mathfrak{{{C}}}$ is the subalgebra of the full 3-dimensional cylindric set algebra $\mathfrak{Sb}\mskip 1.5mu\left({}^{3}{{U}}(\mathfrak{{{B}}})\right)$ completely generated by $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$. Then

[TABLE]

and $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$ is the set of atoms of $\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$.

Proof

Recall from §6, $\left\{\,R^{\mathfrak{{{B}}}}_{{{t}}}:{{t}}\in{{B}}\,\right\}$ is the set of atoms of $\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$ by (MR987611, , Lemma 12(ii)). Relativizing $\mathfrak{{{C}}}$ to ${{V}}(\mathfrak{{{B}}})=\bigcup_{{{t}}\in{{B}}}R^{\mathfrak{{{B}}}}_{{{t}}}$ gives the canonical relativized cylindric set algebra $\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}$,

[TABLE]

We have $\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}\cong\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}$ by Theorem 8.1, and $\mathfrak{{{A}}}\cong\mathfrak{Ra}\mskip 1.5mu\mathfrak{Cm}\mskip 1.5mu\mathfrak{{{B}}}$ by Theorem 9.1, so $\mathfrak{{{A}}}\cong\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rc}\mskip 1.5mu\mathfrak{{{B}}}=\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$ via the isomorphism

[TABLE]

11 Relativized Relational Representation

Suppose $\mathfrak{{{B}}}$ is the suitable structure of a complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$. Given a triple $\langle{{u}},{{v}},{{w}}\rangle\in{}^{3}{{U}}(\mathfrak{{{B}}})$, we say that the ordered pairs in $\{{{u}},{{v}},{{w}}\}\times\{{{u}},{{v}},{{w}}\}$ occur in $\langle{{u}},{{v}},{{w}}\rangle$, and that $\langle{{u}},{{v}},{{w}}\rangle$ carries those pairs. Let ${{V}}_{2}$ be the set of ordered pairs that occur in triples in ${{V}}(\mathfrak{{{B}}})$, i.e., ${{V}}_{2}=\{\langle{{u}},{{v}}\rangle:{{u}}=({{p}}\kappa)^{\mathfrak{{{B}}}},\,{{v}}=({{p}}\lambda)^{\mathfrak{{{B}}}},\,{{p}}\in Tr\mathfrak{{{B}}},\,\kappa,\lambda<3\}.$ Given a trail ${{p}}\in Tr\mathfrak{{{B}}}$, the triple of ${{p}}$ is $\langle({{p}}0)^{\mathfrak{{{B}}}},({{p}}1)^{\mathfrak{{{B}}}},({{p}}2)^{\mathfrak{{{B}}}}\rangle$. For every trail ${{p}}\in Tr(\mathfrak{{{B}}})$ we will define a function $\mathcal{A}_{{p}}:\{{{u}},{{v}},{{w}}\}\times\{{{u}},{{v}},{{w}}\}\to\mathsf{At\,}{\mathfrak{{{A}}}}$, where $\langle{{u}},{{v}},{{w}}\rangle=\langle({{p}}0)^{\mathfrak{{{B}}}},({{p}}1)^{\mathfrak{{{B}}}},({{p}}2)^{\mathfrak{{{B}}}}\rangle$. Let ${{t}}\in{{B}}$ be the end of ${{p}}$. Then ${{t}}_{2}{\mskip 1.5mu;\mskip 1.5mu}{{t}}_{0}\geq{{t}}_{1}\neq 0$ since ${{t}}\in{{B}}$, so ${{{t}}_{2}}^{\mathsf{r}}={{{t}}_{0}}^{\mathsf{d}}$ by (42). From ${{t}}_{2}{\mskip 1.5mu;\mskip 1.5mu}{{t}}_{0}\geq{{t}}_{1}\neq 0$ we get ${{t}}_{0}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{t}}_{1}}\geq\breve{{{t}}_{2}}\neq 0$ and $\breve{{{t}}_{2}}{\mskip 1.5mu;\mskip 1.5mu}{{t}}_{1}\geq{{t}}_{0}\neq 0$ by (33), so ${{{t}}_{0}}^{\mathsf{r}}={\breve{{{t}}_{1}}}^{\mathsf{d}}={{{t}}_{1}}^{\mathsf{r}}$ and ${{{t}}_{2}}^{\mathsf{d}}={\breve{{{t}}_{2}}}^{\mathsf{r}}={{{t}}_{1}}^{\mathsf{d}}$ by (42) and (35). Then $\mathcal{A}_{{p}}$ is defined by

[TABLE]

As a general rule, the end of ${{p}}$ is \big{\langle}\mathcal{A}_{{p}}({{v}},{{w}}),\mathcal{A}_{{p}}({{u}},{{w}}),\mathcal{A}_{{p}}({{u}},{{v}})\big{\rangle}, where $\langle{{u}},{{v}},{{w}}\rangle$ is the triple of ${{p}}$. Further applications of (33), (35), (36), and (42) show that for all ${{x}},{{y}},{{z}}\in\{{{u}},{{v}},{{w}}\}$, $\mathcal{A}_{{p}}({{x}},{{y}}){\mskip 1.5mu;\mskip 1.5mu}\mathcal{A}_{{p}}({{y}},{{z}})\geq\mathcal{A}_{{p}}({{x}},{{z}}).$ Since $\mathcal{A}_{{p}}$ depends only on the end of ${{p}}$, and type (i) reductions do not change the end or the triple of a trail, it follows that $\mathcal{A}_{{p}}=\mathcal{A}_{{q}}$ for any two trails ${{p}}\approx{{q}}$ that are related by condition (i). Suppose two trails ${{p}}\approx{{q}}$ are related by condition (iii). According to the condition, they must agree except possibly at the pointer, so ${{p}}\kappa={{q}}\kappa$ for all $\kappa<3$, hence $\mathcal{A}_{{p}}=\mathcal{A}_{{q}}$. Finally, suppose two trails ${{p}}\approx{{q}}$ are related by a pointer-preserving reduction of type (ii), e.g.,

[TABLE]

Note that $s_{1}={{t}}_{1}$ since ${{p}}$ is a trail. Adding this equation to the parts of the definitions of $\mathcal{A}_{{q}}$ and $\mathcal{A}_{{p}}$ that involve index $1$, we get

[TABLE]

so $\mathcal{A}_{{p}}$ and $\mathcal{A}_{{q}}$ agree where they are both defined. It follows by induction on the generation of $\approx$ from reductions of types (i)–(iii), that if ${{p}}\approx{{q}}$ then $\mathcal{A}_{{p}}$ and $\mathcal{A}_{{q}}$ agree whenever they are both defined. Every pair $\langle{{u}},{{v}}\rangle$ in ${{V}}_{2}$ occurs in the triple of some trail ${{p}}$, and is thereby assigned to the atom $\mathcal{A}_{{p}}({{u}},{{v}})$. If $\langle{{u}},{{v}}\rangle$ also occurs in the triple of some trail ${{q}}$, and is thereby assigned to the atom $\mathcal{A}_{{q}}({{u}},{{v}})$, we have ${{u}}=({{p}}\kappa)^{\mathfrak{{{B}}}}=({{q}}\kappa^{\prime})^{\mathfrak{{{B}}}}$ and ${{v}}=({{p}}\lambda)^{\mathfrak{{{B}}}}=({{q}}\lambda^{\prime})^{\mathfrak{{{B}}}}$, hence ${{p}}\kappa\approx{{q}}\kappa^{\prime}$ and ${{p}}\lambda\approx{{q}}\lambda^{\prime}$. Since the pointer of ${{p}}$ is irrelevant to the definition of $\mathcal{A}_{{p}}$, we have, as a rule, $\mathcal{A}_{{p}}=\mathcal{A}_{{{p}}\kappa}$ for all $\kappa<3,$ so from either of these equivalences we conclude that $\mathcal{A}_{{p}}({{u}},{{v}})=\mathcal{A}_{{q}}({{u}},{{v}})$. Thus there is a function $\mathcal{A}:{{V}}_{2}\to\mathsf{At\,}{\mathfrak{{{A}}}}$ such that $\mathcal{A}({{u}},{{v}})=\mathcal{A}_{{p}}({{u}},{{v}})$ for any trail ${{p}}$ whose triple contains $\langle{{u}},{{v}}\rangle$, with the key property

[TABLE]

Define a binary relation $\Xi({{u}})$ for each ${{u}}\in{{U}}(\mathfrak{{{B}}})$ as follows. If $|{{u}}|=1$ then $\Xi({{u}})=\emptyset$. If $|{{u}}|>1$ then $\Xi({{u}})=\{\langle{{u}},({{p}}\lambda)^{\mathfrak{{{B}}}}\rangle,\langle{{u}},({{p}}\mu)^{\mathfrak{{{B}}}}\rangle\}$ where ${{p}}$ is the unique reduced trail in ${{u}}$, hence ${{u}}={{p}}^{\mathfrak{{{B}}}}$, $\kappa$ is the pointer of ${{p}}$, i.e., ${{p}}={{p}}\kappa$, and $\{0,1,2\}=\{\kappa,\lambda,\mu\}$. In this case, \Xi({{u}})=\{\langle{{u}},({{p}}\lambda)^{\mathfrak{{{B}}}}\rangle:\text{{{p}}={{p}}\kappa\in{{u}} is reduced},\kappa\neq\lambda<3\}. From the definitions it follows that $\Xi({{u}})\subseteq{{V}}_{2}$ and $(\Xi({{u}}))^{-1}\subseteq{{V}}_{2}$, since the ordered pairs in $\Xi({{u}})$ occur in the triple of the reduced trail in ${{u}}$, which is a permutation of $\langle{{u}},({{p}}\lambda)^{\mathfrak{{{B}}}},({{p}}\mu)^{\mathfrak{{{B}}}}\rangle$. Points $({{p}}\lambda)^{\mathfrak{{{B}}}}$ and $({{p}}\mu)^{\mathfrak{{{B}}}}$ (which may coincide) have shorter reduced trails (and hence are distinct from ${{u}}$) because condition (ii) in the definition of $\approx$ applies to ${{p}}\lambda$ and ${{p}}\mu$ (due to $\kappa\neq\lambda,\mu$). In computing their reduced trails, only conditions (ii) and (iii) in the definition of $\approx$ are used. Condition (i) cannot apply because ${{p}}$ is reduced. Thus, if $\langle{{v}},{{w}}\rangle\in\Xi({{u}})$, either ${{v}}$ has a shorter reduced trail than ${{w}}$, i.e., $|{{v}}|<|{{w}}|$, or ${{w}}$ has a shorter reduced trail than ${{v}}$, i.e., $|{{w}}|<|{{v}}|$. We now turn to proving that ${{V}}_{2}$ can be partitioned into three pieces, the set $Id_{{{U}}(\mathfrak{{{B}}})}$ of identity pairs, the set $\Xi({{U}})$ of $\Xi$-pairs, and the set ${{B}}_{2}$ of base-pairs:

[TABLE]

A pair $\langle{{u}},{{v}}\rangle$ with ${{u}},{{v}}\in{{U}}(\mathfrak{{{B}}})$ is an identity pair if ${{u}}={{v}}$, a $\Xi$-pair if $\langle{{u}},{{v}}\rangle\in\Xi({{u}})$ or $\langle{{v}},{{u}}\rangle\in\Xi({{v}})$, and a base-pair if it occurs in the triple of a trail ${{p}}$ of length 1. In connection with the following lemma, it is worth observing that ${{V}}(\mathfrak{{{B}}})$ is not closed under the permutation of triples.

Lemma 8

If $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle\in{{V}}(\mathfrak{{{B}}})$ then $\langle{{u}}_{\pi(0)},{{u}}_{\pi(1)},{{u}}_{\pi(2)}\rangle\in{{V}}(\mathfrak{{{B}}})$ for every non-permutation $\pi\colon\{0,1,2\}\to\{0,1,2\}$.

Proof

Assume $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle\in{{V}}(\mathfrak{{{B}}})$. Then there is a reduced trail

[TABLE]

where ${{t}}_{{n}}={{a}}=\langle{{a}}_{0},{{a}}_{1},{{a}}_{2}\rangle\in{{B}}$ and ${{a}}_{0},{{a}}_{1},{{a}}_{2}\in\mathsf{At\,}{\mathfrak{{{A}}}}$, such that $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle=\langle({{p}}0)^{\mathfrak{{{B}}}},({{p}}1)^{\mathfrak{{{B}}}},({{p}}2)^{\mathfrak{{{B}}}}\rangle.$ For distinct $\kappa,\lambda<3$, define the non-permutation (replacement) $[\kappa/\lambda]\colon 3\to 3$ by $[\kappa/\lambda]=\{\langle k,\lambda\rangle,\langle\lambda,\lambda\rangle,\langle\mu,\mu\rangle\}$ where $\{\kappa,\lambda,\mu\}=\{0,1,2\}$. Let

[TABLE]

Then, as simple computational checks will show,

[TABLE]

Thus the lemma holds for the six listed replacements. The remaining 15 non-permutations can be obtained from these by composition in multiple ways, so the lemma holds for all of them.

Lemma 9

Every triple in ${{V}}(\mathfrak{{{B}}})$ is a permutation of one of the following six triples, for some distinct ${{u}},{{v}},{{w}}\in{{U}}(\mathfrak{{{B}}})$.

1. (i)

$\langle{{u}},{{v}},{{w}}\rangle$* where $\{\langle{{u}},{{v}}\rangle,\langle{{u}},{{w}}\rangle\}=\Xi({{u}})$ and $\langle{{v}},{{w}}\rangle$ is a $\Xi$-pair,* 2. (ii)

$\langle{{u}},{{v}},{{w}}\rangle$* where $\{\langle{{u}},{{v}}\rangle,\langle{{u}},{{w}}\rangle\}=\Xi({{u}})$, and $\langle{{v}},{{w}}\rangle$ is a base-pair,* 3. (iii)

$\langle{{u}},{{v}},{{w}}\rangle$* where $\langle{{u}},{{v}}\rangle$, $\langle{{u}},{{w}}\rangle$, and $\langle{{v}},{{w}}\rangle$ are base-pairs,* 4. (iv)

$\langle{{u}},{{u}},{{v}}\rangle$* where $\langle{{u}},{{v}}\rangle$ is a $\Xi$-pair,* 5. (v)

$\langle{{u}},{{u}},{{v}}\rangle$* where $\langle{{u}},{{v}}\rangle$ is a base-pair,* 6. (vi)

$\langle{{u}},{{u}},{{u}}\rangle$.

Proof

Triples of types (i)–(iii) contain three distinct points in ${{U}}(\mathfrak{{{B}}})$, triples of types (iv) and (v) have two distinct points, while triples of type (vi) contain just a single point (repeated twice). Types (i), (ii), and (iv) are the triples of trails that have length $>1$, while types (iii) and (v) are the triples of trails of length $1$. Triples of type (vi) are the triples of trails of all lengths. We proceed by induction on the length of trails.

Start with a trail ${{p}}=\langle{{t}},\nu\rangle$ of length 1, where ${{t}}\in{{B}}$, and $\nu<3$. If ${{t}}$ is a diversity cycle, then ${{t}}$ is not in ${{E}}_{01}$, ${{E}}_{02}$, or ${{E}}_{12}$ since ${{t}}_{2}\not\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, ${{t}}_{1}\not\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, and ${{t}}_{0}\not\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, respectively. The trails ${{p}}0$, ${{p}}1$, and ${{p}}2$ are reduced, for in the definition of reduced trail, the hypothesis of condition (i) is not met because ${{t}}\notin{{E}}_{01}\cup{{E}}_{02}\cup{{E}}_{12}$, and the hypotheses of condition (ii) and (iii) are not met because $|{{p}}|=1$. The three reduced trails ${{p}}0$, ${{p}}1$, and ${{p}}2$ are distinct, because they have distinct pointers, so their equivalence classes (points in ${{U}}(\mathfrak{{{B}}})$) are distinct, and the triple $\langle({{p}}0)^{\mathfrak{{{B}}}},({{p}}1)^{\mathfrak{{{B}}}},({{p}}2)^{\mathfrak{{{B}}}}\rangle$ of ${{p}}$ has type (iii), i.e., $({{p}}0)^{\mathfrak{{{B}}}}$, $({{p}}1)^{\mathfrak{{{B}}}}$, and $({{p}}2)^{\mathfrak{{{B}}}}$ are distinct.

Assume ${{t}}$ is an identity cycle, i.e., ${{t}}$ has the form $\langle{{a}},{{a}},{{{a}}}^{\mathsf{d}}\rangle$, $\langle{{{a}}}^{\mathsf{r}},{{a}},{{a}}\rangle$, $\langle{{a}},{{{a}}}^{\mathsf{r}},\breve{{{a}}}\rangle$, or $\langle{{{a}}}^{\mathsf{d}},{{{a}}}^{\mathsf{d}},{{{a}}}^{\mathsf{d}}\rangle$, for some atom ${{a}}\in\mathsf{At\,}{\mathfrak{{{A}}}}.$ If ${{t}}=\langle{{a}},{{a}},{{{a}}}^{\mathsf{d}}\rangle$, then ${{t}}\in{{E}}_{01},$ ${{p}}0=\langle{{t}},0\rangle$ and ${{p}}2=\langle{{t}},1\rangle$ are reduced so $({{p}}0)^{\mathfrak{{{B}}}}$ and $({{p}}2)^{\mathfrak{{{B}}}}$ are distinct, ${{p}}0=\langle{{t}},0\rangle\approx\langle{{t}},1\rangle={{p}}1$ since ${{t}}_{2}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$, so $({{p}}0)^{\mathfrak{{{B}}}}=({{p}}1)^{\mathfrak{{{B}}}}$, and the triple of ${{p}}$ has the form $\langle{{u}},{{u}},{{v}}\rangle$, which is type (v). Similarly, if ${{t}}=\langle{{{a}}}^{\mathsf{r}},{{a}},{{a}}\rangle$, then ${{t}}\in{{E}}_{12},$ ${{p}}0$ and ${{p}}1$ are reduced, ${{p}}1\approx{{p}}2$, and the triple of ${{p}}$ has the form $\langle{{v}},{{u}},{{u}}\rangle$, which is type (v). If ${{t}}=\langle{{a}},{{{a}}}^{\mathsf{r}},\breve{{{a}}}\rangle$ then ${{t}}\in{{E}}_{02},$ ${{p}}0$ and ${{p}}1$ are reduced, ${{p}}0\approx{{p}}2$, and the triple of ${{p}}$ has the form $\langle{{u}},{{v}},{{u}}\rangle$, again of type (v). Finally, if ${{t}}=\langle{{{a}}}^{\mathsf{d}},{{{a}}}^{\mathsf{d}},{{{a}}}^{\mathsf{d}}\rangle$ then ${{t}}\in{{E}}_{01}\cup{{E}}_{12}$, ${{p}}0$ is reduced, ${{p}}0\approx{{p}}1\approx{{p}}2$, and the triple of ${{p}}$ has type (vi).

As inductive hypothesis, we assume that the triple of every trail of length ${{n}}$ or less is exactly one of the six types. Let ${{p}}$ be a trail of length ${{n}}+1$, i.e., there is a trail ${{q}}\in Tr(\mathfrak{{{B}}})$ such that

[TABLE]

We may assume ${{p}}$ is not subject to any reductions of type (i) because if it were, it would be $\approx$ to a shorter trail whose triple is the same as that of ${{p}}$, hence subject to the inductive hypothesis, which yields the desired conclusion. Therefore we assume ${{p}}$ is (i)-reduced. Let $\kappa=\kappa_{{{n}}-1}$ and $\{\kappa,\lambda,\mu\}=\{0,1,2\}$. Note that ${{p}}\lambda$ and ${{p}}\mu$ are not reduced, and indeed $({{q}}\lambda)^{\mathfrak{{{B}}}}=({{p}}\lambda)^{\mathfrak{{{B}}}}$ and $({{q}}\mu)^{\mathfrak{{{B}}}}=({{p}}\mu)^{\mathfrak{{{B}}}}$ since, by a (ii)-reduction,

[TABLE]

The inductive hypothesis applies to the trail ${{q}}$ of length ${{n}}$.

Suppose ${{t}}_{{n}}\notin{{E}}_{\kappa\lambda}\cup{{E}}_{\kappa\mu}$. This implies that ${{p}}\kappa$ is reduced, because ${{p}}$ is (i)-reduced and no reduction of types (ii) or (iii) can apply to ${{p}}\kappa$. Hence, by definition, $\Xi(({{p}}\kappa)^{\mathfrak{{{B}}}})=\{\langle({{p}}\kappa)^{\mathfrak{{{B}}}},({{p}}\lambda)^{\mathfrak{{{B}}}}\rangle,\langle({{p}}\kappa)^{\mathfrak{{{B}}}},({{p}}\mu)^{\mathfrak{{{B}}}}\rangle\}.$ If $({{q}}\lambda)^{\mathfrak{{{B}}}}=({{q}}\mu)^{\mathfrak{{{B}}}}$ then the triple of ${{p}}$ is the second kind of type (iv). Assume $({{q}}\lambda)^{\mathfrak{{{B}}}}\neq({{q}}\mu)^{\mathfrak{{{B}}}}$. Depending on the type of the triple of ${{q}}$, the diversity pair $\langle({{q}}\lambda)^{\mathfrak{{{B}}}},({{q}}\mu)^{\mathfrak{{{B}}}}\rangle$ is either a $\Xi$-pair, which means the triple of ${{p}}$ is type (i), or a base-pair and the triple of ${{p}}$ is type (ii). The triple of ${{p}}$ cannot be type (vi).

Suppose ${{t}}_{{n}}\in{{E}}_{\kappa\lambda}$. From this we get ${{p}}\kappa\approx{{p}}\lambda$ by a reduction of type (iii), so $({{p}}\kappa)^{\mathfrak{{{B}}}}=({{p}}\lambda)^{\mathfrak{{{B}}}}$. Recall that $({{q}}\lambda)^{\mathfrak{{{B}}}}=({{p}}\lambda)^{\mathfrak{{{B}}}}$ and $({{q}}\mu)^{\mathfrak{{{B}}}}=({{p}}\mu)^{\mathfrak{{{B}}}}$, so if $({{q}}\lambda)^{\mathfrak{{{B}}}}=({{q}}\mu)^{\mathfrak{{{B}}}}$ then the triple of ${{p}}$ is type (vi). Assume $({{q}}\lambda)^{\mathfrak{{{B}}}}\neq({{q}}\mu)^{\mathfrak{{{B}}}}$. From the inductive hypothesis applied to ${{q}}$ we know the diversity pair $\langle({{q}}\lambda)^{\mathfrak{{{B}}}},({{q}}\mu)^{\mathfrak{{{B}}}}\rangle$ is either a $\Xi$-pair, in which case the triple of ${{p}}$ is type (iv), or it is a base-pair and the triple of ${{p}}$ is type (v). Similarly, if ${{t}}_{{n}}\in{{E}}_{\kappa\mu}$ then the triple of ${{p}}$ is type (iv) or (v).

The following result is the analogue of (MR987611, , Theorem C) for weakly associative relation algebras.

Theorem 11.1

Suppose $\mathfrak{{{B}}}$ is the suitable structure of a complete atomic $\mathfrak{{{A}}}\in\mathsf{WA}$. Then $\mathfrak{{{A}}}\cong\mathfrak{Rl}\mskip 1.5mu_{{{S}}(\mathfrak{{{B}}})}\mathfrak{{{A}}}^{\prime}$ for some set relation algebra $\mathfrak{{{A}}}^{\prime}\subseteq\mathfrak{Re}\left({{U}}(\mathfrak{{{B}}})\right)$, where

[TABLE]

Proof

For brevity, use “$\mathsf{C}_{\kappa}$” and “$\mathsf{D}_{\kappa\lambda}$” in place of “$\mathsf{C}_{\kappa}^{[{}^{3}{{U}}(\mathfrak{{{B}}})]}$” and “$\mathsf{D}_{\kappa\lambda}^{[{}^{3}{{U}}(\mathfrak{{{B}}})]}$”. Let ${{E}}(\mathfrak{{{B}}})=\mathsf{C}_{2}{{V}}(\mathfrak{{{B}}})$. Then, as we will show, $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)\cong\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right)$ via the isomorphism ${{X}}\mapsto\mathsf{C}_{2}{{X}}$, where ${{X}}$ is an element of $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$, i.e., ${{V}}(\mathfrak{{{B}}})\cap\mathsf{C}_{2}{{X}}={{X}}\in{{C}}.$ Indeed, if ${{Y}}=\mathsf{C}_{2}{{X}}$, then ${{X}}$ can be recovered from ${{Y}}$ since ${{X}}={{V}}(\mathfrak{{{B}}})\cap{{Y}}$, and ${{Y}}$ is an element of $\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right)$, i.e., ${{E}}(\mathfrak{{{B}}})\supseteq\mathsf{C}_{2}{{Y}}={{Y}}\in{{C}},$ because ${{Y}}\in{{C}}$ since $\mathfrak{{{C}}}$ is a subalgebra (closed under cylindrifications), $\mathsf{C}_{2}{{Y}}={{Y}}$ since cylindrifying twice is the same as doing it once, and ${{Y}}\subseteq{{E}}(\mathfrak{{{B}}})$ since ${{X}}\subseteq{{V}}(\mathfrak{{{B}}})$, by the monotonicity of cylindrification. The map ${{X}}\mapsto\mathsf{C}_{2}{{X}}$ is thus a bijection. We need to show it preserves relative multiplication and converse. Toward this end, suppose ${{X}},{{Y}}$ are elements of $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$, that is, ${{V}}(\mathfrak{{{B}}})\cap\mathsf{C}_{2}{{X}}={{X}}\in{{C}}$ and ${{V}}(\mathfrak{{{B}}})\cap\mathsf{C}_{2}{{Y}}={{Y}}\in{{C}}.$ For relative product we will prove

[TABLE]

The left side of ($*{**}$) is obtained by applying the map $\mathsf{C}_{2}$ to ${{X}}$ and ${{Y}}$, and then computing the relative product of their images in the target algebra $\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right)$, while the right side is the result of applying the map $\mathsf{C}_{2}$ to the relative product of ${{X}}$ and ${{Y}}$, as computed in the source algebra $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$, and simplifying using $\mathsf{c}_{2}{{x}}\cdot\mathsf{c}_{2}{{y}}=\mathsf{c}_{2}({{x}}\cdot\mathsf{c}_{2}{{y}})$. The right side of ($*{**}$) is included in the left side, by just the monotonicity and idempotence of cylindrification, so we need only show the left side is included in the right side. Suppose $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle$ is in the left side of (${*}{**}$), i.e.,

[TABLE]

Then $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle\in{{E}}(\mathfrak{{{B}}})=\mathsf{C}_{2}\left({{V}}(\mathfrak{{{B}}})\right)$, so by the definitions of cylindrification and ${{V}}(\mathfrak{{{B}}})$, there is some trail ${{p}}$ such that

[TABLE]

so, by the definitions of cylindrification and diagonal elements, there is some ${{v}}\in{{U}}(\mathfrak{{{B}}})$ such that

[TABLE]

hence

[TABLE]

Note that if $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle$ happens to also be in ${{V}}(\mathfrak{{{B}}})$, then $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle$ will be in $\mathsf{C}_{2}\{\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle\}$, and will consequently be in the right hand side of (${**}*$). We can prove $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle\in{{V}}(\mathfrak{{{B}}})$ if ${{u}}_{0},{{u}}_{1},{{v}}$ are not distinct. Since $\langle{{u}}_{0},{{u}}_{1}\rangle$ occurs in the triple of ${{p}}$, we get $\langle{{u}}_{0},u_{1},u_{0}\rangle,\langle{{u}}_{0},{{u}}_{1},{{u}}_{1}\rangle\in{{V}}(\mathfrak{{{B}}})$ by Lemma 8. Consequently, if ${{u}}_{0}={{v}}$ then $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle=\langle{{u}}_{0},u_{1},u_{0}\rangle\in{{V}}(\mathfrak{{{B}}})$, and if ${{u}}_{1}={{v}}$ then $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle=\langle{{u}}_{0},{{u}}_{1},{{u}}_{1}\rangle\in{{V}}(\mathfrak{{{B}}})$. If ${{u}}_{0}={{u}}_{1}$ then $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle=\langle{{u}}_{0},{{u}}_{0},{{v}}\rangle=\langle({{q}}0)^{\mathfrak{{{B}}}},({{q}}0)^{\mathfrak{{{B}}}},({{q}}1)^{\mathfrak{{{B}}}}\rangle\in{{V}}(\mathfrak{{{B}}})$ by Lemma 8 (${{r}}$ can also be used here). We may therefore assume that ${{u}}_{0},{{u}}_{1},{{v}}$ are distinct. By Lemma 9 each of the pairs $\langle{{u}}_{0},{{u}}_{1}\rangle$, $\langle{{u}}_{0},{{v}}\rangle$, and $\langle{{v}},{{u}}_{1}\rangle$ must be either a $\Xi$-pair or a base-pair.

Let ${{w}}\in\{{{u}}_{0},{{u}}_{1},{{v}}\}$ be any point with maximum length, i.e., $|{{w}}|\geq|{{u}}_{0}|,|{{u}}_{1}|,|{{v}}|$, and let $\{{{w}},{{x}},{{y}}\}=\{{{u}}_{0},{{u}}_{1},{{v}}\}$. We will show that the triple of the reduced trail in ${{w}}$ is a permutation of $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle$. There are two cases.

Case 1: Some point in $\{{{u}}_{0},{{u}}_{1},{{v}}\}$ has length $>1$. By our assumption, $|{{w}}|>1$. Since endpoints of base-pairs have length 1, $\langle{{w}},{{x}}\rangle$ must be a $\Xi$-pair, hence either $\langle{{w}},{{x}}\rangle\in\Xi({{w}})$ and $|w|>|x|$, or else $\langle{{x}},{{w}}\rangle\in\Xi({{x}})$ and $|x|>|w|$. But $|x|\not>|w|$ by the choice of ${{w}}$, so $\langle{{w}},{{x}}\rangle\in\Xi({{w}})$, and, similarly, $\langle{{w}},{{y}}\rangle\in\Xi({{w}})$. Since ${{x}}\neq{{y}}$, $\Xi({{w}})$ has at least two elements. By its definition, $\Xi({{w}})$ has at most two elements, so $\Xi({{w}})=\{\langle{{w}},{{x}}\rangle,\langle{{w}},{{y}}\rangle\}$. Since the ordered pairs in $\Xi({{w}})$ occur in the triple of the reduced trail in ${{w}}$, the triple of the reduced trail in ${{w}}$ must be a permutation of $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle$.

Case 2: $|{{u}}_{0}|=|{{u}}_{1}|=|{{v}}|=1$. In this case ${{w}}$ can be any one of the three points. By Lemma 9, $\langle{{u}}_{0},{{u}}_{1}\rangle$, $\langle{{u}}_{0},{{v}}\rangle$, and $\langle{{u}}_{1},{{v}}\rangle$ are base-pairs, i.e., the triple has type (iii) in that lemma. (Types (i) and (ii) require one of the points to have length $>1$. Types (iv), (v), and (vi) have non-distinct triples.) Recall that the points arise from trails ${{p}},{{q}},{{r}}$ by ${{u}}_{0}=({{p}}0)^{\mathfrak{{{B}}}}=({{q}}0)^{\mathfrak{{{B}}}}$, ${{u}}_{1}=({{p}}1)^{\mathfrak{{{B}}}}=({{r}}1)^{\mathfrak{{{B}}}}$, and ${{v}}=({{q}}1)^{\mathfrak{{{B}}}}=({{r}}0)^{\mathfrak{{{B}}}}$, which imply ${{p}}0\approx{{q}}0$, ${{p}}1\approx{{r}}1$, and ${{q}}1\approx{{r}}0$. Therefore ${{p}},{{q}},{{r}}$ all begin at the same cycle, say ${{t}}\in{{B}}$, because $\approx$-reductions only relate trails that begin at the same cycle. In general, the reduced trail of a point ${{u}}$ is always a “subtrail” of any trail that has ${{u}}$ in its triple. Type (i) reductions extract parts in the middle of a trail. Type (ii) reductions just change the pointer. Type (iii) reductions shorten a trail by deleting the end and the ordinal preceding it. The reduced trail of a point ${{u}}$ of length 1 has length 1 (is of the form $\langle{{t}},\kappa\rangle$) and its beginning ${{t}}$ is the beginning of every trail whose triple includes ${{u}}$. In the current case, the trails ${{p}},{{q}},{{r}}$ and the indices $0,1$ produce three distinct points, the most a trail can produce. Since the three points ${{u}}_{0},{{u}}_{1},{{v}}$ have length 1, their reduced trails have length 1, i.e., have the same cycle, say ${{t}}\in{{B}}$, as both beginning and end, and differ only in their pointers, so to get three points we must use all three pointers. Therefore $\{{{u}}_{0},{{u}}_{1},{{v}}\}=\{(\langle{{t}},0\rangle)^{\mathfrak{{{B}}}},(\langle{{t}},1\rangle)^{\mathfrak{{{B}}}},(\langle{{t}},2\rangle)^{\mathfrak{{{B}}}}\}$ and because these points are distinct, ${{t}}$ must be a diversity cycle, i.e., ${{t}}_{0}+{{t}}_{1}+{{t}}_{2}\leq{0\kern-0.39993pt\raise 5.59721pt\hbox{,}\kern 0.39993pt}$. Thus, in this case as well as Case 1, $\langle{{u}}_{0},{{u}}_{1},{{v}}\rangle$ is a permutation of the triple of the reduced trail in ${{w}}$.

By K and the definition of ${{B}}$, we have $\langle\mathcal{A}({{u}}_{1},{{v}}),\mathcal{A}({{u}}_{0},{{v}}),\mathcal{A}({{u}}_{0},{{u}}_{1})\rangle\in{{B}}.$ Let ${{t}}=\langle{{t}}_{0},{{t}}_{1},{{t}}_{2}\rangle\in{{B}}$ be the end of ${{p}}$. Then ${{t}}_{2}=\mathcal{A}_{{p}}({{u}}_{0},{{u}}_{1})=\mathcal{A}({{u}}_{0},{{u}}_{1})\leq\mathcal{A}({{u}}_{0},{{v}}){\mskip 1.5mu;\mskip 1.5mu}\mathcal{A}({{v}},{{u}}_{1}),$ so, letting {{t}}^{\prime}=\Big{\langle}\mathcal{A}({{u}}_{1},{{v}}),\mathcal{A}({{u}}_{0},{{v}}),\mathcal{A}({{u}}_{0},{{u}}_{1})\Big{\rangle}, we have ${{t}}\mathrel{{{T}}_{2}}{{t}}^{\prime}.$ Define a trail ${{p}}^{\prime}$ extending ${{p}}$ by {{p}}^{\prime}={{p}}\odot\big{\langle}{{t}},2,{{t}}^{\prime},0\big{\rangle}. Then ${{u}}_{0}=({{p}}0)^{\mathfrak{{{B}}}}=({{p}}^{\prime}0)^{\mathfrak{{{B}}}}$ and ${{u}}_{1}=({{p}}1)^{\mathfrak{{{B}}}}=({{p}}^{\prime}1)^{\mathfrak{{{B}}}}$ since ${{p}}0\approx{{p}}^{\prime}0$ and ${{p}}1\approx{{p}}^{\prime}1$ by type (iii) reductions. Let ${{z}}=({{p}}^{\prime}2)^{\mathfrak{{{B}}}}$. Then $\langle{{u}}_{0},{{u}}_{1},{{z}}\rangle\in{{V}}(\mathfrak{{{B}}})$ because $\langle{{u}}_{0},{{u}}_{1},{{z}}\rangle$ is the triple of ${{p}}^{\prime}$, hence also $\langle{{u}}_{0},{{z}},{{z}}\rangle\in{{V}}(\mathfrak{{{B}}})$ by Lemma 8.

Since $\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$ is complete and atomic, its element ${{X}}$ must be a join of atoms. Since each atom has the form $R^{\mathfrak{{{B}}}}_{{{s}}}$ for some ${{s}}\in{{B}}$ by Theorem 10.1, from (##) we get $\langle{{u}}_{0},{{v}},{{v}}\rangle\in R^{\mathfrak{{{B}}}}_{{{s}}}\subseteq{{X}}$ for some ${{s}}\in{{B}}$. By the definition of $R^{\mathfrak{{{B}}}}_{{{s}}}$, $\langle{{u}}_{0},{{v}},{{v}}\rangle$ is the triple of some trail ${{p}}^{\prime\prime}$ that ends at ${{s}}$. Therefore {{s}}=\Big{\langle}{\left(\mathcal{A}({{u}}_{0},{{v}})\right)}^{\mathsf{r}},\mathcal{A}({{u}}_{0},{{v}}),\mathcal{A}({{u}}_{0},{{v}})\Big{\rangle}. From $\langle{{u}}_{0},{{z}},{{z}}\rangle\in{{V}}(\mathfrak{{{B}}})$ it follows that $\langle{{u}}_{0},{{z}},{{z}}\rangle$ is the triple of a trail ending at \Big{\langle}\mathcal{A}({{z}},{{z}}),\mathcal{A}({{u}}_{0},{{z}}),\mathcal{A}({{u}}_{0},{{z}})\Big{\rangle}. The triple of ${{p}}^{\prime}$ is $\langle{{u}}_{0},{{u}}_{1},{{z}}\rangle$, so by the relevant definitions,

[TABLE]

It follows that $\langle{{u}}_{0},{{z}},{{z}}\rangle$ is the triple of a trail ending at ${{s}}$, and therefore $\langle{{u}}_{0},{{z}},{{z}}\rangle\in R^{\mathfrak{{{B}}}}_{{{s}}}\subseteq{{X}}$. By a similar argument, $\langle{{z}},{{u}}_{1},{{z}}\rangle\in{{Y}}$. Thus we have $\langle{{u}}_{0},{{z}},{{z}}\rangle\in\mathsf{D}_{12}\cap{{X}}$ and $\langle{{z}},{{u}}_{1},{{z}}\rangle\in\mathsf{D}_{02}\cap{{Y}}$, which imply, together with $\langle{{u}}_{0},{{u}}_{1},{{z}}\rangle\in{{V}}(\mathfrak{{{B}}})$, that \langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle\in\mathsf{C}_{2}\big{(}\{\langle{{u}}_{0},{{u}}_{1},{{z}}\rangle\}\big{)}\subseteq\mathsf{C}_{2}\Big{(}{{V}}(\mathfrak{{{B}}})\cap\mathsf{C}_{1}(\mathsf{D}_{12}\cap{{X}})\cap\mathsf{C}_{0}(\mathsf{D}_{02}\cap{{Y}})\Big{)}, i.e., $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle$ is in the right hand side of ($*{**}$), as desired. Thus relative product is preserved. For converse, we will prove

[TABLE]

The left side of ($*{**}*$) is obtained by applying the isomorphism $\mathsf{C}_{2}$ to ${{X}}$, and then computing the converse of its image in the target algebra $\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right)$. The right side of ($*{**}*$) is obtained by computing the converse of ${{X}}$ in the source algebra $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)$, applying the isomorphism $\mathsf{C}_{2}$, and simplifying using $\mathsf{c}_{2}{{x}}\cdot\mathsf{c}_{2}{{y}}=\mathsf{c}_{2}({{x}}\cdot\mathsf{c}_{2}{{y}})$ and the definition of ${{E}}(\mathfrak{{{B}}})$. The right side of ($*{**}*$) is included in the left side, by just the monotonicity and idempotence of cylindrification, so we need only show the left side is included in the right side. Suppose $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle$ is in the left side of ($*{**}*$). Then

[TABLE]

By the definitions of cylindrification and diagonal elements, we obtain successively

[TABLE]

and, finally, there is some ${{x}}\in{{U}}(\mathfrak{{{B}}})$ such that $\langle{{u}}_{1},{{u}}_{0},{{x}}\rangle\in{{X}}$. We assumed ${{X}}\subseteq{{V}}(\mathfrak{{{B}}})$, so $\langle{{u}}_{1},{{u}}_{0},{{x}}\rangle\in{{V}}(\mathfrak{{{B}}})$, hence $\langle{{u}}_{1},{{u}}_{0},{{u}}_{0}\rangle,\langle{{u}}_{1},{{u}}_{1},{{u}}_{0}\rangle,\langle{{u}}_{0},{{u}}_{1},{{u}}_{0}\rangle\in{{V}}(\mathfrak{{{B}}})$ by Lemma 8. Combining these facts with our assumption that ${{V}}(\mathfrak{{{B}}})\cap\mathsf{C}_{2}{{X}}={{X}}$, we obtain successively

[TABLE]

Together with $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle\in{{E}}(\mathfrak{{{B}}})$, we conclude $\langle{{u}}_{0},{{u}}_{1},{{u}}_{2}\rangle$ is in the right hand side of ($*{**}*$), as desired. This completes the proof that $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)\cong\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right).$ Since $\mathfrak{{{C}}}$ is a subalgebra of $\mathfrak{Sb}\mskip 1.5mu({}^{3}{{U}}(\mathfrak{{{B}}}))$, applying the $\mathfrak{Ra}\mskip 1.5mu$ operator gives $\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\subseteq\mathfrak{Ra}\mskip 1.5mu\mathfrak{Sb}\mskip 1.5mu({}^{3}{{U}}(\mathfrak{{{B}}})).$ According to (HMT85, , 5.3.16(1)), $\mathfrak{Re}\left({{U}}(\mathfrak{{{B}}})\right)\cong\mathfrak{Ra}\mskip 1.5mu\mathfrak{Sb}\mskip 1.5mu\left({}^{3}{{U}}(\mathfrak{{{B}}})\right)$ via the isomorphism ${{f}}$ defined for ${{X}}\subseteq{{U}}(\mathfrak{{{B}}})\times{{U}}(\mathfrak{{{B}}})$ by

[TABLE]

Hence there is an algebra $\mathfrak{{{A}}}^{\prime}\subseteq\mathfrak{Re}({{U}}(\mathfrak{{{B}}}))$ such that $\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\cong\mathfrak{{{A}}}^{\prime}$ via the isomorphism ${{f}}^{-1}$, where, if ${{Y}}\subseteq{}^{3}{{U}}(\mathfrak{{{B}}})$, then

[TABLE]

Let ${{S}}(\mathfrak{{{B}}})={{f}}^{-1}({{E}}(\mathfrak{{{B}}}))$. Then it is easy to prove

[TABLE]

Relativize $\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}$ and its image $\mathfrak{{{A}}}^{\prime}$ under the isomorphism ${{f}}^{-1}$ to the ternary relation ${{E}}(\mathfrak{{{B}}})$ and its image ${{S}}(\mathfrak{{{B}}})$ under ${{f}}^{-1}$, obtaining $\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right)\cong\mathfrak{Rl}\mskip 1.5mu_{{{S}}(\mathfrak{{{B}}})}\left(\mathfrak{{{A}}}^{\prime}\right).$ It was shown above that $\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right)\cong\mathfrak{Rl}\mskip 1.5mu_{{{E}}(\mathfrak{{{B}}})}\left(\mathfrak{Ra}\mskip 1.5mu\mathfrak{{{C}}}\right),$ and from Theorem 10.1 we have $\mathfrak{{{A}}}\cong\mathfrak{Ra}\mskip 1.5mu\mathfrak{Rl}\mskip 1.5mu_{{{V}}(\mathfrak{{{B}}})}\left(\mathfrak{{{C}}}\right),$ so we conclude that $\mathfrak{{{A}}}\cong\mathfrak{Rl}\mskip 1.5mu_{{{S}}(\mathfrak{{{B}}})}\left(\mathfrak{{{A}}}^{\prime}\right).$

12 Elementary Laws of $\mathsf{WA}$

Throughout this section, we assume $\mathfrak{{{A}}}$ is an arbitrary algebra satisfying axioms (1)–(10). Few specific references are made to elementary facts from the theory of Boolean algebras, so there are no references to axioms (1)–(3). Axiom (4) is not needed until (38).

[TABLE]

Proof

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Proof

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Proof

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Proof

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Proof

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Proof

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[TABLE]

Proof

The proof is similar to that of (15), using (16) instead of (5).

[TABLE]

Proof

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[TABLE]

Proof

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Proof

[TABLE]

[TABLE]

Proof

We prove only the first equality in (21). The proof of the second is quite similar.

[TABLE]

[TABLE]

Proof

Propositions (22) and (23) follow by Boolean algebra from (21), and have the following consequences, because of (19) and (18).

[TABLE]

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Proof

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Proof

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Proof

From ${{u}}\leq{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ get $\breve{{{u}}}\leq\breve{{1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}}={1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}$ by (14) and (11). Then ${{u}}\leq\breve{{{u}}}$ by (27) and $\breve{{{u}}}\leq\breve{\breve{{{u}}}}={{u}}$ by (27) and (7), hence we have $\breve{{{u}}}={{u}}$.

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Proof

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Proof

Assuming $\emptyset\neq{{X}}\subseteq{{A}}$ and $\sum{{X}}$ exists, we must show ${\left(\sum{{X}}\right)}\breve{\ }$ is the least upper bound of $\{\breve{{{x}}}:{{x}}\in{{X}}\}$. By (14), $\breve{{{x}}}\leq{\left(\sum{{X}}\right)}\breve{\ }$ for every ${{x}}\in{{X}}$ since ${{x}}\leq\sum{{X}}$, so ${\left(\sum{{X}}\right)}\breve{\ }$ is an upper bound of $\{\breve{{{x}}}:{{x}}\in{{X}}\}$. Assume $\breve{{{x}}}\leq{{y}}$ for every ${{x}}\in{{X}}$ (${{y}}$ is an upper bound). Then ${{x}}\leq\breve{{{y}}}$ for every ${{x}}\in{{X}}$ by (7) and (14), so $\breve{{{y}}}$ is an upper bound of ${{X}}$, hence $\sum{{X}}\leq\breve{{{y}}}$. But then ${\left(\sum{{X}}\right)}\breve{\ }\leq{{y}}$ by (7) and (14). Thus ${\left(\sum{{X}}\right)}\breve{\ }$ is the least upper bound of $\{\breve{{{x}}}:{{x}}\in{{X}}\}$.

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Proof

If ${{X}}=\emptyset$ then $\sum{{X}}=0$ and $\{{{z}}{\mskip 1.5mu;\mskip 1.5mu}{{{x}}}:{{x}}\in{{X}}\}=\emptyset$, so the result follows from (18). Assume $\emptyset\neq{{X}}\subseteq{{A}}$ and $\sum{{X}}$ exists. For any ${{y}}\in{{A}}$, the following statements are equivalent (mostly for Boolean algebraic reasons).

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Since the first statement is true when ${{z}}{\mskip 1.5mu;\mskip 1.5mu}{\sum{{X}}}={{y}}$, ${{z}}{\mskip 1.5mu;\mskip 1.5mu}{\sum{{X}}}$ is itself an upper bound of $\{{{z}}{\mskip 1.5mu;\mskip 1.5mu}{{{x}}}:{{x}}\in{{X}}\}$. Since the last statement implies the first, ${{z}}{\mskip 1.5mu;\mskip 1.5mu}{\sum{{X}}}$ is included in all the upper bounds of $\{{{z}}{\mskip 1.5mu;\mskip 1.5mu}{{{x}}}:{{x}}\in{{X}}\}$. Thus ${{z}}{\mskip 1.5mu;\mskip 1.5mu}{\sum{{X}}}$ is the least upper bound of $\{{{z}}{\mskip 1.5mu;\mskip 1.5mu}{{{x}}}:{{x}}\in{{X}}\}$, i.e., ${{z}}{\mskip 1.5mu;\mskip 1.5mu}{\sum{{X}}}=\sum\{{{z}}{\mskip 1.5mu;\mskip 1.5mu}{{{x}}}:{{x}}\in{{X}}\}$.

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Proof

To show $\breve{{{x}}}$ is an atom it suffices to show that $\breve{{{x}}}\neq 0$ and if $0\neq\breve{{{x}}}\cdot{{y}}$ then $\breve{{{x}}}\leq{{y}}$. Note that if $\breve{{{x}}}=0$ then ${{x}}=\breve{\breve{{{x}}}}=\breve{0}=0$ by (13) and (7). Thus $\breve{{{x}}}$ is not zero because ${{x}}$ is not zero. For the other part, we have

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Proof

Suppose ${{x}},{{y}},{{z}}$ are atoms such that ${{x}}{\mskip 1.5mu;\mskip 1.5mu}{{y}}\geq{{z}}.$ The atom ${{z}}$ is not zero, so ${{x}}{\mskip 1.5mu;\mskip 1.5mu}{{y}}\cdot{{z}}\neq 0,$ from which we get ${{y}}\cdot\breve{{{x}}}{\mskip 1.5mu;\mskip 1.5mu}{{z}}\neq 0$ by (24), then ${{y}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{z}}}\cdot\breve{{{x}}}\neq 0$ by (25), then $\breve{{{z}}}\cdot\breve{{{y}}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{x}}}\neq 0$ by (24), then $\breve{{{z}}}{\mskip 1.5mu;\mskip 1.5mu}{{x}}\cdot\breve{{{y}}}\neq 0$ by (25) and (7), and finally ${{x}}\cdot{{z}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{y}}}\neq 0$ by (24) and (7). Since $\breve{{{x}}},\breve{{{y}}},\breve{{{z}}}$ are also atoms by (32), these last five inequalities are respectively equivalent to the inequalities $\breve{{{x}}}{\mskip 1.5mu;\mskip 1.5mu}{{z}}\geq{{y}}$, ${{y}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{z}}}\geq\breve{{{x}}}$, $\breve{{{y}}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{x}}}\geq\breve{{{z}}}$, $\breve{{{z}}}{\mskip 1.5mu;\mskip 1.5mu}{{x}}\geq\breve{{{y}}}$, and ${{z}}{\mskip 1.5mu;\mskip 1.5mu}\breve{{{y}}}\geq{{x}}$.

Define domain and range operators as follows.

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Proof

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Proof

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Proof

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Proof

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Proof

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Proof

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Proof

If ${{{x}}}^{\mathsf{d}}=0$, then ${{x}}={{{x}}}^{\mathsf{d}}{\mskip 1.5mu;\mskip 1.5mu}{{x}}=0{\mskip 1.5mu;\mskip 1.5mu}{{x}}=0$ by (36) and (19), but ${{x}}\neq 0$ since ${{x}}$ is an atom. Therefore ${{{x}}}^{\mathsf{d}}\neq 0$. Assume ${{y}}\in{{A}}$. We derive ${{{x}}}^{\mathsf{d}}\leq{{y}}$ from ${{{x}}}^{\mathsf{d}}\cdot{{y}}\neq 0$. First we show ${{x}}\leq({1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\cdot{{y}}){\mskip 1.5mu;\mskip 1.5mu}1$.

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Next we use ${{x}}\leq({1\kern-0.80002pt\raise 5.59721pt\hbox{\rm,}\kern 0.80002pt}\cdot{{y}}){\mskip 1.5mu;\mskip 1.5mu}1$ (in the second step below) to show ${{{x}}}^{\mathsf{d}}\leq{{y}}$. The first use of weak associativity (4) occurs in the fourth step.

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Proof

Since ${{{x}}}^{\mathsf{r}}$ and ${{{y}}}^{\mathsf{d}}$ are atoms by (41), they can fail to be equal only by being disjoint. However, if ${{{x}}}^{\mathsf{r}}\cdot{{{y}}}^{\mathsf{d}}=0$ then we get a contradiction as follows.

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Proof

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so ${{u}}={{{x}}}^{\mathsf{d}}$ since ${{u}}$ and ${{{x}}}^{\mathsf{d}}$ are atoms.